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NeX [460]
3 years ago
6

A roller coaster car rapidly picks up speed as it rolls down a slope. As it starts down the slope, its speed is 4m/s. But 3 seco

nds later at the bottom of the slope its speed is 22m/s. What is its average acceleration?
Physics
1 answer:
slamgirl [31]3 years ago
7 0

Answer:

6 m/s^2

Explanation:

The average acceleration of the car is given by:

a=\frac{v-u}{t}

where

v is the final speed

u is the initial speed

t is the time elapsed

For the car in this problem,

v = 22 m/s

u = 4 m/s

t = 3 s

Therefore, the acceleration of the car is

a=\frac{22 m/s-4 m/s}{3 s}=6 m/s^2

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Describe a ball's motion as it rolls up a slanted
emmasim [6.3K]

The ball will decelerate as it moves upwards.

The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

The given parameters;

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As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.

Thus, the ball decelerate as it moves upwards.

The acceleration of the ball is calculate as;

a = \frac{v_f -v_0}{t} \\\\

<em>at the highest point on the incline plane, the final velocity </em>v_f<em> is zero</em>

a = \frac{0-1.25}{4.22} \\\\a = -0.3 \ m/s^2

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

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4 0
3 years ago
Does Ap Physics have anything to do with probablity?
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4 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∈.        

                           

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