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1 year ago

2. What process does water redeposit into a lake in the form of rain?

Physics

1 answer:

weqwewe [10]1 year ago

5 0

The process that water redeposit into a lake in the form of rain is precipitation.

<h3>What is water runoffs?</h3>

Water runoff occurs when there is more water than land can absorb.

The water flows across the surface of the land and into nearby creeks, streams, or lakes.

Runoff can come from both natural processes and human activity.

When rain falls to the earth from clouds and runs downhill into rivers and lakes.

During evaporation, the water turns from liquid into gas, and moves from oceans and lakes into the atmosphere where it forms clouds.

<h3>What is precipitation?</h3>

Precipitation is any liquid (rain) or frozen water that forms in the atmosphere and falls back to the Earth, for example it could fall on land or into lakes and rivers.

Thus, the process that water redeposit into a lake in the form of rain is precipitation.

Learn more about precipitation here: brainly.com/question/1783904

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Which has the greater acceleration, a person going from 0 m/s to 10 m/s in 10 seconds or an ant going from 0 m/s to 0.25 m/s in

monitta

Person-1

Initial velocity=u=0m/s
Final velocity=v=10m/s
Time=10s=t

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}$

$\\ \sf\longmapsto Acceleration=\dfrac{10}{10}$

$\\ \sf\longmapsto Acceleration=1m/s^2$

Person-2

initial velocity=0m/s=u
Final velocity=v=0.25m/s
Time=t=2s

$\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}$

$\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}$

$\\ \sf\longmapsto Acceleration=0.125m/s^2$

Person-1 is accelerating faster.

4 0

3 years ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

4 0

3 years ago

Read 2 more answers

The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,

ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy = -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)² .........................1

dt/dy = -54× 2y / (7 + x² + y²)² .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²

dt/dx = -0.373702

and

dt/dy = -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy = -1.121107

7 0

3 years ago

Starting circuit One battery. 2 light bulbs in parallel; switch What is the voltage across the battery? What is the voltage acro

balu736 [363]

Answer:

The voltage across light bulb 1 and light bulb 2 is the the same i.e V

Explanation:

In a parallel circuit, the Voltage is same across all the components of the circuit and the current flowing through each component is added to get the total current across the circuit.

Let us say, the voltage across the circuit is V. The voltage across light bulb 1 and light bulb 2 is the the same i.e V

5 0

3 years ago

How come in free fall you feel weightless even though gravity is pulling down on you?

AysviL [449]

Feeling of Weight.

When walking, you feel the weight on your feet, therefore, your brain automatically refers to it as a source of weight.

In the air there is no platform to land on, therefore the brain does not have the conscience to register you getting pulled down.

7 0

3 years ago