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HACTEHA [7]
4 years ago
10

If a 2000 kg car is at rest and a 500 N force is applied to it. What acceleration will it experience?

Physics
1 answer:
Komok [63]4 years ago
4 0
Net Force = mass * acceleration.
Therefore, Acceleration = Net force / mass
A=2000/500
A=4 m/s^2
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saitama is the strongest

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what is the magnitude of the electric force between charges of 0.25 C and 0.11 C at a separation of 0.88 m? if the separation be
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three resistors which values of 15 ohm 40 ohm 60 ohm respectively are connected in parallel. what is their equivalent resistance
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1) Equivalent resistance, 1/R = 1/15 + 1/40 + 1/60 = 8+3+2 /120 = 13/120

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7 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.
BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

y_{i}=0m

v_i=80\frac{m}{s}

a=4\frac{m}{s^2}

y_{e}=1000m

g=9.8\frac{m}{s^2}

From Kinematics, the position y as a function of time when the engine still works will be:

y(t)=v_it+\frac{1}{2}at^2

At what time the altitud will be y_{e}=1000m?

v_it+\frac{1}{2}at^2=y_{e} ⇒ \frac{1}{2}at^2+v_it-y_{e}=0

Using the quadratic formula: t_1=10s.

How much time does it take for the rocket to touch the ground? No the function of position is:

y(t)=y_{e}+v_et-\frac{1}{2}gt^2

Where our new initial position is y_{max}, the velocity when the engine breaks is v_e=v_i+at=120\frac{m}{s} and the only acceleration comes from gravity (which points down).

Now, when the rocket tounches the ground:

y_{e}+v_et-\frac{1}{2}gt^2=0

Again, using the quadratic ecuation:

t_2=24.49s

Now, the total time from the moment it takes off and the moment it tounches the ground will be:

t_T=t_1+t_2=34.49s.

6 0
4 years ago
A 0.5 kg air-track car is attached to the end of a horizontal spring of constant k = 20 N/m. The car is displaced 15 cm from its
Dimas [21]
Find the amount of work that the spring does. This can be found using the equation 1/2kx^2. Then, you must set that equal to the amount of kinetic energy the car has. This is possible thanks to the work-energy theorem.

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Solve to find velocity. Remember, the spring is displaced .15 m, not 15!

To find the acceleration, use F = ma. The force being applied to the car is kx, and you know the mass. You do the math.

For problem C I don't know, haven't done that yet in my class. Sorry!
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3 years ago
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