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3 years ago

When a plane minor is placed horizontally on level ground at a distance of om from the foot of tower

Physics

1 answer:

Licemer1 [7]3 years ago

5 0

Answer:

b : 60m

Explanation:

when a plane mirror is placed horizontally on a level ground at a distance of 60m from the top of tower as shown in figure. a/c to question, The top of tower and its image in the mirror subtend an angle of 90° at the eyes. so, angle made by top of tower and horizontal line is 45° as

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4 years ago

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Write the nuclear equation for the alpha decay of astatine-213.

Sonbull [250]

Answer:

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Explanation:

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3 years ago

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3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altidude

Serjik [45]

Answer:

a) 252 ft/s

b) 1076 ft

Explanation:

The equation for motion for uniform acceleration (we can use it because the rocket is affected only by gravity) is as follows:

Y(t) = Y0 + V0 * t + 1/2 * a * t^2

Where

Y(t): altitude at a given time

Y0: initial altitude

V0: initial speed

a: acceleration, in this case -32.2 ft/s^2 (negative because gravity points down)

We set a 1 dimensional coordinate system with Y pointing up and the origin of coordinates at ground level.

We consider t=0 as the moment where powered flight ended (motor ran oou of fuel), at this moment the altitude was

Y(0) = 89.6 ft

Therefore:

Y0 = 89.6 ft

We also know that the rocket fell to ground 16 seconds later, therefore

Y(16) = 0 ft

So we can write

Y(t=16) = Y0 + V0 * t + 1/2 * a * t^2

V0 * t = Y(t=16) - Y0 - 1/2 * a * t^2

V0 =( Y(t=16) - Y0 - 1/2 * a * t^2 )/t

V0 =( 0 - 89.6 - 1/2 * (-32.2) * 16^2 )/16 = 252 ft/s

In the highest point of flight the rocket will have a speed = 0

The first derivative of the equation of motion is the equation of speed:

V(t) = V0 + a * t

If we equate this to zero we eill find the time at which the rocket achieved it's highest altitude.

0 = V0 + a * t

a * t = 0 - V0

t = -V0/a

t = -252/(-32.2) = 7.83 s

Now, we can take this time value andd plug it back into the position equation

Y(7.83) = 89.6 + 252 * 7.83 + 1/2 * (-32.2) * 7.83^2 = 1076 ft

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3 years ago

What would happen to the two balls if one of them were kept positively charged and the charge on the other ball were slowly incr

Whitepunk [10]

Answer:

The force of repulsion between the two balls will increase

Explanation:

The electrostatic force between two charged objects is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the distance between the centres of the two objects

We see that the magnitude of the force is directly proportional to the charges on the two objects. in this problem, we have two positively charged balls (so, there is a force of repulsion between them, since like charges repel each other and unlike charges attract each other), and the positive charge in one of them is slowly increased: this means that either $q_1$ or $q_2$ in the formula is increasing, and so the magnitude of the force is increasing.

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3 years ago