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just olya [345]
3 years ago
12

How does kinect energy affect the stopping distance of a vehicle traveling at 30 mph compared to the same vehicle traveling at 6

0 mph?
Physics
1 answer:
mote1985 [20]3 years ago
8 0

If it is the same vehicle, then the 60mph vehicle has more kinetic energy since it is moving faster. Therefore, it requires more energy to stop, and if it is the same car with the same beak system, the braking distance of the 30mph car will be significantly shorter than the 60mph car.



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3. How much work is done when you pull a 6 N wagon for 5 meters?
8090 [49]

Answer:

<h2>30 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 6 N

distance = 5 m

We have

workdone = 6 × 5 = 30

We have the final answer as

<h3>30 J</h3>

Hope this helps you

3 0
3 years ago
suppose a ball had a potential energy of 5 j when you dropped it. What would be its kinetic energy just as it hit the ground. (i
Andru [333]
Since we are ignoring air resistance which is a non-conservative force, the potential energy will be completely converted into kinetic energy, resulting in a final kinetic energy of 5J.
4 0
3 years ago
There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time t=0 to be at angle θ=0 radians
mario62 [17]

Here is the missing information.

An exhausted bicyclist pedal somewhat erraticaly when exercising on a static bicycle. The angular velocity of the wheels takes the equation ω(t)=at − bsin(ct) for t≥ 0, where t represents time (measured in seconds), a = 0.500 rad/s2 , b = 0.250 rad/s and c = 2.00 rad/s .

Answer:

0.793 rad

Explanation:

From the given question:

The angular velocity of the wheel is expressed by the equation:

\omega (t) =\dfrac{d\theta}{dt}

The angular velocity of the wheels takes the description of the equation ω(t)=at−bsin(ct)

SO;

\dfrac{d \theta}{dt} = at - b \ sin \ ct

dθ = at dt - (b sin ct) dt

Taking the integral of the above equation; we have:

\int \limits^{\theta}_{0} \ d \theta = \int \limits ^{t=2}_{0} at  \ dt - (b \ sin \ ct) \dt

[\theta] ^{\theta}_{0} = a \bigg [\dfrac{t^2}{2} \bigg]^2_0 - \bigg[ -\dfrac{b}{c} \ cos \ ct \bigg] ^2_0

where;

a = 0.500 rad/s2 ,

b = 0.250 rad/s and

c = 2.00 rad/s

\theta = (0.500 \ rad/s^2 ) \bigg [\dfrac{(2s)^2}{2} \bigg] - \bigg[ -\dfrac{0.250 \ rad/s}{2.00 \ rad/s} \ cos \ (2.00 \ rad/s )( 2.00 \ s) \bigg] - \bigg [ \dfrac{0.250 \ rad/s}{2.00 \ rad/s}\bigg ] cos 0^0

\mathbf{\theta = 0.793 \ rad}

Hence, the angular displacement after two seconds = 0.793 rad

3 0
3 years ago
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spin [16.1K]

Answer:

The answer is B.

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blagie [28]
Using the formula KE=1/2mv^2

a: The kinetic energy doubles.
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4 years ago
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