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3 years ago

Questions

Physics

1 answer:

Firdavs [7]3 years ago

6 0

Answer:

People of jallianwala bagh where gathered for a public meeting

1)false

2)true

Explanation:

The number one of true and false is false because the exit was blocked so there was exits it was only blocked

The number two is true because the British general dyar came with the riflemen

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A box experiencing a gravitational force of 600 N is being pulled to the right with force of 250 N. A 25 N frictional force acts

jeka57 [31]

The answer is: 0 Newtons

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3 years ago

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A loading car is at rest on a track forming an angle of 25° with the vertical. The gross weight of the car and its load is 5500

koban [17]

Idk I just need other answers

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3 years ago

Suppose that Lisa walks her dog around the block for a little exercise. The block is 1 mile around. If she walks around the bloc

butalik [34]

We know that displacement is the length of straight line joining the initial position with the final position. as lisa walks around the block once, she comes back to her initial position here she started from, hence the initial and position of lisa is same. so her displacement is zero.

We know that distance is the length of the actual path followed by object between initial and final position. As lisa walks around the block once, she travels 1 mile. hence the distance traveled by lisa is 1 mile.

displacement : 0

distance : 1 mile

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3 years ago

How far (in meters) above the earth's surface will the acceleration of gravity be 21.0 % of what it is on the surface?

weeeeeb [17]

Answer:

7532m

Explanation:

Gravity on the surface of the earth with radius R is given by:

$F_1=\frac{Gm_{earth}m_{object}}{R^2_{earth}}$

Gravity a distance r above the surface:

$F_2=\frac{Gm_{earth}m_{object}}{(R+r)^2}$

How big is r if:

$F_2=0.21\times F_1$

You get the following equation:

$R^2=0.21(R+r)^2=0.21R^2+0.41Rr+0.21r^2$

Solve for r:

$r^2+2Rr-\frac{79}{21} R^2=0$

$r=-R+\sqrt{R^2+\frac{79}{21}R^2}=-R+\sqrt{\frac{100}{21}R^2}=(\frac{10}{\sqrt{21}}-1)R$

8 0

3 years ago

A race car starts from rest on a circular track. The car increases its speed at a constant rate at as it goes 4.25 times around

miskamm [114]

Answer:

Angle = 1.07°

Explanation:

Total acceleration consists of translational acceleration (a) which is tangential and centripetal acceleration which is (radial).

Now, Tangential and radial acceleration are always perpendicular to each other. Thus, in a triangle system they are opposite and adjacent sides. From trigonometric ratios,

Opposite/Adjacent = tanθ

Now centripetal acceleration is given as v²/r

Thus, the angle to the radial is given as;

tanθ = translational acceleration/centripetal acceleration

So, tanθ = a/(v²/r) = ar/v²

Thus, θ = tan^(-1)(ar/v²)

Now, Distance of one round of circular motion of 'r' radius

= circumference of circle = 2πr

N = number of trips car makes around the circle

Thus,

total distance = 2πrN

Now, from equation of motion, we know that v² = u² + 2as

Where s is total distance and u is initial velocity which is zero in this case.

Thus, v² = 0² + 2as

Making s the subject;

s = v²/2a

Thus,

2πrN = v²/2a

So let's simplify to bring out ar/v² which is what we will use to calculate the angle. So,

2πrN x 2a = v²

4πN(ar) = v²

Thus, ar/v² = 1/(4πN)

From the question, N = 4.25

Thus,

ar/v² = 1/(4π x 4.25) = 0.0187

From earlier, we saw that the angle is given by;

θ = tan^(-1)(ar/v²)

Thus, θ = tan^(-1)(0.0187)

θ = 1.07°

6 0

4 years ago