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2 years ago

If a builder of mass 75kg climbs a vertical ladder of 25m how much energy has she gained ?

1 answer:

4 0

So,

GPE (graviational potential energy) = mass x g x height

GPE is depends on where zero height is defined. In this situation, we define h = 0 as the initial height.

$GPE = 75 \ kg*9.8 \ \frac{m}{s^2}*25 \ m$

$GPE = 18,375 \frac{kg*m^2}{s^2}$

$GPE = 18,375 \ joules(J) \ or \ 18.375 \ kilojoules(kJ)$

The builder has gained 18.375 kJ of PE.

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Calculate the force exerted on the wall assuming that force is horizontal and using the data in the schematic representation of

Jlenok [28]

Answer:

1.93 x 10∧3 N

Explanation:

The picture attached shows the calculation

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3 years ago

What affects sounds at their source? MENTION DOPPLER EFFECT AND VIBRATIONS

wel

Answer:

Doppler effect, the apparent difference between the frequency at which sound or light waves leave a source and that at which they reach an observer, caused by relative motion of the observer and the wave source. This phenomenon is used in astronomical measurements, in Mössbauer effect studies, and in radar and modern navigation.

Explanation:

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3 years ago

Air pressure may be represented as a function of height above the surface of the Earth as shown below.. P(h) = P_0 e^-.00012h. .

Rufina [12.5K]

Answer. Second Option: .85p_o=p_o e^-.00012h

Solution:

P(h)=Po e^(-0.00012h)

Air pressure: P(h)

Height above the surface of the Earth (in meters): h

Air pressure at the sea level: Po

Height at which air pressure is 85% of the air pressure at sea level:

h=?, P(h)=85% Po

P(h)=(85/100) Po

P(h)=0.85 Po

Replacing P(h) by 0.85 Po in the formula above:

P(h)=Po e^(-0.00012h)

0.85 Po = Po e^(-0.00012h)

5 0

3 years ago

Read 2 more answers

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

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2 years ago

How do vibrations cause sound waves

Mariulka [41]

<span>sound waves are a type of wave sometimes called compression waves, vibrations with enough of an amplitude can compress and decompress the air adjacent to the object causing the waves to form.</span>

3 0

3 years ago

Read 2 more answers