PHYSICS COLLEGE

A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train moves a distance of meters.

Answers

Answer 1

Answer:

So lets fill out what we have first:

Vi or initial velocity = 20 m/s

Acceleration or a = 4 m/s^2

Time for the motion = 10s

Now, using the four main kinematic equations we can deduce that the best kinematic equation to use in these terms is:

Δx = Vi(t) + 0.5at²

Plug all of our information in:

Δx = (20)(10) + (0.5)(4)(100)

Δx = 400 m

Answer 2

Answer:

Answer:

400 m

Explanation:

answer on ed

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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor
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A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

Answers

Answer:

22.1 m

Explanation:

$v_(o)$ = initial speed of ball = 14.3 m/s

$\theta$ = Angle of launch = 27°

Consider the motion of the ball along the vertical direction.

$v_(oy)$ = initial speed of ball = $v_(o) Sin\theta = 14.3 Sin27 = 6.5 ms^(-1)$

$a_(y)$ = acceleration due to gravity = - 9.8 ms⁻²

$t$ = time of travel

$y$ = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

$y = v_(oy) t + (0.5) a_(y) t^(2) \n- 3.50 = (6.5) t + (0.5) (- 9.8) t^(2)\n- 3.50 = (6.5) t - 4.9 t^(2)\nt = 1.74 s$

Consider the motion of the ball along the horizontal direction.

$v_(ox)$ = initial speed of ball = $v_(o) Cos\theta = 14.3 Cos27 = 12.7 ms^(-1)$

$X$ = Horizontal distance traveled

$t$ = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

$X = v_(ox) t\nX = (12.7)(1.74)\nX = 22.1 m$

A block, initially at rest, has a mass m and sits on aplane inclined at angle (theta). It slides a distance d before hitting a spring and compresses the spring by a maximum distance of xf. If the coefficient of kinetic friction between the plane and block is uk, then what is the force constant of the spring?

Answers

Answer: k = ma + uk×mgcosθ/ xf

Explanation: The body is placed on a frictionless inclined ramp.

The weight of the object has 2 components, horizontal component (mgsinθ) and vertical component (mgcosθ).

The horizontal component of weight is responsible for making tje object slide down the plane even with no applied force.

So from newton's second law of motion

mgsinθ - uk×R = ma

Where uk = coefficient of kinetic friction.

R = normal reaction = mgcosθ

mgsinθ - uk×mgcosθ = ma

mgsinθ = ma + uk×mgcosθ

mgsinθ is the applied force in this case. This applied force compresses a spring.

According to hooke's law,

F =ke

Where F = ma + uk×mgcosθ, e =xf

F = applied force , e = extension and k = spring constant.

k = F/e

k = ma + uk×mgcosθ/ xf

A 2-C charge experiences a force of 40 N when put at a certain location inspace. The electric field at that location is a. 2 N/C.b. 20 N/C. c. 30 N/C. d.
40 N/C. e. 60 N/C.

Answers

Answer:

E = 20 N/C

Explanation:

Given that,

Charge, q = 2 C

Force experience, F = 40 N

We need to find the electric field at that location.

The electric field in terms of electric force is given by :

F = qE

Where

E is the electric field

$E=(F)/(q)\n\nE=(40\ N)/(2\ C)\nE=20\ N/C$

So, the electric field at that location is 20 N/C.

You have 5 cats and they each have a mass of 4kg per cat. What is the mass of all of them together?

Answers

Answer:

It would be 20kg

Explanation:

This would be just 5x4 as there are 5 cats and each are 4kg. You can also add 4, 5 times as well.

I hope Im correct

Instantaneous speed is...a) A speed of 1000 km/h
b) The speed attained at a particular instant in time.
c) The speed that can be reached in a particular amount of time.

PLEASE HURRY

Answers

Answer:

The speed attained at a particular instant in time.

Explanation:

Instantaneous speed is the speed attained at a particular instant in time.

It is given by :

$v=(dx)/(dt)$

It is equal to the rate of change of speed.

It can be also defined as when the speed of an object is constantly changing, the instantaneous speed is the speed of an object at a particular moment (instant) in time.

Hence, the correct option is (b).

Madelin fires a bullet horizontally. The rifle is 1.4 meters above the ground. The bullet travels 168 meters horizont before it hits the ground. What speed did Madelin's bullet have when it exited the rifle?

Answers

The position vector of the bullet has components

$x=v_0t$

$y=1.4\,\mathrm m-\frac g2t^2$

The bullet hits the ground when $y=0$ , which corresponds to time $t$ :

$1.4\,\mathrm m-\frac g2t^2=0\implies t=0.53\,\mathrm s$

The bullet travels 168 m horizontally, which would require a muzzle velocity $v_0$ such that

$168\,\mathrm m=v_0(0.53\,\mathrm s)$

$\implies v_0\approx320\,(\mathrm m)/(\mathrm s)$

Final answer:

In the given physics problem, the bullet travels horizontally 168 meters before hitting the ground from a height of 1.4 meters. By calculating the time it takes for the bullet to fall to the ground due to gravity and then applying that time to the horizontal distance traveled, we find that the speed of the bullet when it exited the rifle was approximately 313.43 m/s.

Explanation:

The scenario defined is a classic Physics problem where an object is fired horizontally and falls to the ground due to gravity. We can calculate the horizontal speed of the bullet using the equations of motion associated with the vertical, free-fall motion of the bullet.

Gravity causes the bullet to fall to the ground. As we know that the height from the ground is 1.4 meters, we can calculate the time taken for the bullet to hit the ground using the equation: time = sqrt(2 * height / g), where g is the gravitational constant (approx. 9.8 m/s^2).

Substituting the given value, we get time = sqrt(2 * 1.4 / 9.8), which is around 0.536 seconds. The bullet travels 168 meters in this time horizontally, therefore its horizontal speed will be distance / time, which is 168 meters / 0.536 seconds = 313.43 m/s. So, Madelin's bullet had a speed of around 313.43 m/s when it exited the rifle.

Learn more about Projectile Motion here:

brainly.com/question/20627626

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