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Molodets [167]
3 years ago
14

When a speeding truck hits a stationary car, the car is deformed and heat is generated. What can you say about the kinetic energ

y of the system after the collision?
Physics
1 answer:
guapka [62]3 years ago
4 0
I can say that the kinetic energy after the collision is less than it was before the collision.  I can say this with confidence because you've said that some energy was used to deform the car, plus there was energy lost from the system in the form of heat.
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Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many
dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

v(t)=t^2e^{-2t}\ m/s

We need to find the distance (x)  traveled by the particle during the first t seconds. It is given by :

x=\int\limits {v.dt}

x=\int\limits {t^2e^{-2t}dt}

Using by parts integration, we get the value of x as :

x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters

Hence, this is the required solution.

6 0
3 years ago
Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the
makvit [3.9K]

Answer:

\mu_k=0.101

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

\dfrac{1}{2}kx^2=\mu_k mgh

\mu_k=\dfrac{kx^2}{2mgh}

\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}

\mu_k=0.101

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

3 0
3 years ago
Please help urgent i need help easy queston
mihalych1998 [28]
I believe it’s B but i’m not sure
3 0
3 years ago
Please help: I don't know how to do these problems
antiseptic1488 [7]
d =2.55.68m and t = 11.36s
In my opinion
3 0
3 years ago
A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
ElenaW [278]

Explanation:

The given data is as follows.

    Length of beam, (L) = 5.50 m

    Weight of the beam, (W_{b}) = 332 N

     Weight of the Suki, (W_{s}) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

                 \sum M_{o} = 0

     -W_{s} \times x + W_{b} \times 1.5 = 0

                x = \frac{W_{b} \times 1.5}{W_{s}}

                   = \frac{332 N \times 1.5}{505 N}

                   = 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0
3 years ago
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