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4 years ago

When you drop a 0.43 kg apple, Earth exerts

Physics

1 answer:

N76 [4]4 years ago

8 0

The magnitude of the Earth's acceleration is $7.0\cdot 10^{-25} m/s^2$

Explanation:

First of all, we start by calculating the magnitude of the force exerted by the apple on the Earth. According to Newton's third law, this is equal to the force exerted by the Earth on the apple, which is the weight of the apple, given by:

$F=mg$

where

m = 0.43 kg is the mass of the apple

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F=(0.43)(9.8)=4.2 N$

Now we can find the Earth's acceleration by applying Newton's second law:

$F=Ma$

where:

F = 4.2 N is the net force exerted by the apple on the Earth

$M=5.98\cdot 10^{24} kg$ is the mass of the Earth

a is the Earth's acceleration

And solving for a, we find:

$a=\frac{F}{M}=\frac{4.2}{5.98\cdot 10^{24}}=7.0\cdot 10^{-25} m/s^2$

Learn more about Newton laws of motion:

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In an eslastic ,the momentum is_______and the mechanical energy is

DIA [1.3K]

Answer:

In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.

Explanation:

There are two types of collisions:

- Elastic collision: in an elastic collision, the total momentum before and after the collision is conserved; also, the total mechanical energy before and after the collision is conserved.

- Inelastic collision: in an inelastic collision, the total momentum before and after the colllision is conserved, while the total mechanical energy is not conserved (in fact, part of the energy is converted into other forms of energy such that thermal energy, due to the presence of frictional forces)

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3 years ago

Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging,

Rzqust [24]

A) A. 380 kHz

To clerly see the image of the fetus, the wavelength of the ultrasound must be 1/4 of the size of the fetus, therefore

$\lambda=\frac{1}{4}(1.6 cm)=0.4 cm=0.004 m$

The frequency of a wave is given by

$f=\frac{v}{\lambda}$

where

v is the speed of the wave

$\lambda$ is the wavelength

For the ultrasound wave in this problem, we have

v = 1500 m/s is the wave speed

$\lambda=0.004 m$ is the wavelength

So, the frequency is

$f=\frac{1500 m/s}{0.004 m}=3.75\cdot 10^5 Hz=375 kHz \sim 380 kHz$

B) B. f(c+v)/c−v

The formula for the Doppler effect is:

$f'=\frac{v\pm v_r}{v\pm v_s}f$

where

f' is the apparent frequency

v is the speed of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative if it is moving away from the source)

$v_s$ is the speed of the source (positive if the source is moving away from the receiver, negative if it is moving towards the receiver)

f is the original frequency

In this problem, we have two situations:

- at first, the ultrasound waves reach the blood cells (the receiver) which are moving towards the source with speed

$v_r = +v$ (positive)

- then, the reflected waves is "emitted" by the blood cells (the source) which are moving towards the source with speed

$v_s = -v$

also

v = c = speed of sound in the blood

So the formula becomes

$f'=\frac{c + v}{v - v_s}f$

C. A. The gel has a density similar to that of skin, so very little of the incident ultrasonic wave is lost by reflection

The reflection coefficient is

$R=\frac{(Z_1 -Z_2)^2}{(Z_1+Z_2)^2}$

where Z1 and Z2 are the acoustic impedances of the two mediums, and R represents the fraction of the wave that is reflected back. The acoustic impedance Z is directly proportional to the density of the medium, $\rho$ .

In order for the ultrasound to pass through the skin, Z1 and Z2 must be as close as possible: therefore, a gel with density similar to that of skin is applied, in order to make the two acoustic impedances Z1 and Z2 as close as possible, so that R becomes close to zero.

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3 years ago

A car moves 8 km from home to school and then back to the home. a) Calculate the Distance covered by this car. b) What is the Di

Sveta_85 [38]

Explanation:

Distance covered = Total distance travelled by the car

Since the car travels to school and back to home again,

Total diatance covered = 8 + 8 = 16km

Displacement = Shortest distance between start point and destination.

Since, the car returns back to home from school,

the start and destination point are same.

So Displacement = 0