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DENIUS [597]
3 years ago
15

If the force that propels the cannonball forward is 500N, how much force will move the cannon backward?

Physics
1 answer:
Licemer1 [7]3 years ago
6 0

Answer:

Well it would be equal to 500N because pushing forward the ball (or whatever maybe a body) would push the canon back an even 500N backwards...

Explanation:

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I need help with part D
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3 years ago
Acceleration is the magnitude of average velocity.​
lina2011 [118]

Answer:

false

Explanation:

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4 0
3 years ago
A 2.3 kg cart is rolling across a frictionless, horizontal track towards a 1.5 kg cart that is initially held at rest. The carts
Inga [223]

Answer:

total momentum = 8.42 kgm/s

velocity of the first cart is 3.660 m/s

Explanation:

Given data

mass m1 = 2.3 kg

mass m2 = 1.5 kg

final velocity V2 = 4.9 m/s

final velocity V3 = - 1.9 m/s

to find out

total momentum  and velocity of the first cart

solution

we know mass and final velocty

and initial velocity of second cart V1 = 0

so now we can calculate total momentum that is m1 v2 + m2 v2

total momentum =  2.3 ×4.9 + 1.5 ×(-1.9)

total momentum = 8.42 kgm/s

and

conservation of momentum  is

m1 V + m2 v1  = m1 v2  + m2 v3

put all value and find V

2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)

V = 8.42 / 2.3

V = 3.660 m/s

so velocity of the first cart is 3.660 m/s

8 0
3 years ago
A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

8 0
3 years ago
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