A ball is dropped from the top of a building.

Susie (45kg) and Troy (65 kg) are sitting 2 meters apart. They smile at each other. What is the gravitational force between them

xxTIMURxx [149]

The gravitational force between Susie and Troy is d) $4.9 \cdot 10^{-8}N$

Explanation:

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where :

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the distance between them

In this problem:

$m_1 = 45 kg$ is Susie mass

$m_2 = 65 kg$ is Troy mass

r = 2 m is the distance between them

Substituting into the equation, we find the gravitational force between Susie and Troy:

$F=(6.67\cdot 10^{-11})\frac{(45)(65)}{(2)^2}=4.9\cdot 10^{-8}N$

Learn more about gravitational force:

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#LearnwithBrainly

4 0

3 years ago

Why do you think we cannot use a common fuel like diesel or gasoline for space travel ?

Ede4ka [16]

Answer:

Explanation:

it take oxygen in the atmosphere to burn it... in space there isn't any air :0

5 0

2 years ago

Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when

nignag [31]

<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

v₁ is the final speed of car X.

$m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s$

So, car X will move with a velocity of -12 m/s.

3 0

3 years ago

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

2 years ago

Two cars are moving in the same direction at the same speed of (72.0 kmh). What is

Licemer1 [7]

Answer:

0 km/h

Explanation:

Relative speed is the speed of a moving body with respect to another.

When two bodies move in the same direction then the relative speed is calculated as difference of their speeds.

In this case;

The two cars have the same speed. The relative speed will be;

72 km/h -72 km/h = 0 km/h

6 0

2 years ago