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lord [1]
3 years ago
7

Calculate the gravitational force between a 10 kg ball and 20 kg ball

Physics
1 answer:
elena-s [515]3 years ago
7 0

Answer:

5.34\cdot 10^{-10} N

Explanation:

The gravitational force is an attractive force exerted between any objects with mass.

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the two objects

In this problem, we have:

m_1 = 10 kg is the mass of the 1st ball

m_2 = 20 kg is the mass of the 2nd ball

r = 5 m is the separation

So, the gravitational force between them is

F=(6.67\cdot 10^{-11})\frac{(10)(20)}{5^2}=5.34\cdot 10^{-10} N

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Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel
krek1111 [17]

Answer:

The correct answer is <u>option (A) that is KEA > KEB .</u>

Explanation:

Let us calculate -

If the object is straighten up and inclined plane , the work done is

W=F_d- F_f_r_id-F_gh

W=F_d-\mu_kmgdcos\theta-mgdsin\theta

The change in kinetic energy is ,

   \Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2

At the top of the inclined plane , the velocity is zero

So,

\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2

\Delta KE=-\frac{1}{2}m\nu_0^2

From the work energy theorem , we have W=-\Delta K in case of friction , so

\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta

KE=Fd-\mu_kmgdcos\theta-mgdsin\theta

For object A-

KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta

For object B

KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta

KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)

Thus , larger mass is going to mean less total work and a lower kinetic energy .

From the above results , we get

KE_A >KE_B

<u>Therefore , option A is correct .</u>

6 0
3 years ago
Describe three electrical appliances that use circuits to convert electrical energy into other forms of energy.
marishachu [46]
We see that around us there are many things that makes use of circuits to make our lives easier Circuits are used to convert electrical energy to other forms of energy. For instance, a heater contains circuits which converts the electrical energy into heat. A circuit in a fan converts the electrical energy into motion. In a doorbell, the circuits are used to convert electrical energy into sound.<span />
3 0
3 years ago
In a local bar, a customer slides an empty beer mug down the counter for a refill. the height of the counter is 1.15 m. the mug
yarga [219]

<span>A.    </span>Let’s say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the components are v0x and v0y. 
<span>v0y = 0 since the customer slides it horizontally so applied force is in the x component only.

<span>The equations for horizontal and vertical projectile motion are:
x = x0 + v0x t 
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>

Setting the origin to be the end corner of the counter so that x0=0 and y0=0, hence:

x = v0x t

y = - 1/2 g t^2 

Given value are: x=1.50m and y=-1.15m (y is negative since mug is going down)

<span>1.50m = v0x t    ---->  v0x= 1.50/t</span>

<span>-1.15m = -(1/2) (9.81) t^2    -----> t =0.4842 s</span>

Calculating for v0x:

v0x = 3.10 m/s

<span>B.    </span>v0x is constant since there are no other horizontal forces so, v0x=vx=3.10m/s

vy can be calculated from the formula:

<span>vy = v0y + at         where a=-g (negative since going down)</span>

vy = -gt = -9.81 (0.4842)

vy = -4.75 m/s

 

Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy

tan(90-Ø )= 3.1/4.75

Ø = 56.87˚<span> below the horizontal</span>

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Artist 52 [7]
8.1577297 Kilograms :)
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When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting
Mrrafil [7]

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

3 0
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