In transverse waves, the medium moves perpendicular to the direction of energy transport? True or false?

A 7.2 kg bowling ball is accelerated with a force of 112,5 N.

Dominik [7]

Answer:

15.625 m/s

Explanation:

112.5/7.2 = 15.625 m/s

6 0

3 years ago

Which wave diffracts the most when encountering an obstacle?

viva [34]

Answer;

-A wave with the longest wavelength.

Explanation;

-Diffraction is the apparent of wave through,around small obstacles and the spreading out of wave past small openings. When thinking of diffraction of a wave think of shining a flashlight around a corner. The light bends around the corner but there is a place where it is dark and the light does not hit. Diffraction of a wave is basically the wave bending around an object then dispersing out.

-The amount of diffraction (the sharpness of the bending) increases with increasing wavelength and decreases with decreasing wavelength. When the wavelength of the waves is smaller than the obstacle, no noticeable diffraction occurs.

8 0

3 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

A weather balloon has a volume of 90.0 l when it is released from sea level. What

frozen [14]

The final atmospheric pressure is $5.19\cdot 10^4 Pa$

Explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

$pV=const.$

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

$p_1 V_1 = p_2 V_2$

where in our problem we have:

$p_1= 1.03\cdot 10^5 Pa$ is the initial pressure (the atmospheric pressure at sea level)

$V_1 = 90.0 L$ is the initial volume

$p_2$ is the final pressure

$V_2 = 175.0 L$ is the final volume

Solving the equation for p2, we find the final pressure:

$p_2 = \frac{p_1 V_1}{V_2}=\frac{(1.01\cdot 10^5)(90.0)}{175.0}=5.19\cdot 10^4 Pa$

Learn more about ideal gases:

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#LearnwithBrainly

3 0

3 years ago

An object moving with a speed or 67m/s and has kinetic energy of 500 J what is the mass of the object

saveliy_v [14]

To solve this equation, simply plug the values into the equation for calculating kinetic energy.

KE = 1/2mv^2
500 = 1/2(m)(67^2)
500 =2244.5m
m = 500/2244.5 = 0.222 kg.

8 0

3 years ago