All categories

3 years ago

The gravitational attraction between two masses of 3kg that are separated by a distance of 1cm is

Physics

1 answer:

Luba_88 [7]3 years ago

6 0

Answer:

6.003×10¯⁶ N

Explanation:

We'll begin by converting 1 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

1 cm = 1 cm × 1 m / 100 cm

1 cm = 0.01 m

Finally, we shall determine the gravitational attraction. This can be obtained as follow:

Mass 1 (M₁) = 3 Kg

Mass 2 (M₂) = 3 Kg

Distance apart (r) = 0.01 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 3 × / 0.01²

F = 6.003×10¯¹⁰ / 1×10¯⁴

F = 6.003×10¯⁶ N

Thus the gravitational attraction is 6.003×10¯⁶ N

You might be interested in

Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?

aniked [119]

An applied force is a force that is applied to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
Normal force is the component, perpendicular to the surface (surface being a plane) of contact.
The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences attractive force from the air it passes through. It also experiences a downward pull because of the normal force.
Solution A.

4 0

4 years ago

Read 2 more answers

An RV is traveling 60 km/h along a highway. A boy sitting near the driver of the RV throws a ball to another boy at the back end

alina1380 [7]

Answer:

Speed of the ball relative to the boys: 25 km/h

Speed of the ball relative to a stationary observer: 35 km/h

Explanation:

The RV is travelling at a velocity of

$v_{RV}=+60 km/h$

Here we have taken the direction of motion of the RV as positive direction.

The boy sitting near the driver throws the ball back with speed of 25 km/h, so the velocity of the ball in the reference frame of the RV is

$v_B = -25 km/h$

with negative sign since it is travelling in the opposite direction relative to the RV. Therefore, this is the velocity measured by every observer in the reference frame of the RV: so the speed measured by the boys is

v = 25 km/h

Instead, a stationary observer outside the RV measures a velocity of the ball given by the algebraic sum of the two velocities:

v = +60 km/h + (-25 km/h) = +35 km/h

So, he/she measures a speed of 35 km/h.

5 0

3 years ago

A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t

emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W = $F^>$ × $x^>$

W = $Fxcos\theta$ ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = $Fxcos($ 0° $)$

W = $Fx$ ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

$W_{net$ = ΔKE

= $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

$Fx$ = $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu²

we make F, the subject of formula

F = $\frac{m}{2x}$ ( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = $\frac{830}{(2)(0.255)}$ ( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0

3 years ago

Match the measurement with the prober SI unit.

padilas [110]

Answer:

Your question is incomplete

8 0

3 years ago

PLEASE HELP LOTTA POINTS

sergiy2304 [10]

There's a nasty wrinkle here that's kind of sneaky, and makes the work harder than it should be.

Look at the first question. There's a number there that's dropped in so quietly that you're almost sure to miss it, but it changes the whole landscape of both of these problems. That's where it says

" ... 20 cm mark (30 cm from the fulcrum) ... " .

That tells us that the yellow bar resting on the pivot is actually a meter stick, but the pictures don't show the centimeter marks on the stick. The left end of the stick is "0 cm", the right end of the stick is "100 cm", and the pivot is under the "50 cm" mark.

When the question talks about hanging a weight, it tells the centimeter mark on the stick where the weight is tied. To solve the problem, we have to first figure out how far that is from the pivot, then calculate how far from the pivot to put the weight on the other side, and finally what centimeter mark that is on the stick.

How to solve the problems:

-- The "moment" of a weight is (the weight) x (its distance from the pivot) .

-- To balance the stick, (the sum of the moments on one side) = (the sum of the moments on the other side).

= = = = = = = = = =

#1). Only one moment on the left side.

(160 gm) x (30 cm from pivot) = 4,800 gm-cm

To balance, we need 4,800 gm-cm of moment on the right side.

(500 gm) x (distance from pivot) = 4,800 gm-cm

Distance from pivot = (4,800 gm-cm) / (500 gm) = 9.6 cm

The 500 gm has to hang 9.6 cm to the right of the pivot. But that's not the answer to the problem. They want to know what mark on the stick to hang it from. The pivot is at the 50cm mark. The 500gm has to hang 9.6 cm to the right of the pivot. That's the 59.6 cm mark on the stick.

= = = = =

#2). There are 2 weights hanging from the left side. We have to find the moment of each weight, add them up, then create the same amount of moment on the right side.

one weight: 120gm, hanging from the 25cm mark.

That's 25cm from the pivot. Moment = (120gm) (25cm) = 3,000 gm-cm

the other weight: 20gm, hanging from the 10cm mark;

That's 40cm from the pivot. Moment = (20gm) (40cm) = 800 gm-cm

Add up the moments on the left side:

(3,000 gm-cm) + (800 gm-cm) = 3,800 gm-cm.

To balance, we need 3,800 gm-cm of moment on the right side.

(500 gm) x (its distance from the pivot) = 3,800 gm-cm

Distance from the pivot = (3,800 gm-cm) / (500 gm) = 7.6 cm

The pivot is at the 50cm mark on the stick. You have to hang the 500gm from 7.6cm to the right of that. The mark at that spot on the stick is (50cm + 7.6cm) = 57.6 cm .

3 0

3 years ago

Read 2 more answers