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raketka [301]
3 years ago
10

A child is sliding on a sled at 1.3 m/s to the right. You stop the sled by pushing on it for 0.80 s in a direction opposite to i

ts motion. Part A If the mass of the child and sled is 34 kg , what is the magnitude of the average force you need to apply to stop the sled?
Physics
1 answer:
Shalnov [3]3 years ago
3 0

Answer:

find acceleration first

a = vf - vi / t

a = 0  - 1.5 / 0.5s   (vf is zero)

a = -3 (negative sign indicates that acceleration is decreasing)

so F=ma

    F = 35 x -3

   F = -105 N (here negative sign indicates that u have to apply a force opposite to the boy's direction i.e from the left or towards the right)

Explanation:

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4 years ago
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If a certain mass has its velocity changed from 6.00 m/s to 7.50 m/s when a 3.00 n force acts for 4.00 seconds, find the mass of
shutvik [7]
By definition we have that the force for time is equal to the product of the mass for the change in speed.
 We have then that
 F * (delta t) = m * (delta v)
 Clearing the mass
 m = (F * (delta t)) / (delta v)
 Substituting the values
 m = ((3.00) * (4.00)) / (7.50-6.00) = 8
 answer
 The mass of the moving object is 8Kg
3 0
4 years ago
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A boy with a mass of 25 kg is sitting in a red wagon of mass 8.5 kg which is at rest. His friend begins pulling him forward, acc
tangare [24]

So, the force that given when the wagon was being pulled is approximately <u>19.1 N (C)</u>.

<h3>Introduction</h3>

Hi ! For intermesso, this question will adopt a lot about the relationship of impulse to change in momentum. <u>Impulse is the total force applied in a certain time interval</u>. Impulses can cause a change of momentum, because momentum itself <u>is a mass that is affected by the velocity of an object</u>. We know that velocity is a vector quantity easy to change its direction. The relationship between impulse and change in momentum is formulated by :

\sf{I = \Delta p}

\sf{F \cdot \Delta t = (m \cdot v') - (m \cdot v)}

\boxed{\sf{\bold{F \cdot \Delta t = m (v' -v)}}}

With the following condition :

  • I = impulse that given (N.s)
  • \sf{\Delta p} = change of momentum (kg.m/s)
  • F = force that given (N)
  • m = mass of the object (kg)
  • v = initial velocity (m/s)
  • v' = final velocity (m/s)
  • \sf{\Delta t} = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • m = mass of the object = 25 kg
  • v = initial velocity = 0 m/s
  • v' = final velocity = 1.8 m/s
  • \sf{\Delta t} = interval of the time = 2.35 s

What was asked :

  • F = force that given = ... N

Step by step :

\sf{F \cdot \Delta t = m (v' -v)}

\sf{F \cdot 2.35 = 25 (1.8 - 0)}

\sf{F = \frac{25 (1.8)}{2.35}}

\boxed{\sf{F = 19.15 \: N \approx 19.1 \: N}}

<h3>Conclusion</h3>

So, the force that given when the wagon was being pulled is approximately 19.1 N (C).

6 0
3 years ago
A proton (1.6726 ? 10-27 kg) and a neutron (1.6749 ? 10-27 kg) at rest combine to form a deuteron, the nucleus of deuterium or "
Alexus [3.1K]

Answer:

Explanation:

The mass of the deuteron = mass of the proton + mass of the neutron + mass equivalent of the energy of 2.2 Mev evolved.

I amu = 931 Mev

2.2 Mev = 2.2 / 931 amu

= ( 2.2 / 931 )x 1.6726 x 10⁻²⁷

= .00395 x 10⁻²⁷

The mass of the deuteron  =( 1.6726 + 1.6749 +  .00395)x 10⁻²⁷ kg

= 3.35145 x 10⁻²⁷ kg

b ) Momentum of gamma ray

= h / λ ( h is plank's constant and λ is wavelength of gamma ray )

= hυ / υλ       (  υ is frequency of gamma ray )

= E / c  ( E is energy of photon and c is velocity o light )

= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹  J / 3 x 10⁸

= 1.173 x 10⁻²¹ Kg m /s

This will be the momentum of deuteron also

Kinetic energy

= p² / 2m ( p is momentum and m is mass of deuteron )

= ( 1.173 x 10⁻²¹ )² / ( 2 x 3.35145 x 10⁻²⁷)

= 1.376 x ⁻¹⁵ J

Energy of gamma ray

= 2.2 x 10⁶ x 1.6 x 10⁻¹⁹  J

= 3.52 x 10⁻¹³ J

So kinetic energy of deuteron is smaller than energy of gamma ray photon .

5 0
3 years ago
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