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Phoenix [80]
1 year ago
9

A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 10 m/s. It is hit straight back a

t the pitcher with a final speed of 24 m/s. Assume the direction of the initial motion of the baseball to be positive.
Physics
1 answer:
Maru [420]1 year ago
4 0

Answer:

-5.1 kg m/s

Explanation:

Impulse is the change in momentum.

Change in momentum= final momentum - initial momentum=mv_{2} +mv_{1}

Plugging in the values= -0.15*24 - (0.15*10) (The motion towards the pitcher is negative as the initial motion is considered to be positive)

Impulse=-5.1 kg m/s (-ve means that it is the impulse towards the pitcher)

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A 1.00 kg ball traveling towards a soccer player at a velocity of 5.00 m/s rebounds off the soccer
matrenka [14]

Answer:

A)   F = - 8.5 10² N,  B)   I = 21 N s

Explanation:

A) We can solve this problem using the relationship of momentum and momentum

          I = Δp

in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction

         v₀ = 8.50 m / s

         v_f = -8.50 m / s

         F t = m v_f -m v₀

         F = m \frac{(v_f - v_o)}{t}

let's calculate

         F = 1.00 \ \frac{(-8.5-8.5)}{2 \ 10^{-2}}

         F = - 8.5 10² N

B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations

         v² = v₀² - 2g (y- y₀)

as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0

         v = \sqrt{2g y_o}

calculate  

         v = \sqrt{2 \ 9.8 \ 10}

         v = 14 m / s

to calculate the momentum we use

         I = Δp

         I = m v_f - mv₀

when it hits the ground its speed drops to zero

we substitute

         I = 1.50 (0-14)

         I = -21 N s

the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is

        I = 21 N s

4 0
1 year ago
Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b
Ainat [17]

Answer: the pair of sunglasses

Explanation:

A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

4 0
1 year ago
You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle
charle [14.2K]

Answer:

a=5.54m/s^{2}

Explanation:

The net force, F_{net} of the box is expressed as a product of acceleration and mass hence

F_{net}=ma where m is mass and a is acceleration

Making a the subject, a= \frac {F_{net}}{m}

From the attached sketch,  

∑ F_{net}=Fcos\theta-F_{f} where F_{f} is frictional force and \theta is horizontal angle

Substituting ∑ F_{net} as F_{net} in the equation where we made a the subject

a= \frac {Fcos\theta-F_{f}}{m}

Since we’re given the value of F as 240N, F_{f} as 41.5N, \theta as 30^{o} and mass m as 30kg

a= \frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}

6 0
2 years ago
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t
Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car at the moment they collided = a_c

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0
2 years ago
A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction
Vaselesa [24]

Answer:

H=1020.12m

Explanation:

From a balance of energy:

\frac{m*Vo^2}{2} -mg*H=-Ff*d   where H is the height it reached, d is the distance it traveled along the ramp and Ff = μk*N.

The relation between H and d is given by:

H = d*sin(30)   Replace this into our previous equation:

\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*N*d

From a sum of forces:

N -mg*cos(30) = 0    =>  N = mg*cos(30)   Replacing this:

\frac{m*Vo^2}{2} -mg*d*sin(30)=-\mu_k*mg*cos(30)*d   Now we can solve for d:

d = 2040.23m

Thus H = 1020.12m

6 0
1 year ago
Read 2 more answers
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