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3 years ago

Claudia throws a baseball to her dog. Which free-body diagram shows the

Physics

1 answer:

chubhunter [2.5K]3 years ago

6 0

Answer:

only the weight of the ball will act on the ball

Explanation: There is no contact force on the ball. Also there is no air resistance on the ball so the friction force on the ball due to air is not shown

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How much force is needed to stretch a spring 1. 2 m if the spring constant is 8. 5 N/m? 7. 1 N 7. 3 N 9. 7 N 10. 2 N.

statuscvo [17]

The amount of force required to stretch or compress the spring is known as the spring force. Its unit is Newton(N). Force is needed to stretch spring is 10.2 N.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

$\rm F_S=Kx \\\\ \rm F_S=8.5 \times 1.2 \\\\ \rm F_S=10.2 N$

Hence the force is needed to stretch the spring is 10.2 N.

To learn more about the spring force refer to the link;

brainly.com/question/4291098

6 0

2 years ago

At a height of ten meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the la

konstantin123 [22]

Answer:

Explanation:

Velocity of sound in air at 20 degree = 343 m/s

Velocity of sound in water at 20 degree = 1470 m/s

Time taken in to and fro movement in air

=( 2 x 10) / 343 = 0.0583 s

Rest of the time is

.171 - .0583 = .1127 s

This time is taken to cover distance in water. If d be the depth of lake

2d / velocity = time taken

2 d / 1470 = .1127

d = 82.83 m

5 0

3 years ago

Please Help!! This is very tough for me. Try you best to answer these problems!!Pleaseee

Andrews [41]

Answer:

everyone else does this to me so lol

Explanation:

7 0

3 years ago

Read 2 more answers

Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of

max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0

3 years ago

Read 2 more answers

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

3 years ago