What’s the resistance of the circuit

A 3600-N Force causes a car to accelerate at a rate of 4m/s2. What is the mass of the car?

Mademuasel [1]

Answer:

14400 kg

Explanation:

According to Newton's second law of motion, F = ma

F = 3600 N

a = 4m/s2

3600 = m/4

3600*4 = m

m = 14400 kg

Therefore the mass of the car is 14400 kg

Hope u understand

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3 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$

Time taken by the rocket to fall from maximum height is 20.29 seconds

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0

3 years ago

What is the voltage in a circuit if the current is 6 Amps and resistance is 6 ohms

Mrrafil [7]

U=R•I=6•6=36 V

............

3 0

4 years ago

It takes a minimum distance of 41.14 m to stop a car moving at 11.0 m/s by applying the brakes (without locking the wheels). Ass

padilas [110]

Answer:

Find the time it took for the car to stop at 11.0m/s

V = deltax/t

t = 41.14/11.0 = 3.74s

Now find at what rate it was decelerating, so find the acceleration during that interval of time.

vf = vi + at

-11.0m/s = a3.74s

a = -2.94m/s^2

The acceleration is negative because is pulling the car towards its opposite direction to make it stop.

Now find how much time it would take for the car to stop at 28.0m/s but with the same acceleration, the car is the same so its acceleration to stop the car will remain the same.

vf = vi + at

0 = 28.0 - 2.94t

t = 9.52

Once the time is obtained, you can find the final position, xf, by plugging the time acceleration and velocity values.

xf = 0 + (28m/s)(9.52s) + 1/2(-2.94)(9.52s)^2

xf = 266.6m - 133.23m = 133m

8 0

4 years ago

Describe the law of conservation of energy. Provide at least one example that illustrates this law. (4 points)

SCORPION-xisa [38]

The law of conservation of energy states that energy is neither created nor destroyed; the amount remains constant. For example, a form of energy-thermal energy, or heat, occurs by convection or conduction

6 0

4 years ago