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olga55 [171]
3 years ago
10

Which object experiences the greatest gravitational force? A)car B)fire truck C)plane D)human

Physics
1 answer:
Svetlanka [38]3 years ago
3 0

Answer: C Plane

Explanation: According to Newton's law, gravitational force is proportional to the product of masses and inversely proportional to the square of distance between them.

Gravitational force depends on mass. The bigger the mass, the more the magnitude of the gravitational force. Since plane is assume to have the highest mass in the options, we can therefore conclude that plane will experience the highest gravitational force.

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A. Describe one meal that you typically order when eating out.
marta [7]
A. “I usually get 10 piece nugget meal with medium fries at McDonald’s with a McFlurry

B. I could replace the 10 piece nuggets with a grilled chicken wrap and replace the medium fries with a small fries then replace the mcflurry with either water or unsweetened tea.
3 0
3 years ago
Read 2 more answers
A soccer ball with mass 0.450 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant fo
Alinara [238K]

Answer:

0.187 m

Explanation:

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Mass (m) = 0.450 Kg

Force (F) = 38 N

Acceleration (a) =?

F = m × a

38 = 0.450 × a

Divide both side by 0.450

a = 38 / 0.450

a = 84.44 m/s²

Finally, we shall determine the distance. This can be obtained as follow:

Initial velocity (u) = 2.20 m/s.

Final velocity (v) = 6 m/s

Acceleration (a) = 84.44 m/s²

Distance (s) =?

v² = u² + 2as

6² = 2.2² + (2 × 84.44 × s)

36 = 4.4 + 168.88s

Collect like terms

36 – 4.84 = 168.88s

31.52 = 168.88s

Divide both side by 168.88

s = 31.52 / 168.88

s = 0.187 m

Thus, the distance is 0.187 m

6 0
3 years ago
Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
3 years ago
Now in "real life," this automobile is cruising at 20.5 m/s (equal to 73.8 km/hr) when it is about to hit a pedestrian stuck in
algol13

Answer:

He needs 1.53 seconds to stop the car.

Explanation:

Let the mass of the car is 1500 kg

Speed of the car, v = 20.5 m/s

He will not push the car with a force greater than, F=2\times 10^4\ N

The impulse delivered to the object is given by the change in momentum as :

F\times t=mv\\\\t=\dfrac{mv}{F}\\\\t=\dfrac{1500\times 20.5}{2\times 10^4}\\\\t=1.53\ s

So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.

5 0
3 years ago
An RC circuit is connected across an ideal DC voltage source through an open switch. The switch is closed at time t = 0 s. Which
steposvetlana [31]

Answer:

e)

Explanation:

In an RC series circuit, at any time, the sum of the voltages through the resistor and the capacitor must be constant and equal to the voltage of the DC voltage source, in order to be compliant with KVL.

At= 0, as the voltage through the capacitor can't change instantaneously, all the voltage appears through the resistor, which means that a current flows, that begins to charge the capacitor, up to a point that the voltage through the capacitor is exactly equal to the DC voltage, so no current flows in the circuit anymore, and the charge in the capacitor reaches to its maximum value.

5 0
2 years ago
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