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3 years ago

The question ask:

Physics

1 answer:

abruzzese [7]3 years ago

4 0

Answer:

$\theta=28.07^{\circ}$

Explanation:

Speed of van, v = 28 m/s

Radius of unbanked curve, r = 150 m

Let $\theta$ is the angle with the vertical. In case of banking of road,

$T\ cos\theta=mg$ .............(1)

And

$T\ sin\theta=\dfrac{mv^2}{r}$ ..........(2)

From equation (1) and (2) :

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

$tan\theta=\dfrac{(28)^2}{150\times 9.8}$

$\theta=28.07^{\circ}$

So, the string makes an angle of 28.07 degrees with the vertical. Hence, this is the required solution.

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II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st

AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

$a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}$

Since we start counting at 0s and the ball departs from rest, this is just $a=\frac{v}{t}$

So we can write:

$F=ma=\frac{mv}{t}=\frac{gmv}{gt}$

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

$F=\frac{Wv}{gt}=\frac{v}{gt}W$

Substituting our values:

$F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N$

Where the average force exerted has been written it terms of the tennis ball's weight W.

8 0

3 years ago

SP1b.

nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

$a = \frac{vf - vi}{t}$

$\frac{15 - 25}{5}$

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0

3 years ago

A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir

mario62 [17]

Answer:

a . 0.35cm

b. 11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

$\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm$

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

$\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm$

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0

3 years ago

11) A man is on a 1/4 on a bridge. A train is coming the same direction he is going. The man can run across the bridge in the sa

lesya692 [45]

15mph

If the man turns and runs toward point A, he will cover

3/8 of the length of the bridge in the time that it takes

the train to reach A.

If the man runs forward toward point B, what part of the bridge

will he cover before the train reaches A? Well, he will cover

3/8 of the bridge, only heading forward toward B. This will put

him 3/8 + 3/8 = 6/8 = 3/4 of the way across the bridge by the

time the train reaches A.

since we know that the man and the train will meet at B, this

means that in the time it takes the man to run the remaining

1/4 of the bridge, the train will cover the entire length of

the bridge.

If it takes the man the same time to cover 1/4 of the bridge

that it takes the train to cover the whole bridge, then the train

must be going four times as fast as the man. Another way of saying

this is that the man runs at 1/4 the speed of the train.

Since the train's speed is known to be 60 mph, this means that

the man runs at (1/4) 60 = 15 mph.

7 0

3 years ago

A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed

malfutka [58]

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here: brainly.com/question/25573309

#SPJ10

5 0

2 years ago