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1 year ago

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51. A balloon starts rising from the ground, vertically upwards, uniformly at the rate of 1 m s-1. At the end of 4 seconds, a bo

dy is released from the balloon. Calculate the time taken by the released body to reach the ground. 51. A balloon starts rising from the ground , vertically upwards , uniformly at the rate of 1 m s - 1 . At the end of 4 seconds , a body is released from the balloon . Calculate the time taken by the released body to reach the ground

1 answer:

strojnjashka [21]1 year ago

5 0

Answer:

4s

Explanation:

when a bOdy rises into the air,the time it takes to reach a particular height is the same as the time it will take the body to fall from that height to the ground.

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As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in

vlada-n [284]

Answer:

$a=-0.33\ m/s^2$

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

$\displaystyle a=\frac{\Delta v}{t}$

Or, equivalently

$\displaystyle a=\frac{v_f-v_o}{t}$

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

$\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2$

The negative sing of a indicates there is deceleration or decreasing speed

7 0

2 years ago

The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the

fomenos

<span>The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the Normal .

When the normal is drawn, the incident ray makes an angle with it known as the angle of incidence and the reflected ray makes an angle with it known as the angle of incidence. These angles are always equal.

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5 0

3 years ago

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

3 years ago

An electroscope is a simple device consisting of a metal ball that is attached by a conductor to two thin leaves of metal foil p

baherus [9]

Answer:

the electroscope separate by the presence of charge carriers

Explanation:

Metal bodies are characterized by having free (mobile) electrons. In the electroscope the plates are in balance; when the external metal ball is touched, a charge is introduced into the device, when the body that touched the ball is separated, an excess charge remains. This charge, being a metal, is distributed over the entire surface, giving a uniform density and an electric force of repulsion is created between the two charged sheets, which tends to separate the sheets. This force is counteracted by the tension component as the sheets are separated at a given angle, the separation reaches the point where

Fe - Tx = 0

Fe = Tx

In summary, the electroscope separate its leaves by the presence of charge carriers

3 0

3 years ago

A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba

Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

The mass of the rubber ball is $m_r = 2 \ kg$

The initial speed of the rubber ball is $u = 3 \ m/s$

The final speed at which it bounces bank $v - 3 \ m/s$

The mass of the clay ball is $m_c = 2 \ kg$

The initial speed of the clay ball is $u = 3 \ m/s$

The final speed of the clay ball is $v = 0 \ m/s$

Generally Impulse is mathematically represented as

$I = \Delta p$

where $\Delta p$ is the change in the linear momentum so

$I = m(v-u)$

For the rubber is

$I_r = 2(-3 -3)$

$I_r = -12\ kg \cdot m/s$

=> $|I_r| = 12\ kg \cdot m/s$

For the clay ball

$I_c = 2(0-3)$

$I_c = -6 \ kg\cdot \ m/s$

=> $| I_c| = 6 \ kg\cdot \ m/s$

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

8 0

3 years ago