7. A 0.5 kg soccer ball is kicked at 10 m/s. A goalie catches the ball and brings it to rest in 0.25 seconds. What
is the force exerted on the ball by the goalie? (Hint: Apply two formulas to solve this problem)
A. 5 N
B. 10 N
C. 20 N
D. 25 N

Answers

Answer 1

A 0.5 kg soccer ball is kicked at 10 m/s. A goalie catches the ball and brings it to rest in 0.25 seconds then the force exerted on the ball by the goalie is 20N. Option C is correct.

Here, we must determine the change in momentum of the soccer ball. The momentum of an object is stated by the product of its mass and velocity. The mass of the soccer ball is 0.5 kg, and its initial velocity is 10 m/s. Therefore, the ball is conducted to be constant, and its final velocity is 0 m/s.

The change in momentum is computed by reducing the final momentum from the initial momentum. In this concern, the initial momentum is 0.5 kg × 10 m/s = 5 kg·m/s, and the final momentum is 0.5 kg × 0 m/s = 0 kg·m/s. Now, the change in momentum is 5 kg·m/s - 0 kg·m/s = 5 kg·m/s.

Next, we separate the change in momentum by the time taken to bring up the ball to rest, which is 0.25 seconds. Thus, the goalie's force exerted on the ball is 5 kg·m/s / 0.25 s = 20 N.

Therefore, the correct answer is C. 20 N.

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Answer 2

The correct Option is C. The force exerted on the ball by the goalie is 20 N.

The formula for the force exerted on an object is given by F = ma, where F is the force, m is the mass of the object and a is the acceleration.

The formula for acceleration is a = (v-u)/t, where v is the final velocity, u is the initial velocity and t is the time taken.

The acceleration is negative if the object is brought to rest.

So, for the given problem, the initial velocity of the soccer ball is 10 m/s and the final velocity is 0.

The time taken to bring it to rest is 0.25 s.

Therefore, the acceleration is given by:a = [tex](0 - 10)/0.25 = - 40 m/s^{2}[/tex]

Now, we can calculate the force exerted by the goalie using the formula: [tex]F = maF = 0.5 kg $\times$ (- 40 m/s^{2} ) = - 20 N[/tex]

We get a negative value for the force, which means that the force exerted is in the opposite direction to the motion of the ball.

However, the magnitude of the force is given by |-20 N| = 20 N.

So, the answer is option (C) 20 N.

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Related Questions

Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you. The lines of magnetic force at any point in the magnetic field will act in
Question 17 options:

A)

a clockwise direction.

B)

a counterclockwise direction.

C)

the direction opposite to the current.

D)

the same direction as the current.

Answers

Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you. The lines of magnetic force at any point in the magnetic field will act in option c)  the direction opposite to the current.

Lenz's law is the law that governs the direction of magnetic force.According to Lenz's law, magnetic fields induced by an electric current have a polarity such that the current's magnetic field opposes any change in current flow. Based on this law, the induced current must produce a magnetic field that opposes the current that produced it.

If the current is flowing towards us, the induced magnetic field must flow in the opposite direction to the current. Therefore, the direction of the lines of magnetic force at any point in the magnetic field will act in the direction opposite to the current.Hence, the correct option is C) the direction opposite to the current.

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With a force of 200 N a body is lifted 20 meters in 20 seconds. Calculate the weight of this body. Use the formula for distance as a function of acceleration with initial velocity equal to zero.

Answers

Answer:

The weight of the body is 3,924 N.

Explanation:

To solve this problem, we can use the formula for distance as a function of acceleration with initial velocity equal to zero:

distance = (1/2) x acceleration x time^2

We know that the distance the body is lifted is 20 meters, the time taken is 20 seconds, and the force applied is 200 N. We can use this information to calculate the weight of the body.

First, we need to calculate the acceleration:

distance = (1/2) x acceleration x time^2

20 = (1/2) x acceleration x (20)^2

acceleration = 0.5 m/s^2

Now that we know the acceleration, we can use the formula for weight:

force = mass x acceleration

We can rearrange this formula to solve for mass:

mass = force / acceleration

mass = 200 N / 0.5 m/s^2

mass = 400 kg

Finally, we can calculate the weight of the body using the formula:

weight = mass x gravity

Assuming a standard acceleration due to gravity of 9.81 m/s^2, we can calculate the weight:

weight = 400 kg x 9.81 m/s^2

weight = 3,924 N

Therefore, the weight of the body is 3,924 N.

A 400 kg bomb sitting at rest on a table explodes into three pieces. A 150 kg piece moves off to the East with a velocity of 150 m/s. A 100 kg piece moves off with a velocity of 200 m/s at a direction of south 60° West.
What is the velocity of the third piece?

Answers

The velocity of the third piece is (81.25 m/s, -43.3 m/s).

To determine the velocity of the third piece, we can use the principle of conservation of momentum.

Given:

Mass of the first piece (m1) = 150 kg

Velocity of the first piece (v1) = 150 m/s (to the East)

Mass of the second piece (m2) = 100 kg

Velocity of the second piece (v2) = 200 m/s at a direction of south 60° West

Let's break down the velocities into their respective horizontal (x) and vertical (y) components.

For the first piece:

v1x = 150 m/s (since it's moving to the East)

v1y = 0 m/s (no vertical component)

For the second piece:

v2x = 200 m/s * cos(60°) = 200 m/s * 0.5 = 100 m/s (horizontal component)

v2y = -200 m/s * sin(60°) = -200 m/s * 0.866 = -173.2 m/s (vertical component, negative since it's moving downward)

Now, let's calculate the momentum of the first and second pieces:

The momentum of the first piece (p1) = m1 * v1

= 150 kg * 150 m/s

= 22,500 kg·m/s

The momentum of the second piece (p2) = m2 * v2

= 100 kg * (100 m/s, -173.2 m/s)

= (10,000 kg·m/s, -17,320 kg·m/s)

To find the total momentum after the explosion, we can add the momenta of the individual pieces:

Total momentum after the explosion = p1 + p2

= (22,500 kg·m/s, 0 kg·m/s) + (10,000 kg·m/s, -17,320 kg·m/s)

= (32,500 kg·m/s, -17,320 kg·m/s)

The total momentum after the explosion should also be equal to the momentum of the third piece:

The momentum of the third piece (p3) = m3 * v3

Given:

Mass of the third piece (m3) = 400 kg (calculated from the given mass of the bomb)

Let's assume the velocity of the third piece is (v3x, v3y).

Therefore, we have the equation:

(32,500 kg·m/s, -17,320 kg·m/s) = 400 kg * (v3x, v3y)

By equating the x and y components separately, we can solve for the velocity components of the third piece:

32,500 kg·m/s = 400 kg * v3x

-17,320 kg·m/s = 400 kg * v3y

Solving these equations, we find:

v3x = 81.25 m/s

v3y = -43.3 m/s

Therefore, the velocity of the third piece is approximately (81.25 m/s, -43.3 m/s).

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Look at this graphic organizer of requirements to apply to become an astronaut.
Requirements for Astronauts
What does the graphic organizer most suggest about the job of an astronaut?
It is technical and potentially tedious.
It is detailed and potentially exhausting.
It is confidential and potentially exciting.
○ It is complex, demanding, and involves flight.
Save and Exit
Next

Answers

The graphic organizer suggests that the job of an astronaut is complex, demanding, and involves flight.

This conclusion can be drawn by examining the nature of the requirements listed in the graphic organizer. Firstly, the requirements listed in the organizer are numerous and encompass various aspects. They include educational qualifications, such as having a bachelor's degree in a relevant field, as well as specific experience, like piloting an aircraft.

These requirements highlight the complexity of the job and indicate that astronauts need to possess a diverse set of skills and knowledge. Additionally, the requirements for physical fitness and health demonstrate the demanding nature of the job.

Astronauts are expected to undergo rigorous physical training to ensure they can handle the physical stresses associated with space travel and the conditions they will encounter in space. This indicates that the job can be physically exhausting and requires individuals to be in excellent health.

Lastly, the inclusion of flight-related requirements, such as the need to pass a long-duration spaceflight physical and participate in aircraft flights, implies that the job of an astronaut involves actual flight experiences. This indicates that astronauts are directly involved in piloting spacecraft and are expected to have practical experience in flying.

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An ideal refrigerator, which is Carnot engine operating in reverse, operates between a freezer temperature of -9 °C and a room temperature at 25 °C. In a period of time, it absorbs 120 J from the freezer compartment. How much heat is rejected to the room? ​

Answers

The amount of heat rejected to the room by the ideal refrigerator can be calculated using the Carnot efficiency. With the given temperatures and heat absorbed, the heat rejected to the room is 225 J.

To calculate the amount of heat rejected to the room by the ideal refrigerator, we can use the Carnot efficiency, which is given by the formula:

Efficiency = 1 - ([tex]T_c_o_l_d[/tex] / [tex]T_h_o_t[/tex])

where[tex]T_c_o_l_d[/tex]is the temperature of the cold reservoir (freezer compartment) and [tex]T_h_o_t[/tex] is the temperature of the hot reservoir (room temperature).

Given:

[tex]T_c_o_l_d[/tex] = -9 °C (converted to Kelvin: 264 K)

[tex]T_h_o_t[/tex]= 25 °C (converted to Kelvin: 298 K)

Heat absorbed from the freezer compartment ([tex]Q_c_o_l_d[/tex] = 120 J

First, we calculate the Carnot efficiency:

Efficiency = 1 - (264 K / 298 K)

Efficiency ≈ 0.1134

The Carnot efficiency represents the ratio of heat transferred from the cold reservoir to the work done by the refrigerator. Since the refrigerator is operating in reverse, the work done is equal to the heat absorbed from the freezer compartment ([tex]Q_c_o_l_d[/tex]).

[tex]Q_c_o_l_d[/tex] = 120 J

Now, we can calculate the heat rejected to the room ([tex]Q_h_o_t[/tex]) using the equation:

[tex]Q_h_o_t[/tex] = Efficiency * [tex]Q_c_o_l_d[/tex]

[tex]Q_h_o_t[/tex] ≈ 0.1134 * 120 J

[tex]Q_h_o_t[/tex] ≈ 13.61 J

Therefore, the amount of heat rejected to the room by the ideal refrigerator is approximately 13.61 J.

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A spacecraft is in a circular orbit around the planet Mars at a height of 140km.
A small part of the spacecraft falls off and eventually lands on the surface of the Mars
The small part has a mass of 1.8kg
During its fall, the small part loses 0.932 MJ of gravitational potential energy.
Calculate the gravitational field strength of Mars ​

Answers

Answer:

3.79 m/s^2

Explanation:

We know the small part loses 0.932 MJ of gravitational potential energy during its fall.

Potential energy = mass x gravitational field strength x height

Re-arranging to solve for gravitational field strength:

g = Potential energy/(mass x height)

Plugging in the given values:

g = 0.932 MJ / (1.8kg x 140km)

= 0.932 x 10^6 J / (1.8 x 1000kg x 140 x 1000m)

= 3.79 m/s^2

Therefore, the gravitational field strength of Mars is calculated to be 3.79 m/s^2.

what type of force

a child on a sled slides down the hill

Answers

Answer:

Gravity.

Explanation:

Gravity causes the child on a sled to slide down the hill.

Hope this helps!

deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W. Refer to appropriate Newton’s laws of motion.

Answers

The expression for the contribution of P to the pressure exerted on W is P = mV/(c^2t), derived using Newton's laws of motion and the definition of pressure.

In order to deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W, we can use the appropriate Newton’s laws of motion. Specifically, we can use the equation F = ma, where F represents force, m represents mass, and a represents acceleration.We know that pressure (P) is defined as force per unit area, or P = F/A. Rearranging this equation, we can solve for force: F = PA.Substituting this into the equation F = ma, we get PA = ma. Rearranging this equation, we can solve for pressure in terms of mass and acceleration: P = ma/A. Finally, we know that acceleration can be expressed in terms of velocity (V) and time (t): a = V/t.Substituting this into our equation for pressure, we get P = mV/(At). Since c represents the speed of sound, we can express A as [tex]A = c^2[/tex]. Therefore, our final expression for the contribution of P to the pressure exerted on W is:[tex]P = mV/(c^{2t})[/tex]In summary, we used the equation F = ma, the definition of pressure (P = F/A), and the relationship between acceleration (a), velocity (V), time (t), and the speed of sound (c) to deduce an expression for the contribution of P to the pressure exerted on W in terms of m, c, and V.

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An object of mass M = 14.0 kg is attached to a cord that is wrapped around a wheel of radius r = 12.0 cm (see figure). The acceleration of the object down the frictionless incline is measured to be a = 2.00 m/s2 and the incline makes an angle = 37.0° with the horizontal. Assume the axle of the wheel to be frictionless. Answer parts a-c.

Answers

a.  the tension in the rope is  91.5 N.

b.   the moment of inertia of the wheel is  0.1008 kg⋅m².

c.  the angular speed of the wheel 2.30 s after it begins rotating is  38.34 rad/s.

How do we calculate?

(a)

The tension in the rope can be found by considering the forces acting on the object.

ma = mg*sin(θ) - T

(14.0 kg)(2.00 m/s²)

= (14.0 kg)(9.8 m/s²)*sin(37°) - T

T = (14.0 kg)(9.8 m/s²)*sin(37°) - (14.0 kg)(2.00 m/s²)

T =  91.5 N

(b)

The moment of inertia of a wheel:

I = (1/2)MR²

I = (1/2)(14.0 kg)(0.12 m)²

I = 0.1008 kg⋅m²

(c)

The angular acceleration of the wheel:

α = a/R

α = angular acceleration,

a = linear acceleration of the object,

R =  radius of the wheel.

α = (2.00 m/s²)/(0.12 m)

α = 16.67 rad/s²

The angular speed (ω) of the wheel after time t is :

ω = ω₀ + αt

ω = 0 + (16.67 rad/s²)(2.30 s)

ω = 38.34 rad/s

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What are the six digit grid coordinates for the windtee?

Answers

The six digit grid coordinates for the windtee  should be 100049.

How do we we calculate?

The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.

When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.

If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.

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A rock with a mass of 0.2 kg with a velocity of 5 m/s strikes a stationary 1 kg wooden ball. After the
collision the rock flies back with a velocity of -2 m/s. What is the velocity of the wooden ball after the
collision?
A. -0.4 m/s
B. -1 m/s
C. 0.4 m/s
D. 1.4 m/s

Answers

Answer:

D.  1.4 m/s

Explanation:

forward direction is +

back direction is -

Momentum = P = mass x velocity = mv

let v =  velocity of ball after collision

Law of Conservation of Momentum:  total momentum before the collision must equal the total momentum after the collision

(0.20 kg)(5 m/s) + (1 kg)(0 m/s) = (0.2 kg)(-2 m/s) + (1 kg)v

1 kg·m/s + 0 = -0.4 kg·m/s + (1 kg)v    

1 kg·m/s + 0.4 kg·m/s =  (1 kg)v    rearrange the equation and solve for v

(1 kg)v  = 1.4 kg·m/s

v = (1.4 kg·m/s) / (1 kg) = 1.4 m/s

What force acts on a projectile in the horizontal direction?

Answers

The force that acts on a projectile in the horizontal direction is Gravitational force.


A projectile is an object upon which the only force is gravity. Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.

Due to the absence of horizontal forces, a projectile remains in motion with a constant horizontal velocity. Horizontal forces are not required to keep a projectile moving horizontally. Hence, The only force acting upon a projectile is gravity.


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.An electron of charge 1.6 x 10-19is situated in a uniform electric filed strength of 120 vm-1 Calculate the force acting on it​

Answers

The force acting on the electron is 1.92 x 10^-17 N.

The problem states that an electron of charge 1.6 x 10^-19 is located in a uniform electric field of 120 Vm^-1, and it asks us to determine the force acting on it.

We can use Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. If the charges are of opposite signs, the force is attractive, while if the charges are of the same sign, the force is repulsive.

The formula for Coulomb's law is F = kq1q2/r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.

Since the electron has a charge of 1.6 x 10^-19 C, and the electric field strength is 120 Vm^-1, we can use the equation F = qE to find the force acting on it.

F = qE = (1.6 x 10^-19 C)(120 Vm^-1) = 1.92 x 10^-17 N.

Therefore, the force acting on the electron is 1.92 x 10^-17 N.

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how can i write answers to get points

Answers

If you’re using the app, go to the “give answers” tab on the bottom, and you can see questions that people have asked, once you find one you want to respond to, click on it and then click on the “answer” button at the bottom

If a 9000kg water flows in a minute through a pipe of cross sectional area 0.3m², what is the speed of water in the pipe? ​

Answers

Answer:

5 m/s

Explanation:

We are given that 9000 kg of water flows through the pipe in 1 minute. Mass flow rate = mass/time

So, mass flow rate = 9000 kg / 1 minute = 150 kg/s

We know the cross sectional area of the pipe is 0.3 m2. From continuity equation, mass flow rate = density * area * velocity

So, 150 = 1000 * 0.3 * v (Density of water is approximately 1000 kg/m3)

Solving for v (velocity):

v = 150/(1000*0.3) = 5 m/s

Therefore, the speed of water in the pipe is 5 m/s.

WHOEVER ANSWERS IS THE BRAINLIEST!!! PLS HELP!!

Answers

Based on the information, we can infer that the temperature on the west and east coasts of the United States is higher than in the central part at latitude 35° North.

What do we see in the image?

In the image you can see the map of the United States and two latitudinal lines of 35° and 45° North. Additionally we see the different temperatures that exist in various cities or locations in the United States.

Based on this information, we can infer that the temperatures on the east and west coasts are higher than the temperatures recorded in the central part. For example, at 35° latitude, the coasts register temperatures of more than 60°F while the central zone registers lower temperatures between 36 and 59°F.

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Consider the figure below. (a) Find the tension in each cable supporting the 524-N cat burglar. (Assume the angle of the inclined cable is 34.0°.) (b) Suppose the horizontal cable were reattached higher up on the wall. Would the tension in the inclined cable increase, decrease, or stay the same?

Answers

(a) The tension in the inclined cable (T1) and horizontal cable (T2) supporting the cat burglar is equal. The tension in the vertical cable (T3) is 524 N.

(b) If the horizontal cable is reattached higher up, the tension in the inclined cable (T1) would increase.

(a) To find the tension in each cable supporting the 524-N cat burglar, we'll consider the forces acting on the system. Let's denote the tension in the inclined cable as T1, the tension in the horizontal cable as T2, and the tension in the vertical cable as T3. The angle between the inclined cable and the vertical cable is given as θ.

In the vertical direction, the tension in the vertical cable T3 balances the weight of the cat burglar:

T3 - 524 N = 0

T3 = 524 N

In the horizontal direction, the tension in the inclined cable T1 can be expressed as:

T1 * cos(θ) = T2

Now, we need to determine the value of θ to calculate T1 and T2. Let's assume that θ is the given angle of θ = 0.

Substituting the angle and rearranging the equation, we have:

T1 = T2 / cos(θ)

T1 = T2 / cos(0)

T1 = T2 / 1

T1 = T2

So, the tension in the inclined cable (T1) is equal to the tension in the horizontal cable (T2).

Therefore, the tension in each cable is as follows:

T1 (inclined cable) = T2 (horizontal cable)

T1 = T2

T3 (vertical cable) = 524 N

(b) If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable (T1) would increase.

The correct answer is option A.  

This is because reattaching the horizontal cable at a higher point on the wall would increase the horizontal component of the tension, resulting in a larger tension in the inclined cable. The tension in the vertical cable (T3) would remain the same as it is independent of the position of the horizontal cable.

In summary, the tension in the inclined cable (T1) and the horizontal cable (T2) are equal, and their value depends on the angle θ. The tension in the vertical cable (T3) is 524 N. If the horizontal cable were reattached higher up on the wall, the tension in the inclined cable would increase, while the tension in the vertical cable would remain the same.

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D 4.8
This is a harder question based on the Law of Conservation of Momentum. Take the time to work
your way through it. Start with a diagram.
A 400 kg bomb sitting at rest on a table explodes into three pieces. A 150 kg piece moves off to the
east with a velocity of 150 m s². A 100 kg piece moves off with a velocity of 200 m s at a direction of
south 60° west. What is the velocity of the third piece?

It is possible

Answers

The velocity of the third piece is v₃ = -12500 kg·m/s / m₃

How do we calculate?

The law of conservation of momentum states that the total momentum before the explosion is equal to the total momentum after the explosion.

velocity of the third piece =  v₃.

The total initial momentum before the explosion = 0

The total final momentum after the explosion= 0

Initial momentum = 0 kg·m/s (since the bomb is at rest)

Final momentum = m₁v₁ + m₂v₂ + m₃v₃

m₁ = mass of the first piece = 150 kg

v₁ = velocity of the first piece = 150 m/s (to the east)

m₂ = mass of the second piece = 100 kg

v₂ = velocity of the second piece = 200 m/s (south 60° west)

m₃ = mass of the third piece = unknown

v₃ = velocity of the third piece = unknown

0 = (150 kg)(150 m/s) + (100 kg)(200 m/s)(cos(60°)) + (m₃)(v₃)

final momentum = 0 and hence  v₃ is found as :

0 = 22500 kg·m/s - 10000 kg·m/s + (m₃)(v₃)

-12500 kg·m/s = (m₃)(v₃)

v₃ = -12500 kg·m/s / m₃

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When white light reflects off of a green surface, which of the following occurs?

1. All wavelengths of light are absorbed.
2. Only the green wavelengths of light are absorbed.
3. Only the green wavelengths of light are reflected.
4. All wavelengths of light are reflected.

Answers

When white light reflects off of a green surface, only the green wavelengths of light are reflected (option d).

1. White light is a combination of all visible wavelengths of light, including red, orange, yellow, green, blue, indigo, and violet.

2. When white light hits a green surface, the surface absorbs some wavelengths of light and reflects others.

3. The color we perceive as "green" is the result of the green wavelengths of light being reflected by the surface.

4. In this case, the green surface absorbs all the wavelengths of light except for the green wavelengths, which are reflected back.

5. As a result, our eyes detect the reflected green light and interpret it as the color green.

6. This phenomenon occurs because the green surface selectively absorbs and reflects different wavelengths of light based on its molecular structure and the interactions between light and matter.

7. The absorption and reflection of specific wavelengths of light give objects their perceived color.

8. Therefore, when white light reflects off of a green surface, only the green wavelengths of light are reflected, while the other wavelengths are absorbed by the surface.

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Two objects with masses of m1 = 3.70 kg and m2 = 5.70 kg are connected by a light string that passes over a frictionless pulley, as in the figure below. Answer parts a-c.

Answers

(a) The tension in the string is determined as 19.6 N.

(b) The acceleration of each object is 5.3 m/s².

(c) The distance each object will move in the first second if it started from rest is 2.65 m.

What is the tension in the string?

(a) The tension in the string is the resultant weight of the masses and magnitude is calculated as follows;

T = ( 5.7 kg - 3.7 kg ) x 9.8 m/s²

T = 19.6 N

(b) The acceleration of each object is calculated as follows;

a = T / m

where;

m is the mass T is the tension

a = 19.6 N / 3.7 kg

a = 5.3 m/s²

(c) The distance each object will move in the first second if it started from rest is calculated as;

s = ut + ¹/₂at²

where;

u is the initial velocity = 0

s = 0 + ¹/₂(5.3)(1²)

s = 2.65 m

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Two blocks, M1 and M2, are connected by a massless string that passes over a massless pulley as shown in the figure. M2, which has a mass of 19.0 kg,
rests on a long ramp of angle theta=25.0∘.
Ignore friction, and let up the ramp define the positive direction.
If the actual mass of M1 is 5.00 kg and the system is allowed to move, what is the acceleration of the two blocks?
What distance does block M2 move in 2.00 s?

Answers

The acceleration of the two blocks is[tex]2.14 m/s^{2[/tex]} and the distance does block M2 move in 2.00 s is 4.27 m.

Now we need to find the acceleration of the two blocks and the distance does block M2 move in 2.00 s.

We know that: mass of M1, m1 = 5.00 kg mass of M2, m2 = 19.0 kgθ = 25.0°Taking upward direction as positive for block M1 and downwards as positive for block M2.

Therefore, we can write the following equation of motion for the two blocks:

For M2: m2g - T = m2a ...(1)

For M1: T - m1g = m1a ...(2)

We can see from the figure that M2 is on an inclined plane making an angle θ with the horizontal.

We can resolve the weight of M2 into two components:

Perpendicular to the plane = m2gcosθParallel to the plane = m2gsinθ

The component parallel to the plane will tend to make the block move downwards.

Therefore, the effective weight will be:

mg = m2gsinθ ...(3)

From equation (1) we can write:

T = m2g - m2a ...(4)

Substituting equation (4) in equation (2), we get:

m2g - m2a - m1g = m1a ...(5)

On solving equation (5), we get the acceleration as:

a = g(m2sinθ - m1) / (m1 + m2)

On substituting the given values, we get:

[tex]a = 2.14 m/s^{2}[/tex]

The distance moved by M2 in 2 seconds can be found out using the formula:[tex]s = ut + \frac{1}{2} at^{2}[/tex]

Here, initial velocity, u = 0m/s Time, t = 2s Acceleration, [tex]a = 2.14 m/s^{2}[/tex]

On substituting these values, we get the distance travelled by M2 as: s = 4.27 m

Therefore, the acceleration of the two blocks is [tex]2.14 m/s^{2}[/tex]. And the distance does block M2 move in 2.00 s is 4.27 m.

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RHETORICAL ANALYSIS: How does Robinson use language in effective and engaging ways to develop his argument to his younger self-and, in the process, to young readers in the present? In your response, consider such techniques as metaphor, repetition, and sentence structure.

Answers

In "The Argonauts," Robinson effectively utilizes language techniques such as metaphor, repetition, and sentence structure to develop his argument to his younger self and engage young readers in the present. Through these techniques, Robinson creates a powerful and relatable narrative that resonates with his audience.

Robinson employs metaphors to convey complex ideas in a compelling and accessible manner. For instance, he compares his struggle with identity and gender to the mythical journey of the Argonauts, making it relatable and captivating for young readers. This metaphorical language enables readers to grasp the profound emotions and challenges he faced during his own personal journey.

Repetition is another technique Robinson employs to reinforce key ideas and create a rhythmic and memorable reading experience. By repeating certain phrases or concepts, he emphasizes their significance and invites readers to reflect on them. This repetition serves to engage young readers by encouraging them to contemplate their own experiences and perspectives.

Furthermore, Robinson carefully structures his sentences to create a sense of rhythm and flow, enhancing the overall readability and impact of his argument. Short, concise sentences create moments of clarity and emphasis, while longer, more descriptive sentences evoke a contemplative and introspective tone. This varied sentence structure adds depth and nuance to his narrative, captivating young readers and keeping them engaged throughout.

In conclusion, through the effective use of metaphor, repetition, and sentence structure, Robinson engages and captivates young readers, inviting them to reflect on their own identities and experiences. His language choices not only develop his argument to his younger self but also establish a connection with present-day young readers, making his work both impactful and relatable.

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Look at the velocity versus time graph below. What is the magnitude of the
displacement of the object after it travels for five seconds?
Velocity (m/s)
Time (s)
A. 30 m
OB. 20 m
OC. 25 m
OD. 35 m

Answers

The magnitude of displacement of the object after five seconds, calculated from the velocity-time graph, is 32.5 m. The correct answer is option E.

Given the velocity versus time graph below, we are required to find the magnitude of the displacement of the object after it travels for five seconds. Velocity-time graph imageThe area under the velocity-time graph corresponds to the displacement of the object. The magnitude of displacement is given by the formula: Displacement = area under a velocity-time graphIf we look at the given graph, it can be seen that the graph is a trapezium. Therefore, we need to split it into two parts: a rectangle and a triangle. The displacement is given by the sum of the area of both parts. To find the area of a rectangle, we use the formula: Area of rectangle = base × height = (10 s − 0 s) × 2 m/s = 20 mTo find the area of a triangle, we use the formula: Area of triangle = 1/2 × base × height = 1/2 × (15 s − 10 s) × 5 m/s = 12.5 mTherefore, the magnitude of displacement of the object, after it travels for five seconds, is given by: Displacement = Area of rectangle + Area of triangle= 20 m + 12.5 m= 32.5 mHence, the correct answer is option E. 32.5 m.

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As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force F exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of 1.50 m and a mass of 52.0 kg. If this fish is moving upward at 3.00 m/s as its head first breaks the surface and has an upward speed of 6.80 m/s after two-thirds of its length has left the surface, assume constant acceleration and determine the following. find a - the salmon's acceleration (answer in m/s^2 upward), find b - the magnitude of the force F during this interval (direction is N).

Answers

Answer:

To solve this problem, we need to use some principles of physics, specifically Newton's second law (F=ma) and the equations of motion. Here are the steps:

1. Calculate the acceleration (a)

We can use the equation of motion to find the acceleration:

v_f^2 = v_i^2 + 2a*d

where:

v_f = final velocity = 6.80 m/s

v_i = initial velocity = 3.00 m/s

d = distance = 2/3 of the length of the fish = 2/3 * 1.50 m = 1.00 m

a = acceleration (which we are trying to find)

Rearranging the equation to solve for a gives us:

a = (v_f^2 - v_i^2) / (2*d)

2. Calculate the magnitude of the force F

Once we have the acceleration, we can use Newton's second law (F=ma) to calculate the force. The net force acting on the fish as it jumps out of the water is the difference between the upward force F exerted by the tail fin and the downward force due to gravity (mg). The net force is also equal to the product of the mass of the fish and its acceleration (ma). Therefore, we have:

F - mg = ma

Rearranging this equation to solve for F gives us:

F = ma + mg

Now let's plug in the numbers and do the calculations.

First, let's find the acceleration:

a = (v_f^2 - v_i^2) / (2*d)

a = (6.80 m/s)^2 - (3.00 m/s)^2) / (2*1.00 m)

a = (46.24 m^2/s^2 - 9.00 m^2/s^2) / 2 m

a = 37.24 m^2/s^2 / 2 m

a = 18.62 m/s^2

The salmon's acceleration is 18.62 m/s^2 upward.

Next, let's find the force F. We know the mass of the fish is 52.0 kg, and the acceleration due to gravity is approximately 9.8 m/s^2. So,

F = ma + mg

F = (52.0 kg)(18.62 m/s^2) + (52.0 kg)(9.8 m/s^2)

F = 969.24 N + 509.6 N

F = 1478.84 N

So, the magnitude of the force F exerted by the salmon's tail fin during this interval is approximately 1479 N.

When a piece of wood is put in a graduated cylinder containing water the level of water rises from 17.7cm cubic to 18.5cm cubic calculate the total volume of the piece of wood given that it's relative density is 0.60

Answers

The total volume of the piece of wood is 1.33[tex]cm^3[/tex].

To calculate the total volume of the piece of wood, we can use the principle of displacement.

1. First, we need to find the difference in volume between the two water levels. The initial volume is 17.7 [tex]cm^3[/tex], and the final volume is 18.5 cm^3. The difference is 18.5 [tex]cm^3[/tex] - 17.7 [tex]cm^3[/tex] = 0.8 [tex]cm^3[/tex].

2. Now, we need to find the volume of water displaced by the piece of wood. Since the relative density of the wood is 0.60, it means that the wood is 0.60 times denser than water.

3. The volume of water displaced by the wood is equal to the difference in volume divided by the relative density of the wood. So, the volume of water displaced is 0.8 cm^3 / 0.60 = 1.33 [tex]cm^3[/tex].

4. Finally, the total volume of the piece of wood is equal to the volume of water displaced. Therefore, the total volume of the piece of wood is 1.33 [tex]cm^3[/tex].

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Suppose that the mirror is moved so that the tree is between the focus point F and the mirror. What happens to the image of the tree?

A. The image moves behind the curved mirror.

B. The image appears shorter and on the same side of the mirror.

C. The image appears taller and on the same side of the mirror.

D. The image stays the same.

Answers

Answer:

C

Explanation:

If the tree is placed between the focus point F and the mirror in a concave mirror, the image of the tree will appear taller and on the same side of the mirror. Therefore, the correct answer is C. The image appears taller and on the same side of the mirror.

A person walks 3.30 km south and then 2.00 km east, all in 3.20 hours. Answer parts a-c.

Answers

(a) the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.

(b) Magnitude of average velocity ≈ 3.85 km and Direction of average velocity ≈ -59.04° (south of east)

(c) the average speed during the given time interval is approximately 1.66 km/h.

(a) To find the magnitude and direction of the person's displacement, we can use the Pythagorean theorem and trigonometry.

Displacement in the x-direction = 2.00 km east

Displacement in the y-direction = -3.30 km south (negative because it is in the opposite direction of the positive y-axis)

Using the Pythagorean theorem:

Magnitude of displacement = √((2.00 km)^2 + (-3.30 km)^2)

Magnitude of displacement ≈ 3.85 km

To find the direction, we can use trigonometry:

θ = tan^(-1)(opposite/adjacent)

θ = tan^(-1)(-3.30 km / 2.00 km)

θ ≈ -59.04° (measured counterclockwise from the positive x-axis)

Therefore, the magnitude of the displacement is approximately 3.85 km, and the direction is approximately 59.04° south of east.

(b) Average velocity is defined as displacement divided by time. The magnitude and direction of average velocity will be the same as the magnitude and direction of displacement.

Magnitude of average velocity ≈ 3.85 km

Direction of average velocity ≈ -59.04° (south of east)

(c) Average speed is defined as total distance traveled divided by time. The total distance traveled is the sum of the magnitudes of the individual displacements.

Total distance = 3.30 km + 2.00 km = 5.30 km

Average speed = Total distance / Time

Average speed ≈ 5.30 km / 3.20 hours

Average speed ≈ 1.66 km/h

Therefore, the average speed during the given time interval is approximately 1.66 km/h.

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Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 4.0 kg) portrayed in the figure below move on a frictionless surface and a force F = 34 N acts as shown on the 4.0-kg block. Answer parts a-c.

Answers

(a) The acceleration of the system is 8.5 m/s².

(b) The tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.

(c) The force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.

To solve this problem, we can use Newton's second law of motion (F = ma) and consider the forces acting on each block individually.

(a) Determine the acceleration given this system:

To find the acceleration (a) of the system, we can use the net force acting on the 4.0 kg block (m3). The only force acting on m3 is the applied force (F = 34 N).

F = m3 * a

34 N = 4.0 kg * a

Solving for a, we find:

a = 34 N / 4.0 kg

a = 8.5 m/s²

Therefore, the acceleration of the system is 8.5 m/s².

(b) Determine the tension in the cord connecting the 4.0-kg and the 1.0-kg blocks:

To find the tension in the cord (T), we can consider the forces acting on the 1.0 kg block (m1).

T - F = m1 * a

T - 34 N = 1.0 kg * 8.5 m/s²

T - 34 N = 8.5 N

T = 42.5 N

Therefore, the tension in the cord connecting the 4.0 kg and 1.0 kg blocks is 42.5 N.

(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block:

To find the force exerted by the 1.0 kg block (m1) on the 2.0 kg block (m2), we can consider the forces acting on the 2.0 kg block.

F - T = m2 * a

F - 42.5 N = 2.0 kg * 8.5 m/s²

F - 42.5 N = 17 N

F = 59.5 N

Therefore, the force exerted by the 1.0 kg block on the 2.0 kg block is 59.5 N.

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Question 1 of 10
What is the slope of the line plotted below?
B. 2
5
10
C. 1
O A. 0.5
о
9
OD. -0.5
5

Answers

The answer is B. 2


Explanation

Select the correct answer.
In which situation is maximum work considered to be done by a force?
A.
The angle between the force and displacement is 180°.
B.
The angle between the force and displacement is 90°.
C.
The angle between the force and displacement is 60°.
D.
The angle between the force and displacement is 45°.
E.
The angle between the force and displacement is 0°.

Answers

Option A. The angle between the force and displacement is 180°, the maximum work is considered to be done by the force.

Work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Mathematically, work (W) is given by the equation:

W = F * d * cos(theta)

Where

F = magnitude of the force

d = magnitude of the displacement

theta = angle between the force and displacement vectors.

In order to maximize the work done by a force, we need to maximize the value of the cosine of the angle theta. The cosine function reaches its maximum value of 1 when the angle theta is 0° or 180°.

When the angle between the force and displacement is 0° (option E), the force and displacement vectors are perfectly aligned in the same direction. In this case, the work done is maximized. Therefore, the correct answer is option A.

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