Ask question

Ask question

All categories

2 years ago

5

A pitcher throws a softball toward home plate. When the ball hits the catcherâ€™s mitt, its horizontal velocity is 32 meters/sec

ond. The softballâ€™s velocity goes to 0 meters/second in 0. 8 seconds when caught. If the softball has a mass of 0. 2 kilograms, whatâ€™s the force of the impact? Use F = ma, where. The force of the impact is newtons.

1 answer:

vladimir1956 [14]2 years ago

7 0

The force of impact of the softball at the given conditions is 8 N.

The given parameters:

<em>Horizontal velocity of the ball, v = 32 m/s</em>
<em>Change in time of motion of the ball, Δt = 0.8 s</em>
<em>Mass of the ball, m = 0.2 kg</em>

The force of impact of the softball is calculated by applying Newton's second law of motion as follows;

$F = ma\\\\F = m \times \frac{\Delta v}{\Delta t} \\\\F = 0.2 \times \frac{32 - 0}{0.8 - 0} \\\\F = \frac{0.2 \times 32}{0.8} \\\\F = 8 \ N$

Thus, the force of impact of the softball at the given conditions is 8 N.

Learn more about Newton's second law of motion here: brainly.com/question/25307325

You might be interested in

The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag

CaHeK987 [17]

The speed of tsunami is a.0.32 km.

Steps involved :

The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?

Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32

As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.

To learn more about tsunami refer : brainly.com/question/11687903

#SPJ4

6 0

2 years ago

What would you do to improve the precision of an experiment?

GenaCL600 [577]

Explanation:

Precision represents that how close the different measurements of the sample one take are to one another.

One can increase the precision in lab by paying attention to each and every detail.
Usage of the equipment properly and also increasing the sample size.
Ensuring that the equipment is calibrated properly. They should be clean and functioning. Using equipment which is not functioning correctly can cause results to swing wildly and also bits of the debris stuck to the equipment can influence the measurements of the mass and the volume.
Each measurement must be taken multiple times, especially if experiments in which combining of the substances in specific amounts is involved.

5 0

3 years ago

Read 2 more answers

A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t

vladimir1956 [14]

a) $\omega=\frac{-mvR}{I+MR^2}$

b) $v=\frac{-mvR^2}{I+MR^2}$

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

$L_1=0$

Later, after the girl throws the rock, the angular momentum will be:

$L_2=(I_M+I_g)\omega +L_r$

where:

$I$ is the moment of inertia of the merry-go-round

$I_g=MR^2$ is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

$\omega$ is the angular speed of the merry-go-round and the girl

$L_r=mvR$ is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

$L_1=L_2$

So we find:

$0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}$

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

$\omega=\frac{-mvR}{I+MR^2}$ is the angular speed

$r=R$ is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

$v=\omega R=\frac{-mvR^2}{I+MR^2}$

5 0

3 years ago

A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s

PolarNik [594]

Answer:

The value of radiation pressure is $2.933 \times 10^{-8}$ Pa

Explanation:

Given:

Intensity $I = 8.8$ $\frac{W}{m^{2} }$

Area of piece $A = 2.1 \times 10^{-4}$ $m^{2}$

From the formula of radiation pressure in terms of intensity,

$P = \frac{I}{c}$

Where $P =$ radiation pressure, $c =$ speed of light

We know value of speed of light,

$c = 3 \times 10^{8}$ $\frac{m}{s}$

Put all values in above equation,

$P = \frac{8.8}{3 \times 10^{8} }$

$P = 2.933 \times 10^{-8}$ Pa

Therefore, the value of radiation pressure is $2.933 \times 10^{-8}$ Pa

8 0

3 years ago

The power in an electrical circuit is given by the equation P= RR, where /is

Rainbow [258]

Answer:D

Explanation:Electric power=I*I*R

=12*12*100

=14400watts

6 0

3 years ago