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3 years ago

A force acting on an object does no work If

Physics

2 answers:

Vitek1552 [10]3 years ago

6 0

It is not in movement good luck

Irina-Kira [14]3 years ago

3 0

If the object doesn't move, no work is done.

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If a planet has 3 times the radius of the Earth, but has the same density as the Earth, what is the gravitational acceleration a

xz_007 [3.2K]

Answer:

The Gravitational Acceleration of the Big planet is G = 3g

where G is the gravitational acceleration of the Big Planet and g is the gravitational acceleration of the Earth

So, it becomes G = 3 x 9.8 $ms^{-2}$

G = 29.4 $ms^{-2}$

Explanation:

Let's start by calculating the <u>Density of the Earth</u> of mass m anad Radius r

We know that mass is density times the volume

ρ = $\frac{m}{V}$

m = ρV

we know the volume is V = $\frac{4}{3}$ $πr^{3}$ (please ignore the symbol of Pi)

m = ρ $\frac{4}{3}$ $πr^{3}$

Calculating the <u>Density of Big Planet</u>

→For Big Planet we know that radius is 3-times the radius of Earth, that is 3r

M = ρ $\frac{108}{3}$ $π(r)^{3}$

So in conclusion, the mass of Earth and the mass of Big planet are related as M = 27m

Now let's come to the Gravitational Force, we know that gravitational force is directly proportional to the mass of the body and inversely proportional to the radius.

→Suppose for Earth: Mass = m, Radius = r

For Big Planet: Mass = M, Radius = R

$\frac{m}{r^{2} }$ : $\frac{M}{R^{2} }$ = g : G

$\frac{Gm}{r^{2} }$ = $\frac{gM}{R^{2} }$

G = $\frac{gMr^{2}}{mR^{2}}$

→ Putting the value of R = 3r and M = 27m

G = $\frac{g27mr^{2}}{m(3r)^{2}}$

By cutting the terms, we get

G = 3g

value of g= 9.8 $ms^{-2}$

G = 3 x $9.8ms^{-2}$

G = 29.4 $ms^{-2}$

5 0

4 years ago

A pilot wishes to fly on a course of 290° with an air speed of 300 knots when the wind blows from 224° at 18 knots. Find the dri

kow [346]

The answer is 3.18 degrees

4 0

4 years ago

Read 2 more answers

Two wires, both with current out of the page, are next to one another. The wire on the left has a current of 1 A and the wire on

Soloha48 [4]

Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

$B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\$

Next, you use the formula for the magnetic force produced by the wires:

$\vec{F_B}=I\vec{L}\ X \vec{B}$

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

$\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}$