A boy runs 100m due North and 100m due east. what is displacement?

Planet A has mass 3M and radius R, while Planet B has mass 4M and radius 2R. They are separated by center-to-center distance 8R.

Aleksandr-060686 [28]

Answer:

Explanation:

In Newton's law of universal gravitation

F = Gm₁m₂/r²

Where G is a gravitational constant = 6.674e-11m³/kgs²

m₁ and m₂ are the masses of the two bodies or objects in question, in kilogram (kg)

r is the distance in meters between them

From the question, the rock is placed halfway between the planets

So, it's distance from planet A is 8R/2 = 4R

And it's distance from planet B is also 8R/2 = 4R

Using F = Gm₁m₂/r²

To Planet A

r = 4R,

m₁ = mass of Rock = m

m₂ = mass of planet A = 3M

So Fa = G mm₂/r² = Gm(3M) / (4R)²

To Planet B,

r = 4R,

m₁ = mass of Rock = m

m₂ = mass of planet B = 4M

Fb = G mm₂/r² = Gm(4M) / (4R)²

Comparing both forces together, we realise that Planet B has the largest force,

so take we F = Fb – Fa

F = Fb – Fa = Gm(4M) / (4R)² – Gm(3M) / (4R)²

F = GmM/16R²)(4–3)

F = GmM/16R²

Note that Force = Mass * Acceleration

So, F = ma

So, ma = GmM/16R² ------- Divide through by m

a = GM/16R²

From the question

M = 7.3×10^23kg

R = 5.8×10^6 m

So, a = (6.674 * 10^-11 * 7.3×10^23)/16(5.8×10^6)²

a = (48.7202 * 10^12)/16(33.64 * 10^12)

a = (48.7202 * 10^12)/(538.4 * 10^12)

a = 48.7202/538.4

a = 0.090517612960760

a = 0.091m/s² ----------Approximated

5 0

2 years ago

Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.

Alecsey [184]

Time taken by the water balloon to reach the bottom will be given as

$h = \frac{1}{2} gt^2$

here we know that

$h = 18 m$

$g = 9.8 m/s^2$

now by the above formula

$18 = \frac{1}{2}*9.8* t^2$

$18 = 4.9 t^2$

$t = 1.92 s$

now in the same time interval we can say the distance moved by it will be

$d = v_x * t$

$d = 8.2 * 1.92 = 15.7 m$

so it will fall at a distance 15.7 m from its initial position

5 0

2 years ago

High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 43.7 m/s just before it strikes a 45

Helga [31]

Answer:

41.27m/s

Explanation:

According to law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the velocity after impact

Given

m1 = 0.2kg

u1 = 43.7m/s

m2 = 45.9g = 0.0459kg

u2 = 30.7m/s

Required

Velocity after impact v

Substitute the given parameters into the formula

0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v

8.74+1.409 = 0.2459v

10.149 = 0.2459v

v = 10.149/0.2459

v = 41.27m/s

Hence the speed of the golf ball immediately after impact is 41.27m/s

8 0

2 years ago

To determine the y-component of a projectile’s velocity, what operation is performed on the angle of the launch?

koban [17]

To determine the y component of velocity of a projectile sine operation is performed on the angle of launch.

Answer: sine

Explanation:

Thus $a_x=0,a_y=g$

The initial velocity u can be resolved along two directions.

Along the X direction initial velocity = u cos θ

Along y direction initial velocity= u sin θ

From the equation of motion $v= u+at$

Thus velocity along x direction $v_x$ =u cos θ

Velocity along y direction $v_y$ = u sinθ -gt

Sign of g is negative.

3 0

2 years ago

Read 2 more answers

A 3 kg rubber block is resting on wet concrete. The coefficient of static friction is 0.3. What is the minimum force that must b

mrs_skeptik [129]

Answer:

You would have to find the friction force of the rubber block which would be found with the equation of Normal force (mass*gravity) times cooeficient of friction which would give 8.82 N for the amount of friction and because you need more force than 8.82 N (assuming gravity is 9.8)

8 0

1 year ago

A boy runs 100m due North and 100m due east. what is displacement?​

A boy runs 100m due North and 100m due east. what is displacement?