Two horizontal forces, 230 N and 120 N, are exerted in opposite direction on a crate. What is the horizontal acceleration of the crate if the mass of the crate is 20 kg?
5.5 m/s2
20 m/s2
17.5 m/s2
72 m/s2

Answers

Answer 1

Answer:

a = 5.5 [m/s²]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the force is equal to the product of mass by acceleration.

ΣF =m*a

where:

F = Force [N] (units of Newtons)

m = mass = 20 [kg]

a = acceleration [m/s²]

Let's take the force of 230 [N] as positive, in this way the other force will be negative, by pointing in the opposite direction.

[tex]230 - 120 = 20*a\\110 = 20*a\\a=5.5 [m/s^{2} ][/tex]


Related Questions

When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) what is the effective force constant of the springs? in N/m (b) The four people get out of the car and bounce it up and down. What is the frequency of the car's vibration?

Answers

Answer:

Explanation:

F=kx

x=F/k

F=2000 kg

x=100 cm=9*10^-3

effective spring constant=k=F/x

k=2000/9*10^-3=2.2*10^-5

now frequency

f=1/2π√k/m

f=1/2*3.14√2.2*10^-5/310

f=1/6.28√7.097*10^-8

f=1/6.28*2.7*10^-4

f=0.16*2.7*10^-4

f=4.32*10^-5

The effective spring constant of the springs is 33755.55 N/m.

The frequency of the car's vibration is 2.07 Hz.

What is force?

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

Weight of the four people: F = 310 × 9.80 N = 3038 Newton.

The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.

Hence, the effective spring constant of the springs: k= force/compression

= 3038 N/0.90 × 10⁻² m

= 33755.55 N/m.

The frequency of the car's vibration is: f = 1/2π√(k/m)

=1/2π√(33755.55/2000)

= 2.07 Hz.

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a bus with a mass of 5000kg is uniformly acceleration from rest. The net force acting on the bus is

Answers

Answer:

50,000N

Explanation:

According to Newton's second law of motion;

Net Force = Mass * acceleration

Given

Mass = 5000kg

Let the acceleration = 10m/s²

Net force = 5000 * 10

Net force = 50,000N

Hence the net force acting on the bus is 50000N

If a dog has a mass of 2.5 kg, what is its weight and what is the normal force that it feels.
I

Answers

Answer:

Weight = normal force = 24.5 N

Explanation:

Given that,

Mass of a dog, m = 2.5 kg

We need to find its weight and the normal force that it feels.

The weight of an object is given by :

W = mg

Where g is the acceleration due to gravity

[tex]W=2.5\times 9.8\\\\=24.5\ N[/tex]

The normal force is balanced by the weight of an object. So,

Weight = normal force = 24.5 N

If Earth’s Moon were replaced with a typical neutron star, what would the angular diameter of the neutron star be as seen from Earth?

Answers

Answer:

[tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex]

Explanation:

d = Diameter of typical neutron star = 20 km = 20000 m

D = Distance between Earth and Moon = [tex]384.4\times 10^6\ \text{m}[/tex]

Here, [tex]D>>d[/tex] so we use small angle approximation

[tex]\delta=\dfrac{d}{D}\\\Rightarrow \delta=\dfrac{20000}{384.4\times 10^6}\\\Rightarrow \delta=0.00005202\ \text{rad}=\dfrac{0.00005202\times 180}{\pi}=0.003^{\circ}[/tex]

The angular diameter of the neutron star would be [tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex] from Earth.

How far out from the sun is the rock line?

Answers

Answer:

5 million miles

Explanation:

A plane is flying due west at 34 m/s. It encounters a wind blowing at 19 m/s south. Find the resultant veloci

Answers

Answer:

The resultant velocity has a magnitude of 38.95 m/s

Explanation:

Vector Addition

Given two vectors defined as:

[tex]\vec v_1=(x_1,y_1)[/tex]

[tex]\vec v_2=(x_2,y_2)[/tex]

The sum of the vectors is:

[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]

The magnitude of a vector can be calculated by

[tex]d=\sqrt{x^2+y^2}[/tex]

Where x and y are the rectangular components of the vector.

We have a plane flying due west at 34 m/s. Its velocity vector is:

[tex]\vec v_1=(-34,0)[/tex]

The wind blows at 19 m/s south, thus:

[tex]\vec v_2=(0,-19)[/tex]

The sum of both velocities gives the resultant velocity:

[tex]\vec v =(-34,-19)[/tex]

The magnitude of this velocity is:

[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]

[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]

d = 38.95 m/s

The resultant velocity has a magnitude of 38.95 m/s

David Wetterman drops a 5 kg watermelon from the top of a 30 m building. What is the velocity of the watermelon as it smashes
into the ground (neglecting air resistance)?
-(1)
A)
24.25 m/s
B)
32.45 m/s
C)
60 m/s
D)
588 m/s

Answers

Answer:

A. 24.25 m/s

Explanation:

velocity = [tex]\sqrt{2 * g * d}[/tex]

velocity = sqr 2 * 9.8 * 30 = sqr 588 = 24.25 m/s

The velocity of the watermelon as it smashes into the ground will be 24.2 m/s

State the third equation of motion?

The third equation of motion is -

v² - u² = 2aS

Given David Wetterman drops a 5 kg watermelon from the top of a 30 m building.

Height of building [S] = 30 m

Mass of watermelon [M] = 5 Kg

Initial velocity [v] = 0 m/s

acceleration [g] = 9.8 m/s²

Using the third equation of motion -

v² - u² = 2aS

v² = 2aS

v² = 2 x 9.8 x 30

v² = 588

v = 24.2 m/s

Therefore, the velocity of the watermelon as it smashes into the ground will be 24.2 m/s.

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What is magnet made of

Answers

Answer:

metals like iron or nickel

Explanation:

It is more difficult to climb a vertical staircasethan a slanted one​ give reason

Answers

Answer:

The reason is similar to the reason why it is difficult to roll an object on a surface with a positive incline than rolling it on the ground

The more the path becomes vertical, the more force we have to apply to oppose the force of gravity

but when we are moving horizontally, we don't have to move against the gravity and hence, it is less difficult than going vertically upwards

An object with a mass of 3.0 kg has a
force of 9.0 newtons applied to it. What
is the resulting acceleration of the
object?

Answers

[tex] \LARGE{ \underline{ \tt{Required \: answer:}}}[/tex]

We have:

Mass of the object = 3 kgForce on the object = 9 N

We need to find:

Resulting accleration of the object?

Solution:

According to Newton's 2nd law of motion, or quantitative measure of Force:

Force = Mass × Accleration

Using this,

➝ F = ma

➝ 9N = 3 kg × a

➝ a = 9/3 m/s²

➝ a = 3 m/s²

Hence,

The resulting accleration of the object is 3 m/s². And we are done! :D

⛱️ [tex] \large{ \blue{ \bf{FadedElla}}}[/tex]

If mass (3.0 kg) multiplying (*) acceleration gives you force (newtons), force dividing mass gives you acceleration; 9/3 =


3 m/s^2

Two kids are roller skating. Amy, with a mass of 55 kg, is traveling forward at 3 m/s. Jenny, who has a mass of 40 kg, is traveling in the opposite direction at 5 m/s. They crash into each other and hold onto each other so that they move as one mass. How fast are they traveling?

Answers

Answer:

-7/19

Explanation:

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. The design specifications call for it to have a rotational kinetic energy of 330 J after it has turned through 30.0 revolutions.

What should be the moment of inertia of the flywheel about its rotation axis?
Express your answer with the appropriate units.

Answers

Answer: 1.14 kg*m/s

Explanation:

The first person explained everything right, they just forgot to convert the angular acceleration to rads/sec^2 from revs/sec^2. Once that is converted, your answer should come out right.

Another small thing, the answer there has an extra unnecessary step. It tells you to find the square root of w^2 to find w but that is unnecessary since the final equation asked for w^2. Hope this helps! :)

The moment of inertia I of the flywheel about its rotation axis is

[tex]1.39Kgm^2[/tex]

Given

Angular displacement,

[tex]\theta = 30rev \\\\\theta = (30) * 2\pi rad \\\\\theta = 188.495rad[/tex]

Therefore, Final angular velocity (w) will be:

[tex]w^2 = 2\alpha\theta\\\\w^2 = 2 * (0.200 * 2\pi) * (188.49)\\\\w^2 = 473.73\\\\w = 21.76 rad/s[/tex]

Therefore,

moment of inertia

[tex]I = 2 * K / w^2[/tex]

[tex]I = 2 * 330 / 473.73[/tex]

[tex]I = 1.39kgm^2[/tex]

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. A car going initially with a velocity 15 m/s accelerates at a rate of 2 m/s2 for 10 seconds. It then accelerates at a rate of -1.5 m/s until stop. Find the car’s maximum speed. Calculate the total distance traveled by the car.

Answers

Answer:

The maximum speed of the car is 35 m/s

The total distance traveled by the car is 658.33 m

Explanation:

Given;

initial velocity of the car, u = 15 m/s

acceleration of the car, a = 2 m/s²

time of car motion, t = 10 s

(i)

Initial distance traveled by the car is given by;

d₁ = ut + ¹/₂at²

d₁ = (15 x 10) + ¹/₂(2)(10)²

d₁ = 150 + 100

d₁ = 250 m

The maximum speed of the car during this is given by;

v² = u² + 2ad₁

v² = (15)² + (2 x 2 x 250)

v² = 1225

v = √1225

v = 35 m/s

(ii)

The final distance cover by the car during the deceleration of 1.5 m/s².

Note: the final or maximum speed of the car becomes the initial velocity during deceleration.

v² = u² + 2ad₂

where;

v is the final speed of the car when it stops = 0

0 = u² + 2ad₂

0 = (35²) + (2 x - 1.5 x d₂)

0 = 1225 - 3d₂

3d₂ = 1225

d₂ = 1225 / 3

d₂ = 408.33 m

The total distance traveled by the car is given by;

d = d₁ + d₂

d = 250 m + 408.33 m

d = 658.33 m

A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?

Answers

Answer:

599.245km/hr

Explanation:

A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?

We solve the above question using vectors

In vector form Air speed is -540i + 0j Wind speed is (-80/√2)i + (80/√2)j

Vector notation wind speed is given as: -56.5685 i + 56.5685j

The vector for the ground speed of the plane =

-540i + 0j -56.5685i + 56.5685j

= -596.56854249i + 56.5685j

The the ground speed of the plane √[(596.56854249)² + (56.5685)²]

= √359094.021081 km/hr

= 599.24454197 km/hr

Approximately

= 599.245km/hr

Consider a block sliding down a ramp whose motion is opposed by frictional forces. The total energy of this system is modeled by the equation:
Etotal = 1/2mv^2 + mgh + Ff(f is underscore)d

Which part of the equation represents the amount of energy converted to thermal energy?

A. mg
B. Ffd
C. mgh
D. 1/2 mv^2

Answers

Answer:

Energy Flows Quick check answers:

1. Ffd.

2. The kinetic energy decreases, and gravitational potential energy increases.

3. The internal energy of the system increases.

4. KEbox= Etotal-mgh

5. Etotal = 1/2m1(v1)^2+1/2m^2(v2)^2+U

The part of the equation that represents the amount of energy converted to thermal energy is [tex]F_f d[/tex].

The given equation for the total energy of a system;

[tex]E_{total} = \frac{1}{2} mv^2 \ +\ mgh\ + \ \ F_fd[/tex]

The definition of the various terms in the energy equation is given as;

[tex]E_{total}[/tex]: this is the total mechanical energy of the system[tex]\frac{1}{2} mv^2[/tex]: this is the kinetic energy of the system[tex]mgh[/tex]:  this is the potential energy of the system[tex]F_f d[/tex]:  this is the energy lost due to friction.

The energy lost due to friction is equal to the energy converted to thermal energy.

Thus, the part of the equation that represents the amount of energy converted to thermal energy is [tex]F_f d[/tex].

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A household refrigerator consumes electrical energy at the rate of 200 W. lf electricity costs 5 k per kWh, calculate the cost of operating the appliance for 30 days

Answers

Answer:

= 720000 [k]

Explanation:

The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.

[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]

= 720000 [k]

someone help please


waves disturb ____, but do not transmit it.

a. energy
b. matter
c. sound
d. none of the above

Answers

Answer:

b. matter

Explanation:

Waves disturb matter but do not transmit it.

Waves are disturbances that transmit energy from one point to another. Although they cause disturbances, they do not transfer the matters in the medium.

Energy is propagated by a wave. When for example, sound waves are produced, the disturbance is propagated via particle - particle interaction But after the wave train moves, the particles remain.

Write Radar gun summary

Answers

Answer:

A radar gun is a device for measuring the speed of moving objects. ... The radar gun is a Doppler radar unit that can be static, vehicle-mounted or hand-held. It measures the

Explanation:

Which describes the genetic disorder that causes neurons in the brain to break down? scribes the genetic disorder that causes neurons in the brain to break down need help will mark brainliest

Answers

Genetic disorders and Parkinson's disease.

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s.

Answers

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  [tex]X = 38.3 \ m[/tex]

Explanation:

From the question we are told that

The acceleration along the x axis is  [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  [tex]v_{0}x = 7.10 m/s[/tex]

Generally from the equation for acceleration along x axis we have that

     [tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]

=>   [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]

=>   [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

At  t =0  s   and  [tex]v_{0}x = 7.10 m/s[/tex]

=>   [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]

=>   [tex]K_1 = 7.10[/tex]      

So  

      [tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

=>  [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]

=>  [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]

At  t =0  s   and   x = -14.0 m

  [tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]

=>   [tex]K_2 = -14[/tex]

So

     [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]

At  t = 10.0 s

      [tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]

=>   [tex]X = 38.3 \ m[/tex]

             

     

Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?

Answers

Answer:

Acceleration = 18g

Explanation:

Given the following data;

Initial velocity, u = 26m/s

Final velocity, v = 0

Time = 0.15 secs

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation, we have;

[tex]a = \frac{0 - 26}{0.15}[/tex]

[tex]a = \frac{26}{0.15}[/tex]

Acceleration = 173.33m/s2

To express it in magnitude of g;

Acceleration = 173.33/9.8

Acceleration = 17.7 ≈ 18g

Acceleration = 18g

A submarine sends out a sonar signal (sound waves) in a direction directly downward it take 2.3 s for the sound waves to travel from the submarine to the ocean bottom and back to the submarine how high (approx) up from the ocean floor is the submarine?speed of sound in water is 1490 m/s

Answers

Explanation:

Using the formula;

2x = vt

x is the distance up from the ocean floor the submarine is

v is the speed of sound in water

t is the time

Given

t = 2.3s

v = 1490m/s

Required

how high (approx) up from the ocean floor is the submarine x

From the formula;

x = vt/2

x = 1490(2.3)/2

x = 745(2.3)

x = 1,713.5m

Hence the submarine is 1713.5m high up from the ocean floor

A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.

Answers

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Magnitude of normal force ( object at rest );  n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x  after object start moving   x = 25 N

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

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A radio wave transmits 38.5 W/m2 of power per unit area. A flat surface of area A is perpendicular to the direction of propagation of the wave. Assuming the surface is a perfect absorber, calculate the radiation pressure on it.

Answers

Answer:

[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

[tex]P=\dfrac{2I}{c}[/tex]

Where

c is speed of light

Putting all the values, we get :

[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]

So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].

A spaceship is accelerating at 1000 m/sec2 . How much force is required from the backthrusters to completely stop the spaceship?

Answers

Answer:

1000x Newton

Explanation:

Step one

given data

acceleration= 1000 m/s²

The question did not specify the mass of the mass of the space ship.

So, let's assume the mass is x kg

Step two:

Required is the force F in Newton

From Newtons first law, it states that a body will continue to be at rest or uniform motion unless acted upon by a force.

F=mass x Acceleration

F=ma

Substituting our given data we have

F=1000x Newton

a current of 200mA through a conductor converts 40 joules of electrical energy into heat in 30 seconds determine the potential drop across the conductor

Answers

Answer:

V = 6.65 [volt]

Explanation:

We must first find the power generated, power is defined as the amount of energy consumed or generated in a given time.

[tex]P=\frac{E}{t}[/tex]

where:

P = power [w]

E = energy = 40 [J]

t = time = 30 [s]

[tex]P =40/30\\P = 1.33[w][/tex]

Now we can calculate the voltage or potential drop by means of the power, the power is calculated by means of the product of the voltage by the current.

[tex]P =V*I[/tex]

where:

V = voltage [volts]

I = current = 200mA = 0.2 [A]

[tex]V = P/I\\V = 1.33/0.2\\V = 6.65 [Volt][/tex]

it is the question 12 part okay​

Answers

Answer:

Yeah it's ok I think. Also, I can't see the answer you gave so maybe updating the question would be nice.

So if you fall into icy water your gonna freeze that’s a given but you can stop it if you get out fast and get in a real hot bath and then get a towel on and you get Hypothermia because your body physically can’t take the cold so it decides to freeze your insides until your heart eventually just stops because it’s frozen.

Can I pls have brainliest :)

A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.

(a) Neglecting air resistance, how long does it take the skier to reach the ground?

(b) How far horizontally does the skier travel in the air before landing?

m

Answers

Explanation:

Given

Velocity v = 23.0m/s

Distance S = 3.45m

Required

Time it will take the skier to reach the ground;

Using the equation of motion;

S = ut + 1/2gt²

3.45 = 23t + 1/2(9.8)t²

3.45 = 23t + 4.9t²

4.9t²+23t-3.45 = 0

Factorize;

t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)

t = -23 ±√529+67.62/9.8

t = -23±√596.62/9.8

t = -23±24.43/9.8

t = 1.43/9.8

t = 0.146 secs

Hence take the skier 0.146 secs to reach the ground.

b) Horizontal distance covered is the range;

Range = U√2H/g

Range = 23√2(3.45)/9.8

Range = 23√6.9/9.8

Range = 23√0.7041

Range = 23(0.8391)

Range = 19.29m

Hence the horizontal distance travelled in air is 19.29m

(a).The time taken by the skier to reach the ground is 0.145 second.

(b).The skier travel in the air before landing is 19.29 meter.

a. Given that A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.

Using equation of motion.

       [tex]S=ut+\frac{1}{2}gt^{2}[/tex]

Where S is vertical distance , u is initial velocity and g is gravitational acceleration.

Substitute S = 3.45 m, u = 23m/s  and g = 9.8 in above equation.

        [tex]3.45=23t+\frac{1}{2} (9.8)t^{2}\\\\4.9x^{2} +23t-3.45=0\\\\t=0.145,-4.83[/tex]

Since, time can not be negative.

So that,  [tex]t=0.145s[/tex]

b. The horizontal distance travel before landing is known as Range.

           Horizontal distance  ,

                                             [tex]=v*\sqrt{\frac{2S}{g} }[/tex]

Substitute v = 23m/s , S = 3.45 and g = 9.8 meter per second square.

                         [tex]Distance=23*\sqrt{\frac{2*3.45}{9.8} } \\\\Distance=23*\sqrt{0.7041} \\\\Distance=23*0.8391=19.29m[/tex]

Thus, The skier travel in the air before landing is 19.29 meter.

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A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

Answers

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .

Answers

Complete Question

A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex]  to  [tex]t_2 = 6.0s[/tex]

What is its average velocity over this time interval?

Answer:

The velocity is  [tex]v = 3.903 \ m/s[/tex]

Explanation:

From the question we are told that

    The first position of the ball is  [tex]x_1 = 8.0 \ cm[/tex]

    The second position of the ball is  [tex]x_2 = - 4.1 \ cm[/tex]

Generally the average velocity is mathematically represented as

       [tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]

=>    [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]

=>    [tex]v = 3.903 \ m/s[/tex]

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