Galileo was a contemporary of

A 60 kg student is standing in the train station next to her 10 kg suitcase when her train is called. (A) Estimate how much work

ASHA 777 [7]

Answer

given,

mass of student = 60 Kg

mass of suitcase = 10 Kg

a) Work done by picking of Suitcase is equal to zero

b) acceleration = 0.1 m/s²

distance = 2 m

Using second law conservation

F = m a

F = 10 x 0.1 = 1 N

c) Work done

W = F x s

W = 1 x 2 = 2 J

d) When the are moving with constant speed acceleration is equal to zero

F = m a

F = 10 x 0 = 0 N

W = F x s = 0 x s = 0 J

e) work done = change in kinetic energy

K.E = 2 J

5 0

3 years ago

Under what condition is the instantaneous acceleration of a moving body equal to its average acceleration over time?

Rzqust [24]

If the acceleration is constant (negative or positive) the instantaneous acceleration cannot be

Average acceleration: [final velocity - initial velocity ] /Δ time

Instantaneous acceleration = d V / dt =slope of the velocity vs t graph

If acceleration is increasing, the slope of the curve at one moment will be higher than the average acceleration.

If acceleration is decreasing, the slope of the curve at one moment will be lower than the average acceleration.

If acceleration is constant, the acceleration at any moment is the same, then only at constant accelerations, the instantaneuos acceleration is the same than the average acceleration.

Constant zero acceleration is a particular case of constant acceleration, so at constant zero acceleration the instantaneous accelerations is the same than the average acceleration: zero. But, it is not true that only at zero acceleration the instantaneous acceleration is equal than the average acceleration.

That is why the only true option and the answer is the option D. only at constant accelerations.

3 0

3 years ago

Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th

alukav5142 [94]

Answer:

The altitude of geostationary satellite is $3.58\times10^{7}\ m$

Explanation:

Given that,

Radius of moon's orbit $r=3.84\times10^{8}\ m$

Time period $T=2.36\times10^{6}\ sec$

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

$T=\sqrt{\dfrac{4\pi^2}{GM}a^3}$

$a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}$

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

$a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}$

$a=4.217\times10^{7}\ m$

We need to calculate the altitude of geostationary satellite

Using formula of altitude

$h = a-R_{e}$

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

$h =4.217\times10^{7}-6.38\times10^{6}$

$h =35790000\ m$

$h=3.58\times10^{7}\ m$

Hence, The altitude of geostationary satellite is $3.58\times10^{7}\ m$

4 0

3 years ago

A gas expands and does PV work on its surroundings equal to 319 J. At the same time, it absorbs 136 J of heat from the surroundi

LiRa [457]

Answer:

The change in energy of the gas during the process is $-1.83\times 10^{2}$ joules.

Explanation:

We can represent this process by the First Law of Thermodynamics, in which gas does work on its surroundings and absorbs heat from there to describe its change in energy. In other words:

$Q_{in} - W_{out} = \Delta E$