All categories

2 years ago

A stone is dropped from the

Physics

1 answer:

ICE Princess25 [194]2 years ago

8 0

Height=h=500m
Acceleration=g=10m/s^2
Initial velocity=u=0
Speed of sound=c=340m/s
TIME TAKEN BY STONE TO HIT WATER=t
Time taken by sound to hear back=T

Now

$\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2$

$\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2$

$\\ \sf\longmapsto 500=5t^2$

$\\ \sf\longmapsto t^2=100$

$\\ \sf\longmapsto t=10s$

Now

$\\ \sf\longmapsto h=cT$

$\\ \sf\longmapsto T=\dfrac{h}{c}$

$\\ \sf\longmapsto T=\dfrac{500}{340}$

$\\ \sf\longmapsto T=1.47\approx 1.5s$

Total time:-

$\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s$

You might be interested in

Hi i just need some points and i want to give some to you

xz_007 [3.2K]

Answer:

ok thanks

Explanation:

8 0

3 years ago

The diagram shows what happens to a system undergoing an adiabatic process.

posledela

The answer is:
B. X: Work is done to the system and temperature increases.
Y: Work is done by the system and temperature decreases.

5 0

4 years ago

Read 2 more answers

Two electrons travel towards each other at 0.2 c parallel to the laboratory x-axis. What is the relative velocity of one electro

Leya [2.2K]

1) In the reference frame of one electron: 0.38c

To find the relative velocity of one electron with respect to the other, we must use the following formula:

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

where

u is the velocity of one electron

v is the velocity of the second electron

c is the speed of light

In this problem:

u = 0.2c

v = -0.2c (since the second electron is moving towards the first one, so in the opposite direction)

Substituting, we find:

$u'=\frac{0.2c+0.2c}{1+\frac{(0.2c)(0.2c)}{c^2}}=\frac{0.4c}{1+0.04}=0.38c$

2) In the reference frame of the laboratory: -0.2c and +0.2c

In this case, there is no calculation to be done. In fact, we are already given the speed of the two electrons; we are also told that they travel in opposite direction, so their velocities are

+0.2c

-0.2c

5 0

3 years ago

The magnet has an unchanging magnetic field: very strong near the magnet, and weak far from the magnet. How did the magnetic fie

STatiana [176]

Answer:

The magnetic field through the coil at first increases steadily up to its maximum value, and then decreases gradually to its minimum value.

Explanation:

At first, the magnet fall towards the coils; inducing a gradually increasing magnetic field through the coil as it falls into the coil. At the instance when half the magnet coincides with the coil, the magnetic field magnitude on the coil is at its maximum value. When the magnet falls pass the coil towards the floor, the magnetic field then starts to decrease gradually from a strong magnitude to a weak magnitude.

This action creates a changing magnetic flux around the coil. The result is that an induced current is induced in the coil, and the induced current in the coil will flow in such a way as to oppose the action of the falling magnet. This is based on lenz law that states that the induced current acts in such a way as to oppose the motion or the action that produces it.

3 0

4 years ago

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

8 0

3 years ago