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

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

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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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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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.

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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.

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.

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.

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

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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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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(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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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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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.

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

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) 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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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.

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(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.

(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;

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;

s = 0 + ¹/₂(5.3)(1²)

s = 2.65 m

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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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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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Answer:

Gravity.

Explanation:

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

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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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(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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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.

(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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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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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.

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.

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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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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