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3 years ago

Which of the following frictionless ramps (A, B, or C) will give the ball the greatest speed at the bottom of the ramp? Explain.

Physics

1 answer:

Elenna [48]3 years ago

8 0

Let the ball is initially at point P1. Finally it reaches at point P2.( bottom most). Since only conservative force is acting on the ball throughout the motion, mechanical energy will be conserved. Therefore conserving energy from P1 to P2, Initially ball is at rest(P1). So initial kinetic energy is 0. Let ball is at height h1 initially, so initial potential energy = mgh ( m= mass of the ball). At bottom most point (P2), potential energy of ball changes into kinetic energy. So final kinetic energy =1/2 mv^2. Final potential energy = 0. Equating energies at P1 and P2, Mgh= 1/2 mv^2. Therefore, v= (2gh)^1/2. Since all the initial and final points are at same height, velocity will be same for all the paths(irrespective of the path taken by ball). So velocity will be same.

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A horizontal circular curve on a highway is designed for traffic moving at 60 km/h. If the radius of the curve is 150 m, and if

KengaRu [80]

Answer:

The value of the correct angle of banking for the road is $\theta = 67.76$ °

Explanation:

Given data

Velocity (v) = 60 $\frac{m}{s}$

Radius = 150 m

The velocity of the car in this case is given by

$v = \sqrt{r g \tan \theta}$

$v^{2} = r g \tan \theta$

$\tan \theta = \frac{v^{2} }{rg}$

Put all the values in above formula we get

$\tan \theta = \frac{60^{2} }{(150)(9.81)}$

$\tan \theta =$ 2.446

$\theta = 67.76$ °

Therefore the value of the correct angle of banking for the road is $\theta = 67.76$ °

4 0

2 years ago

A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed

Tems11 [23]

Answer:

The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.

Explanation:

5 0

3 years ago

Read 2 more answers

Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N

nadezda [96]

First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say $F_{1}$ , and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
$F_{1}$ = $7i$ , where $i$ = Unit vector in the x-direction.

2) Take the second force, say $F_{2}$ , and assume that it is making an angle $\alpha$ with the first force $F_{1}$ .

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

$F_{2} = (15*cos \alpha)i + (15*sin \alpha )j$

Now from simple vector addition, we know that,
$F_{R} = F_{1} + F_{2}$ --- (A)

Where $F_{R}$ = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find $F_{1}+F_{2}$ first!
$F_{1}+F_{2}$ =   $7i+(15*cos \alpha)i + (15*sin \alpha )j$

=> $F_{1}+F_{2}$ =   $(7+15*cos \alpha)i + (15*sin \alpha )j$

Now the magnitude of $F_{1}+F_{2}$ is,
| $F_{1}+F_{2}$ | = $\sqrt{ (7+ 15*cos \alpha)^{2} + (15*sin \alpha )^{2}}$

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}$

Since $(sin \alpha)^{2} + (cos \alpha)^{2} = 1$ , therefore,

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Since  | $F_{1}+F_{2}$ | = | $F_{R}$ |, and the magnitude of the resultant force is 20N, therefore,

| $F_{R}$ | = | $F_{1}+F_{2}$ |
20 = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Take square on both sides,
400 = $49 + 225 + 210*(cos \alpha)$
$(cos \alpha) = \frac{3}{5}$

$\alpha = 53.13^{o}$

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0

2 years ago

How much distance does it take to stop a car going 30 m/s (67 mph) if the brakes can apply a force equal to one half the car’s w

denpristay [2]

Answer:

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1 year ago

A car traveling at 30 m/s north slows to a stop in 15 seconds. What is the acceleration? PLEASE HELP

Lapatulllka [165]

The answer is C..........

6 0

2 years ago

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