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4 years ago

How is the pressure exerted and the area on which it is exerted are related to each other?

Physics

1 answer:

andrezito [222]4 years ago

3 0

Pressure = force/area

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An apple from the top branch of the tree and an apple from the bottom branch of a tree fall at the same time. Which apple will h

Vedmedyk [2.9K]

B. The apple from the bottom will hit the ground earlier. This is because an increase in height causes an increase in the time that the object will fall, and therefore will affect the final velocity of the falling object. Moreover, the reduction in velocity due to friction from the air should also be considered.

3 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

Part B 

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t

const2013 [10]

Answer:

$\theta=12.19^{\circ}$

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

$sin\theta =\dfrac{u}{v}$

Now by putting the values in the above equation we get

$sin\theta =\dfrac{30}{142}$

$sin\theta=0.21$

$\theta=12.19^{\circ}$

Therefore the angle will be 12.19° .

4 0

3 years ago

Here is a force diagram of an object in water. The weight of the object is 15N and the buoyancy force is 17N. Will the object fl

lawyer [7]

Answer:

Object will float.

Explanation:

Total force on the body = Weight of body + Buoyancy force on body.

Weight of body = 15 N downwards = 15 N

Buoyancy force on body = 17 N upwards = -17 N

Total force on body = 15 - 17 = -2 N = 2 N upwards

So, the body will float.

Object will float.

8 0

3 years ago

You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You disc

FrozenT [24]

Answer:

The balloon hit the ground with velocity -15.34 m/s

Explanation:

Lets explain how to solve the problem

You found that the best height to pitch a water balloon in order for it to

burst when it hits the ground is 12 meters.

We consider that the 12 meters is the maximum height, so the velocity

at this height is zero.

To find the velocity when the balloon hits the ground lets use the rule

v² = u² + 2gh, where v is the final velocity, u is the initial velocity, g is

the acceleration of gravity and h is the height.

u = 0 , h = 12 m , g = 9.8 m/s²

Substitute these values in the equation above

v² = 0 + 2(9.8)(12)

v² = 235.2

Take square root for both sides

v = ± $\sqrt{235.2}$

The velocity is downward, then it's a negative value

Then v = -15.34 m/s

The balloon hit the ground with velocity -15.34 m/s

6 0

3 years ago