A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.

Answers

Answer 1

Answer:

Answer:

At the highest point the velocity is zero, the acceleration is directed downward.

Explanation:

This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.

I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.

At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.

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A physics cart has a projectile launcher mounted on top. While traveling on a straight track at 0.500 m/s, a projectile is fired. It lands back in the same place on top of the launcher after the cart has moved a distance of 2.30 m. In the frame of reference of the cart, (a) at what angle was the projectile fired and (b) what was the initial velocity of the projectile? (c) What is the shape of the projectile as seen by an observer on the cart? A physics student is watching the demonstration from a classroom seat. According to the student, (d) what is the shape of the projectile’s path, and (e) what is its initial velocity?

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mirror. What is the magnification of the image? Where is the car located, and what is its height? Is the image real or virtual? Is the image upright or inverted? Draw a ray diagram to show where the image forms and how large it is with respect to the object

Answers

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$(1)/(v)=(1)/(f)+(1)/(u)$

Put the value into the formula

$(1)/(0.24)=(1)/(0.25)+(1)/(u)$

$(1)/(u)=(1)/(0.24)-(1)/(0.25)$

$(1)/(u)=(1)/(6)$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-(v)/(u)$

Put the value into the formula

$m=-(0.24)/(-6)$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=(h')/(h)$

$h=(0.080)/(0.04)$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Answer:

Distance of the object = 6 m

Height of the object = 2 m

Explanation:

Thinking process:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Therefore, using formula of lens:

$(1)/(u) = (1)/(f) + (1)/(u)$

$(1)/(u) = (1)/(6)$

solving, gives u = 6

The magnification is calculated as follows:

m = -0.24/-6

= 0.04

The height = 2 m

The diagram yields an image behind the mirror which is upright.

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact with the ground. Determine the weight of the automobile.

Answers

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, $P=2.2* 10^5\ Pa$

Area of each tire, $A=0.023\ m^2$

Area of 4 tires, $A=0.092\ m^2$

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

$P=(F)/(A)$

$F=P* A$

$F=2.2* 10^5* 0.092$

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

Final answer:

The vehicle's weight can be calculated by rearranging the definition of pressure (Pressure = Force/Area) to solve for Force (Force = Pressure * Area), then multiplying by four to account for all four tires. Remember that the result will be in newtons, so to convert it to kilograms, divide by gravitational acceleration.

Explanation:

Your question revolves around the concept of pressure. Tire pressure is a type of air pressure which is a part of physics. To determine the weight of the automobile in which each of the four tires has an area of 0.023 m2 and is inflated to a gauge pressure of 2.2 x 105 Pa, we need to utilize the fundamental equation of pressure:

P = F/A

Where P is the pressure, F is the force (which in this case will be the weight of the car), and A is the area of each of the tires where they are in contact with the ground. Solving for the weight (F) results in:

F = P * A

In your case, because there are four tires we multiply the result by four, therefore:

F = 4 * (2.2 x 105 Pa) * (0.023 m2)

We have to multiply this by 4 to account for all four tires. Finally, your weight will be in newtons, to convert it to kg you will divide by gravitational acceleration (approx 9.8 m/s2).

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A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?

Answers

Final answer:

The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.

Explanation:

To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

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Impulse is the______of the force and time of contact

Answers

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

A^^\-> points in the -x direction with a magnitude of 21. What is the y component of A^^\->

Answers

Answer:

$A_y=-36.37^(\circ)$

Explanation:

Given that,

Vector A points in the -x direction with a magnitude of 21.

Let the x component is making an angle of 60 degrees with negative x axis. The x component of a vector is given by :

$A_x=A\ cos\theta$

$-21=A\ cos(60)$

$A=(-21)/(cos(60))$

A = -42 units

The y component of a vector is given by :

$A_y=A\ sin\theta$

$A_y=-42\ sin(60)$

$A_y=-36.37^(\circ)$

So, the y component of vector A is (-36.37) degrees. Hence, this is the required solution.

A ball with radius .15 m is rolling with an angular velocity of 6.5 rad/s. (a) What linear distance will the ball roll in 5.0 seconds? The ball then slows down with a linear acceleration of -1.2 rad/s2. (B) How fast will it be rolling rad/s after .65 second? (C) If a piece of gum is stuck on the outer part of the ball, what is its linear speed? (4.9, 5.72, .858)