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AfilCa [17]
2 years ago
11

A car and a truck are moving next to each other on the highway. The truck has a mass of 910 kg and the car’s mass is 227 kg. IF

their momentums are equal and the truck is moving at 18 m/s, what is the speed of the car?
Physics
1 answer:
eimsori [14]2 years ago
3 0

Answer:

<h2>the car must move with the speed</h2><h2>v = 72.16 m/s</h2>

Explanation:

As we know that the momentum of the car and truck is same

so as per the formula of momentum we have

P = mv

so we have

m_1v_1 = m_2v_2

so we will have

910 \times 18 = 227 v

v = 72.16 m/s

So the car must move with the speed

v = 72.16 m/s

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A child sleds down a hill with an acceleration of 2.94 m/s2. If her initial speed is 0.0 m/s and her final speed is 17.5 m/s, ho
hram777 [196]

Answer:

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

Explanation:

Given that the child accelerates uniformly and that both initial (v_{o}) and final speeds (v_{f}), measured in meters per second, and acceleration (a), measured in meters per square second, are known, we proceed to use the following kinematic equation to determine the time taken to travel from the top of the hill to the bottom (t), measured in seconds, is:

t = \frac{v_{f}-v_{o}}{a} (1)

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 17.5\,\frac{m}{s} and a = 2.94\,\frac{m}{s^{2}}, then the time taken is:

t = \frac{17.5\,\frac{m}{s}-0\,\frac{m}{s} }{2.94\,\frac{m}{s^{2}} }

t = 5.952\,s

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

8 0
3 years ago
A truck is traveling 20m/s accelerates 3 m/s^2 for 4 seconds how far did it travel while it was accelerating. Using guess method
stellarik [79]
<h3>\huge\underline\bold\blue{ƛƝƧƜЄƦ}</h3><h3>Given</h3>

\blue\star v = 20m\s

\blue\star a = 3m\s^2

\blue\star t = 4sec

Firstly we have to find u

\star a = \dfrac{v - u}{t}

\star 3m\s =\dfrac{20 - u}{4}

\star12m\s = 20 - u

\star20 - u = 12m\s

\star- u = -8

\star u = 8

Now we can easily find distance by using second equation of motion

\red\stars = ut + 1\2 at^2

\red\stars = 8(4) + 1\2(3)(16)

\red\stars = 32 + 24

\red\stars = 56

So distance is 56 m\s hope it helps

5 0
3 years ago
In order to increase the amount of work completed, it is necessary to decrease the force applied to an object. decrease the time
wlad13 [49]
<span> <span> The answer to your question is: increase the force applied to the object.

Two items are provided as a basis for that conclusion:
1. According to Newton's Second Law of Motion, the formula for finding force is: F = ma
where F is the force,
m is the mass of an object,
and a is the acceleration of the object.

And 2: work = force x distance or W = F x d.</span></span>
5 0
3 years ago
Read 2 more answers
A 65kg person throw a 0.045kg snowball forward with a ground speed of 30m/s. A second person, with a mass of 60kg, catches the s
Kobotan [32]
Well, st first we should find <span>initial momentum for the first person represented in the task which definitely must be :
</span>(65+0.045)*2.5
And then we find the final one :  65*x + 0.045*30
Then equate them together : x=2.48 m/s 
So we can get the velocity, which is is 2.48 m/s
In that way, according to the main rules of <span>conservation of momentum you can easily find the solution for the second person.
Regards!</span>
6 0
3 years ago
Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out
blondinia [14]

Answer:

\theta = 20.98 degree

Explanation:

As we know that the speed of the sound is given as

v = 332 + 0.6 t

now at t = 273 k = 0 degree

v = 332 m/s

so we have

a sin\theta = N\lambda

a sin\theta = N(\frac{v_1}{f})

now when temperature is changed to 313 K we have

t = 313 - 273 = 40 degree

now we have

v = 332 + (0.6)(40)

v_2 = 356 m/s

a sin\theta' = N(\frac{v_2}{f})

now from two equations we have

\frac{sin19.5}{sin\theta} = \frac{332}{356}

so we have

sin\theta = 0.358

\theta = 20.98 degree

7 0
3 years ago
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