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diamong [38]
3 years ago
6

A car is moving 5.82M/S when it accelerates at 2.35M/S^2for 3.25S. What is its final velocity?

Physics
1 answer:
rodikova [14]3 years ago
3 0

Answer:

Explanation:

vf=vi+at

vi=5.82 m/s

a=2.35 m/s2

t=3.25 s

vf=5.82+2.35*3.25

vf=5.82+7.64

vf=13.46

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What are some examples of gravitational forces that keep an object in orbit? Pls give at least 2-3 example if you can.​
Lelu [443]

Answer:

Without any external forces a moving object will continue to move in a straight line. The gravitational force between the two objects will provide the centripetal force to keep the objects moving around one another.

1. satellite in orbit around the earth             (motion of earth is negligible)

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3 0
3 years ago
An 80-kg man is skating northward and happens to suddenly collide with a 20-kg boy who is ice skating toward the east. Immediate
Fantom [35]

Answer:

6.25 m/s

Explanation:

mass of man (m1) = 80 kg

mass of boy (m2) = 20 kg

mass of man and boy after collision (m12)= 20 + 80 = 100 kg

velocity of man and boy after collision (v) = 2.5 m/s

angle θ = 60 °

How fast was the boy moving just before the collision ?

  • From the diagram attached, the first image shows the man and the boys motion while the second diagram shows their motion rearranged to form a triangle. With the momentum of the man and the boy forming the sides of the triangle.
  • M₁₂ =  total momentum after collision = m12 x v = 100 x 2.5 = 250
  • Mboy = momentum of the boy before collision = m2 x Velocity of boy
  • Mman = momentum of the man before collision = m1 x velocity of man  
  • from the triangle, cos θ = \frac{Mboy}{M₁₂}

        cos 60 = \frac{Mboy}{250}

        Mboy = 250 x cos 60 = 125

  • recall that momentum of the boy (Mboy) also = m2 x Velocity of boy

        therefore

        125 = 20 x velocity of boy

         velocity of boy = 125 / 20 = 6.25 m/s

4 0
3 years ago
The gradual reduction of the amplitude
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3 years ago
A transformer connected to a 120-v(rms) at line is to supply 13,000V(rms) for a neon sign.
lyudmila [28]

Answer:

(a) 108

(b) 110.500 kW

(c) 920.84 A

Solution:

As per the question:

Voltage at primary, V_{p} = 120\ V          (rms voltage)

Voltage at secondary, V_{s} = 13000\ V  (rms voltage)

Current in the secondary, I_{s} = 8.50\ mA  

Now,

(a) The ratio of secondary to primary turns is given by the relation:

\frac{N_{s}}{N_{p}} = \frac{V_{s}}{V_{s}}

where

N_{p} = No. of turns in primary

N_{s} = No. of turns in secondary

\frac{N_{s}}{N_{p}} = \frac{13000}{120} ≈ 108

(b) The power supplied to the line is given by:

Power, P = V_{s}I_{s} = 13000\times 8.50 = 110.500\ kW

(c) The current rating that the fuse should have is given by:

\frac{V_{s}}{V_{p}} = \frac{I_{p}}{I_{s}}

\frac{13000}{120} = \frac{I_{p}}{8.50}

I_{p} = \frac{13000}{120}\times 8.50 = 920.84\ A

 

6 0
3 years ago
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