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fiasKO [112]
3 years ago
6

What is the acceleration of a 5kg rock that is thrown with a force of 18N?

Physics
1 answer:
natita [175]3 years ago
8 0

Answer:

F = m×a

a = F /m

a = 90 m/s²

hope it's helps you

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What is the difference in the speed of the generator with a small magnet and a generator with a large magnet?
Delvig [45]
With a small magnet with a generator it will be taken up quickly because how small it is while with a big generator it would take more force for it for the generator to attach because the larger the magnet that heavier it will be because it is attached to the North Pole magnet
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A 2,000 kg car is moving at 15m/s when it collides with a 1200kg car sitting still. Briefly compare the impulse imparted on the
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impulse = F × t

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Question 24
Verizon [17]
Q 24: Newton's first law of motion states that anything that stands still will stay still unless unbalanced force is acted upon that object. Newton's second law of motion states that the velocity of an object changes when subjected to external Force. Newton's third law of motion says that for every action there is an equal and opposite reaction. when bumper cars crash into each other one car goes one way and the other goes the other way.

Q 25: if the bumper cars doubled their Mass the motion would be halfed but if the net force is doubled the motion will double.
8 0
3 years ago
Marys airplane trip took 5.8 hours for one-half of that time, the airplane flew at a constant speed of 640 miles per hour and fo
Olenka [21]
Distance is speed x time.  Half of the trip is 5.8/2 = 2.9hrs.
640 x 2.9 = 1856mi
580 x 2.9 = 1682mi
1856mi+1682mi=3538mi.

You could also calculate her average speed.  This is easy since it was divided in two equal time slices.  Average Speed = (640+580)/2 = 610mi/hr
Now 610mi/hr x 5.8hrs = 3538mi
7 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

5 0
3 years ago
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