Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horizontally with a constant speed of 3.80 m/s relative to observer 2 standing on the sidewalk.a) What is the speed of the ball 1.00 s after it is released, as measured by observer 2?

b) What is the direction of travel of the ball 1.00 s after it is released, as measured relative to the horizontal by observer 2?

Answers

Answer 1
Answer:

a) 10.5 m/s

While for observer 1, in motion with the car, the ball falls down straight vertically, according to observer 2, which is at rest, the ball is also moving with a horizontal speed of:

v_x = 3.80 m/s

As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

v_y = u_y + gt

where

u_y =0 is the initial vertical speed

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Therefore, after t = 1.00 s, the vertical speed is

v_y = 0 + (9.8)(1.00)=9.8 m/s

And so the speed of the ball, as observed by observer 2 at rest, is given by the resultant of the horizontal and vertical speed:

v=√(v_x^2 +v_y^2)=√((3.8)^2+(9.8)^2)=10.5 m/s

b) \theta = -68.8^(\circ)

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

\theta = tan^(-1) ((v_y)/(v_x))=tan^(-1) ((-9.8)/(3.8))=-68.8^(\circ)

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so v_x has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so v_y is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

\theta = -68.8^(\circ)


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Determine whether or not each of the following statement is true. If a statement is true, prove it. If the statement is false, provide a counterexample and explain how it constitutes a counterexample. Diagrams can be useful in explaining such things. If the electric potential in a certain region of space is constant, then the charge enclosed by any closed surface completely contained within that region is zero.

Answers

Answer:

True

Explanation:

This is a representation of Gauss law.

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Look at the figure below and calculate the length of side y.A. 8.5
B. 17
12
y
C. 6
O D. 12
45
Х

Answers

Answer:

I want to say a because you want to subtract and simplify

You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, ???? = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)

Answers

Answer:

 μ = 0.423

Explanation:

To solve this exercise we must use Newton's second law and kinematics together, let's start using expressions of kinematics to find the acceleration of the body

Let's fix a reference system where the x axis is parallel to the inclined plane, but the acceleration is only on this axis

            x = v₀ t + ½ a t²

The body starts from rest so its initial speed is zero

            a = 2 x / t²

            a = 2 0.5 /0.5²

            a = 4 m / s²

Taking the acceleration of the body, we use Newton's second law, we take the direction up the plane as positive

  X axis

                fr - Wₓ = m a          (1)

  Y Axis  

               N- W_(y) = 0

                N = W_{y}

We use trigonometry to find the components of the weight

            sin 45 = Wₓ / W

           cos 45 = W_{y} / W

           Wₓ = W sin 45

           W_{y} = W cos 45

The out of touch has the expression

             fr = μ N

             fr = μ W_{y}

We substitute in 1

             μ mg cos 45 - mg sin 45 = m a

             W_{y} = (a + g sin 45) / g cos 45

              μ = a / g cos 45 + 1

We calculate

Acceleration goes down the plane, so it is negative

           a = -4 m / s²

            μ = 1- 4 / (9.8 cos 45)

            μ = 0.423

Answer:

The μ = 0.422

Explanation:

The distance travelled by the mass is equal to:

L=ut+(1)/(2)at^(2)  \n0.5=(0*5)+(1)/(2) a(0.5^(2) )\na=4m/s^(2)

The sum of forces in y-direction equals zero:

∑Fy = 0

N - (m * g * cosθ) = 0

N - (1 * 9.8 * cos45) = 0

N = 6.93 N

The sum of forces in x-direction is equal to:

∑Fx = ma

(m * g * sinθ) - fk = m * a

(1 * 9.8 * sin45) - fk = 1 * 4

fk = 2.93 N

fk = μ * N

2.93 = μ * 6.93

μ = 0.422

A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

Answers

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

\mu mg = (mv^2)/(R)

here we have

\mu g = (v^2)/(R)

now we know that

v = √(\mu Rg)

here we have

\mu = 0.600

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

v = √((0.600)(35)(9.81))

v = 14.35 m/s

A force of 240.0 N causes an object to accelerate at 3.2 m/s2. What is the mass of the object?

Answers

the mass would be 75kg

What does a planet need in order to retain an atmosphere? How does an atmosphere affect the surface of a planet and the ability of life to exist?

Answers

Answer:

Explained

Explanation:

In order to retain atmosphere a planet needs to have gravity. A gravity sufficient enough to create a dense atmosphere around it, so that it can retain heat coming from sun. Mars has shallow atmosphere as its gravity is only 40% of the Earth's gravity. Venus is somewhat similar to Earth but due to green house effect its temperature is very high. Atmosphere has a huge impact on the planets ability to sustain life. Presence of certain kind gases make the atmosphere poisnous for life. The atmosphere should be such that it allows water to remain in liquid form and maintain an optimum temperature suitable for life.