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Nikolay [14]
4 years ago
15

Calculate the pressure of water at the bottom of well if the depth of water is 10m (g=9.8 m/s​

Physics
1 answer:
Dominik [7]4 years ago
8 0
Pressure = height x density x gravity = 10 x 1000 x 10 = 100 000 Pascals
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By copying their genomes, they retain the tool kit and at the same time generate a garage full of spare parts. Gene duplication can provide the raw material for expression changes to occur, and polyploidy itself can trigger epigenetic changes

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3 years ago
Mass is constant, but weight can change with location . Explain.
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Weight is the force of gravity and gravity changes due to location like on the moon there is a different force of gravity
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4 years ago
A greyhound has a mass of 35 kg and can run at a speed of 20 m/s. What is the kinetic energy that a greyhound can achieve
prisoha [69]

Hi there!

Recall the equation for kinetic energy:

\large\boxed{KE = \frac{1}{2}mv^2}

KE = Kinetic energy (J)

m = mass (kg)

v = velocity (m/s)

Plug in the given values:

KE = \frac{1}{2}(35)(20^2) = \boxed{7000 J}

7 0
3 years ago
The driver of a car traveling at 23.1 m/s applies the brakes and undergoes a constant
Vsevolod [243]

Answer:

The tires make 125 revolutions before the car stops

Explanation:

Circular and Linear Motion

A tire rotates around a fixed point and the tire when in contact with the ground, drives a vehicle in a linear path. This is an example of a relationship between both types of movements that can be taking place simultaneously.

The car is moving with an initial speed of v_o=23.1\ m/s and then breaks at a=-1.03\ m/s^2 until it stops. We can compute the time take to stop by using

\displaystyle v_f=v_o+a.t

Solving for t

\displaystyle t=\frac{v_f-v_o}{a}

Putting in numbers

\displaystyle t=\frac{0-23.1}{-1.03}

\displaystyle t=22.427\ sec

Now, let's transfer this information to the circular motion. We know the tangent speed is

\displaystyle v_t=w.r

Being w the angular speed and r the radius of the circle, in this case, the tires. The tangent speed is the same as the speed of motion of the car. It gives us the initial angular speed

\displaystyle w_o=\frac{v_t}{r}

\displaystyle w_o=\frac{23.1}{0.33}=70\ rad/s

When the circular motion is not uniform, i.e. there is angular acceleration \alpha, the angular speed is a function of time

\displaystyle w=w_o+\alpha t

We can compute the angular acceleration knowing the final angular speed is zero when the car stops.

\displaystyle \alpha=\frac{w-w_o}{t}=\frac{0-70}{22.427}

\displaystyle \alpha=-3.121\ rad/s^2

The rotation angle is also a function of time as shown

\displaystyle \theta=w_o\ t+\frac{\alpha t^2}{2}

Using the given and computed values

\displaystyle =70(22.427)-\frac{3.121(22.427)^2}{2}

\displaystyle \theta =784.95\ rad

Knowing each revolution is 2\pi radians, the number of revolutions is

\displaystyle n=\frac{\theta }{2\pi}=125\ rev

The tires make 125 revolutions before the car stops

8 0
4 years ago
The peak intensity of a radiation from star beta is 350nm in what spectral bands is this
Olenka [21]

good job i hope you do good



5 0
3 years ago
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