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3 years ago

Your friend's car is stuck in the snow. You push very hard and become very 1 tired but the car won't move. Did you do work?

Physics

1 answer:

crimeas [40]3 years ago

7 0

Answer:

Yes you worked

Explanation:

You may have not seen a good outcome but you did put effort in.

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago

Imagine an infinite earth with a hole dripped through it. You fall in and accelerate at g~10m/s/s. How long until you reach the

Soloha48 [4]

An object with non-zero mass (even negligible mass is non-zero) will never reach the speed of light. Due to relativistic effects, each "unit" of acceleration becomes less effective at increasing your velocity (relative to some other object, of course) as your relative velocity approaches the speed of light.

And even if there was a way, If you would accelerate to the 99,99% of the speed light in just 1 second, you would experience a G-force of aprox. 30,600,000 g's which is enough to kill you in a few seconds

6 0

3 years ago

a transmission-line cable, of length 3 km, consists of 19 strands of identical copper conductors, each 1.5 mm in diameter. becau

Ivan

the resistance of the cable is 582.9 ohms

we are given the length of the cable which is 3 km, of 1.5 mm in, the diameter and resistivity of copper which is 1.72 m

The formula we are referring to for calculating the resistance of the cable is

R = ρl/A.

As there are 19 strands of copper conductors, so the resistance will be

R = 19( ρl/A)

Here ρ is the resisitivity = 1.72 , l is the length = 3(1+0.05)*10³3= 3150 m

A=pie/4(1.5 x 10⁻³)^2 =1.766 x 10⁻⁶ =1.766 x 10^-6

Substituting the values in the formula we get

R = 19 ( 1.72*3150 )/1.766 x 10⁻⁶

= 582.9 ohm

To know more about resistance refer to the linkhttps://brainly.com/question/14547003?referrer=searchResults.

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6 0

1 year ago

How much energy from the sun actually reaches the corn answer?

Serhud [2]

The energy from the sun that reaches the corn is about two billionths.

3 0

3 years ago

Read 2 more answers

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

7 0

3 years ago