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anastassius [24]

3 years ago

10

The earth has a mass ME = 5.98 · 1024 kg and the moon has a mass MM = 7.36 · 1022 kg. The distance from the center of the earth

to the center of the moon is r = 3.82 · 105 km. What is the gravitational force between the earth and moon?

1 answer:

ivann1987 [24]3 years ago

6 0

Answer:

2 x 10^20 N

Explanation:

Me = 5.98 x 10^24 kg

Mm = 7.36 x 10^22 kg

r = 3.82 x 10^5 km = 3.82 x 10^8 m

The gravitational force between earth and moon is

F = G Me x Mm / r^2

F = (6.67 x 10^-11 x 5.98 x 10^24 x 7.36 x 10^22) / (3.82 x 10^8 x 3.82 x 10^8)

F = 2 x 10^20 N

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In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a

Ghella [55]

Answer:

m = B²qR² / 2 V

Explanation:

If v be the velocity after acceleration under potential difference of V

kinetic energy = loss of electric potential energy

1/2 m v² = Vq ,

v² = 2 Vq / m ----------------------- ( 1 )

In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force

magnetic force = centripetal force

Bqv = mv² / R

v = BqR / m

v² = B²q²R² / m² ------------------------- (2)

from (1) and (2)

B²q²R² / m² = 2 Vq / m

m = B²q²R² / 2 Vq

m = B²qR² / 2 V

7 0

3 years ago

A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo

guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s $(1km\h=\frac{5}{18} m/s)$

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

$33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)$

$a = 1.83 m/s^2$

$Force=mass\times acceleration$

$Force=710 kg\times 1.83 m/s^2$

$Force=1,299.3 N$

The net force acting on the car 1,299.3 N.

8 0

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

group of students is making model cars that will be propelled by model rocket engines. These engines provide a nearly constant t

iris [78.8K]

Answer:

increase.

Explanation:

According to the newton’s second law of motion force is expressed as product of mass and acceleration.

F = m a

If the force acting is constant, then.

m∝ $\frac{1}{a}$

That is if the mass of object increases the acceleration decreases and vice versa. The above equation is used when the force acting on the body is constant.

As the thrust force from the rocket engine is constant throughout there will be a variation in the mass or acceleration.

Thus, it won't stay the same.

As the weight of the car is maximum at the start because of the fuel present in the rocket engine and minimum at the end as the fuel burns throughout the journey of the car. Weight will be minimum at the end and hence acceleration is maximum at the end.

Thus, it won't decrease.

As the acceleration is going from minimum at the start to maximum at the end, therefore it is continuously increases throughout its journey.

Thus, it will increase.

7 0

3 years ago

A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .

timofeeve [1]

The solution would be like this for this specific problem:

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

6 0

3 years ago