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Airida [17]
3 years ago
12

A 833 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 561 kg. The two cars lock up

and slide together with a speed of 16.8 km/h. What was the speed of the first car just before the collision?
Physics
1 answer:
V125BC [204]3 years ago
5 0

Answer:

The speed of the first car just before the collision is 7.841 m/s.

Explanation:

Given that,

Mass of the automobile, m = 833 kg

Mass of the parked car, m' = 561 kg

The final speed of the cars lock up and slide together with a speed of 16.8 km/h, V = 16.8 km/h = 4.67 m/s

If two cars lock up and slide together, then it is the case of inelastic collision. The momentum will remains conserved such as :

mv+m'v'=(m+m')V

v is the speed of the first car

v' is the speed of the second car that is at rest

So,

mv=(m+m')V

v=\dfrac{(m+m')V}{m}

v=\dfrac{(833+561)\times 4.67}{833}

v=7.81\ m/s

So, the speed of the first car just before the collision is 7.841 m/s. Hence, this is the required solution.

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Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

6 0
2 years ago
What is the final velocity of a car that starts at 22 m/s and accelerates at 3.78 m/s for distance of 45 m
Pepsi [2]

v^2 = v0^2 +2ad v^2 = 22^2 + 2*3.78*45 = 824.2 v= √824.2 = 28.7 m/s

5 0
3 years ago
Describe how engineers designed a parachute to create the forces needed to slow down the falling person.
shepuryov [24]

Answer:

Due to the resistance of air, a drag force acts on a falling body (parachute) to slow down its motion. Without air resistance, or drag, objects would continue to increase speed until they hit the ground. The larger the object, the greater its air resistance. Parachutes use a large canopy to increase air resistance. Also, Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down. Sorry if not helpful.

8 0
3 years ago
Which of the following has zero acceleration? an object...
tangare [24]

Answer:

B object at rest

Explanation:

object at rest means it wouldn't be moving, like a parked car or sleeping person therefore, B is the correct answer

8 0
2 years ago
Did I do these questions correctly?
SOVA2 [1]
Yes, they seem right to me.
4 0
3 years ago
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