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2 years ago

If a girl is running along a straight road with a uniform velocity 1.5m/s find her acceleration

Physics

1 answer:

Jlenok [28]2 years ago

7 0

Answer:

Dear user,

Answer to your query is provided below

Acceleration is zero because of no change in velocity.

Explanation:

Remember that velocity is a vector quantity and a vector can change in 3 ways

•Magnitude only

•Direction only

•Both magnitude and direction.

Now the magnitude of velocity (speed) can stay constant while the direction is changing. This is the case in circular motion.

In the question above, it is mentioned that the girl is moving along a straight road. Therefore no change in direction of velocity.

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Discuss how the hardness or softness of the landing surface is related to the time required to stop the egg

monitta

Answer

To understand this concept it is necessary to understand Newton's Second Law

According to Newtons Second law applied force is equal to rate of change of momentum of a body.

Mathematically,

$F=\frac{dp}{dt}$

here, $\frac{dp}{dt}$ is rate of change of momentum with respect to time

It means If two eggs fall from same height,one on softer surface and other on hard surface that time the momentum of both eggs will remain equal at both the surfaces. But, impact time will be different. On hard surface egg will stop almost instantly so impact time will be small and hence the force on egg will be large therefore the egg will breakup.

On the other hand on the soft surface like a cotton, egg will not stop instantly but it will slow down for a few seconds and then stop due to which the time of impact will increase. Therefore the force on egg will be less and it won't break up.

3 0

2 years ago

A police siren of frequency fsiren is attached to a vibrating platform. The platform and siren oscillate up and down in simple h

babunello [35]

Answer:

he maximum frequency occurs when the denominator is minimum

f’= f₀ $\frac{343}{343 + v_s}$

Explanation:

This is a doppler effect exercise, where the sound source is moving

f = fo $\frac{v}{v-v)s}$ when the source moves towards the observer

f ’=f_o $\frac{v}{v+v_{sy}}$ Alexandrian source of the observer

the maximum frequency occurs when the denominator is minimum, for both it is the point of maximum approach of the two objects

f’= f₀ $\frac{343}{343 + v_s}$

8 0

2 years ago

Devonte pushes a wheelbarrow with 830 W of power. How much work is required to get the wheelbarrow across the yard in 11 s? Roun

xxMikexx [17]

Answer: 9130 joules

Explanation:

Workdone by wheelbarrow = ?

Time = 11 seconds

Power = 830 watts

Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.

i.e Power = (workdone/time)

830 watts = Workdone / 11 seconds

Workdone = 830 watts x 11 seconds

Workdone = 9130 joules

Thus, 9130 joules of work is required to get the wheelbarrow across the yard.

8 0

2 years ago

Read 2 more answers

A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

$s=1.63 m$

5 0

2 years ago

Two tiny, spherical water drops, with identical charges of -6.19 × 10-16 C, have a center-to-center separation of 1.22 cm. (a) W

Lynna [10]

Answer:

$F = 2.32*10^{-17} \ N$

$n =3869 \ electrons$

Explanation:

From the question we are told that

The charge on each water drop is $q_1=q_2=q = - 6.19*10^{-16} \ C$

The distance of separation is $d = 1.22\ cm = 0.0122 \ m$

Generally the electrostatic force between the water drops is mathematically represented as

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

Here k is the coulombs constant with value $k = 9*10^9 \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

$F = \frac{9*10^9 * -6.19 *10^{-16} * (-6.19*10^{-16}) }{ 0.0122^ 2}$

$F = 2.32*10^{-17} \ N$

Generally the quantity of charge is mathematically represented as

$q = n * e$

Here n is the number of electron present

and e is the charge on one electron with value $e = 1.60*10^{-19} \ C$

$n = \frac{6.19 *10^{-16}}{1.60*10^{-19}}$

$n =3869 \ electrons$

4 0

2 years ago

If a girl is running along a straight road with a uniform velocity 1.5m/s find her acceleration ​

If a girl is running along a straight road with a uniform velocity 1.5m/s find her acceleration