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trapecia [35]
2 years ago
5

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
2 answers:
Aleks04 [339]2 years ago
8 0

Answer:

The kinetic energy of the system after the collossion is less than the kinetic energy of the system before the collision. This is because the heat generated from the collision which is a direct consequence of the friction between both vehicles and possibly the road too. This heat energy due to friction is not a conservative form of energy and cannot be regained in another form. So after the collision this energy is lost, thereby reducing the total kinetic energy the system had before the collision.

Explanation:

Collisions as these are inelastic collisions. For inelastic collisions, the total momentum is conserved before and after the collision. The kinetic energy on the other hand is not conserved. This is because in most cases some of the energy is lost due friction in the form of heat.

Elastic collisions on the other hand generally have the total momentum and kinetic energy conserved before and after collision.

Thank you for reading.

nadya68 [22]2 years ago
7 0
The Kinetic energy was transformed immediately into potential and thermal energy. 
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For metalloids on the periodic table, how do the group number and the period number relate?
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A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a
o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet m=5 gm

mass of wood block M=1 kg

Length of string L=2 m

Center of mass rises to an height of 0.38 cm

initial velocity of bullet u=450 m/s

let v_1 and v_2 be the velocity of bullet and block after collision

Conserving momentum

mu=mv_1+Mv_2 -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

\frac{Mv_2^2}{2}=Mgh_{cm}

v_2=\sqrt{2gh_{cm}}

v_2=\sqrt{2\times 9.8\times 0.38}

v_2=0.272 m/s

substitute the value of v_2 in equation 1

5\times 450=5\times v_1+1000\times 0.272

v_1=395.6 m/s

4 0
2 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

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6 0
2 years ago
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melisa1 [442]

Answer:

B. Trial 2

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7 0
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Answer:

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R_{E} = 6371\ km = radius of earth

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e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
2 years ago
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