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olganol [36]
3 years ago
7

If a biker starts at rest and accelerates to 15 m/s in 7.5 seconds, what is his acceleration? *

Physics
1 answer:
sergeinik [125]3 years ago
6 0

Answer:

2 m/s^2

Explanation:

from the question

v=15 m/s

t=7.5

a=?

from the first equation of motion

v=u+at

where,

v=final velocity

u=initial velocity

a=acceleration

t=time

from the question (u) will be zero because the body started at rest

v=u+at

15=(0)+a×7.5

15=7.5a

a=15/7.5

a=2 m/s^2

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Answer:

Thus, the initial velocity of the bullet is 507.5 m/s.

Explanation:

mass of bullet, m = 0.0085 kg

mass of block, M = 0.99 kg

Height raised, h = 0.95 m

Let the initial velocity of bullet is u and the final velocity of block and bullet system is v.

Use conservation of energy

Potential energy of bullet bock system = kinetic energy of bullet block system

(M+m)\times g\times h = \frac{1}{2}\times \left ( M+m \right )v^{2}

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 0.95}=4.32 m/s

Now use conservation of linear momentum

mu = (M+m) v

0.0085 x u = (0.99 + 0.0085) x 4.32

0.0085 u = 4.314

u = 507.5 m/s

Thus, the initial velocity of the bullet is 507.5 m/s.  

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3 years ago
What is correlation of productivity and positive attitude
Ivahew [28]

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3 0
3 years ago
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A 6.0 kg block, starting from rest, slides down a frictionless incline of length 2.0 m. When it arrives at the bottom of the inc
erica [24]
Since 
<span>Vf^2 = 2*a*S </span>
<span>Given S=3.6m, thus </span>
<span>a = Vf^2/(2*3.6) </span>
<span>a = Vf^2/7.2 </span>

<span>Let d be the distance along the slope at which the velocity is 0.5Vf, then </span>
<span>(0.5Vf)^2 = 2*a*d </span>
<span>or </span>
<span>d = (0.5*Vf)^2/(2*a) </span>
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8 0
3 years ago
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block
kotegsom [21]

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  v  = 11.76  \ m/ s

Explanation:

From the question we are told that

   The  total distance traveled is  d =  1.2  \ m

    The mass of the block is  m_b  =  0.3 \ kg

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   PE  =  KE

Where  PE  is the potential energy which is mathematically represented as

      PE  =  m * g  *  h

substituting values

     PE  =   3 *  9.8  *  0.60

      PE  =  17.64 \  J

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          KE  =  \frac{1}{2}  *  m v^2

So

      \frac{1}{2}  *  m* v ^2  =  PE

substituting values  

  =>    \frac{1}{2}  *  3 * v ^2  = 17.64

=>       v  = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }

=>    v  = 11.76  \ m/ s

4 0
3 years ago
Give an example of a measurement that is precise to the nearest tenth of a gram.
Svet_ta [14]

Answer:

So we have a measure in grams, we can start with something like:

143.523 grams.

Now we want this measurement to be precise to the nearest tenth of a gram.

The nearest tenth of a gram is the first digit after the decimal point, then the digits that come after this are not useful, because they are outside our precision range.

Then we must write our measurement as:

143.5 grams

Where the digit that came after the 5 was a 2, so we rounded down.

4 0
3 years ago
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