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Vinil7 [7]
3 years ago
15

What is the change in momentum of a ball if a force of 5.0 N acts on it for a brief time of 4 ms?

Physics
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

0.02 Ns

Explanation:

From the question given above, the following data were obtained:

Force (F) = 5 N

Time (t) = 4 ms

Change in momentum =?

Next, we shall convert 4 ms to s. This can be obtained as follow:

1 ms = 1×10¯³ s

Therefore,

4 ms = 4 ms × 1×10¯³ s / 1 ms

4 ms = 4×10¯³ s

Thus, 4 ms is equivalent to 4×10¯³ s

Change in momentum = Impulse

Impulse (I) = Force (F) × time (t)

Change in momentum = Force (F) × time (t)

Force (F) = 5 N

Time (t) = 4×10¯³ s

Change in momentum =?

Change in momentum = 5 × 4×10¯³

Change in momentum = 0.02 Ns

Therefore, the change in the momentum of the ball is 0.02 Ns.

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The magnitude of the resultant force is to be 200 N, directed along the positive ݒaxis. Determine the magnitude of the force Fin
Ilia_Sergeevich [38]

Answer: F = 776.18N θ = 29.41°

Question:

The magnitude of the resultant force is to be 200 N, directed along the positive y-axis. Determine the magnitude of the force F Find the angle θ.

Attached is the image.

Explanation:

To solve this question we need to resolve the forces to the x and y axis.

For the x axis;

Fcosθ = 700cos15....1

For the y axis;

200 = Fsinθ - 700sin15

Fsinθ = 200 + 700sin15 ...2

Dividing equation 2 by 1

Fsinθ/Fcosθ = (200 + 700sin15)/700cos15

Tanθ = (200 + 700sin15)/700cos15

θ = Taninverse(200 + 700sin15)/700cos15)

θ = 29.41°

From equation 1;

F = 700cos15/cosθ

F = 700cos15/cos29.41

F = 776.18N

3 0
4 years ago
A = F/m is an explanation of<br> in  Physics
Semmy [17]

Answer:

The equation a=F/m or the acceleration is equal to the net force of an object divided by that object's mass, is an equation derived and explained by Sir Issac Newton's second law of motion. Newton's second law of motion states that the force of an object is equal to the mass times the acceleration of that object.

7 0
3 years ago
How many complete revolutions are needed to draw the angle 725°?
uysha [10]
C. A complete revolution is 360 degree. two revolution is 720.
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A pitcher throws a 0.144-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just
Solnce55 [7]

(a) 12.8 kg m/s

The impulse delivered by the bat on the baseball is equal to the change in momentum of the baseball:

I=\Delta p = m(v-u)

where we have

m = 0.144 kg is the mass of the ball

v = -47 m/s is the final velocity of the ball

u = 42 m/s is the initial velocity of the ball

Substituting into the equation, we find

I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s

And since we are interested in the magnitude only,

I=12.8 kg m/s

(b) 2.78 kN

The impulse exerted on the ball is also equal to the product between the average force and the contact time:

I=F\Delta t

where

F is the average force exerted on the ball

\Delta t=0.0046 s is the contact time

Solving the formula for F, we find

F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN

(c) The force exerted on the ball is much larger (1988 times more) than the weigth of the ball

The weight of the ball is given by

W=mg

where

m = 0.144 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the equation for W, we find

W=(0.144 kg)(9.8 m/s^2)=1.4 N

So as we see, the force exerted on the ball (2783 N) is almost 2000 times larger than the weight of the ball (1.4 N):

\frac{F}{W}=\frac{2783 N}{1.4 N}=1988

8 0
3 years ago
A (nonconstant) harmonic function takes its maximum value and its minimum value
AURORKA [14]

Answer:

Explanation:

Consider that F (any function) <0 .

u(x,y) is a coontinuous function in the closed interval or region R.

Let us consider a point (p,q) that is inside the region and it is a maximum point.

Then it should be must

uxx (p,q) <0 where uxx means double differentiation

and uy(p,q) >0

Since ux(p,q) = 0 = uy(p,q) where ux and uy means single differentiation with respect to x and y respectively.

Say, Maximum limits of the region is T

therefore q<T

then uy (p,q) = 0 if q<T

if q = T then

point (p,q) = (p,T) will be on the boundary of R then we claim that

uy(p,q) >0

Similarly for the minimum also it will work

3 0
3 years ago
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