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12345 [234]
3 years ago
11

Why is the outer surface of a teapot slightly polished

Physics
1 answer:
evablogger [386]3 years ago
6 0

Answer:

Well teapots are metal. If it were plastic, putting it on top of an oven it would melt. And metals are usually shiny.

Explanation:

If you have any questions feel free to ask in the comments - Mark

Also when you have the chance please mark me brainliest.

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Two gliders collide on a frictionless air track that is aligned along the x axis. Glider A has an initial velocity of +4.0 m/s a
DedPeter [7]

Answer:

As collision is elastic,thus we can use conservation of momentum equation

mA=0.2 kg

(vB)1=0 m/s.......................as it is on rest before collision

(vA)1=4 m/s

(vA)2=-1 m/s

(vB)2=2 m/s

using equation

(mA*vA+mB*vB)1= (mA*vA+mB*vB)2

Where 1 and 2 represents before and after collision

(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)

0.8=-0.2+(2mB)

mass of object B=mB=0.3 Kg

6 0
3 years ago
What is the inverse square law and how does it relate to gravity?
Nesterboy [21]

Answer:

Inverse Square Law Newton proposed the Inverse Square Law. The effect of gravity (and also on forces such as sunlight) works like this. If say we have a half-mass Earth, it would produce a gravity of not half but a quarter (the square of 2).

7 0
3 years ago
A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
1. When you have different masses for each sphere, how does the force that the larger mass sphere exerts on the smaller mass sph
aleksandrvk [35]

1) The forces are equal (Newton's third law of motion)

2) The force between the spheres will quadruple

3) The force of gravity exerted by the notebook on you is negligible

Explanation:

1)

In this part of the problem, we want to compare the gravitational force exerted by the larger mass sphere on the smaller mass sphere to the force exerted by the smaller mass sphere to the larger mass sphere.

We can do this by using Newton's third law of motion, which states that:

<em>"When an object A exerts a force (called </em><em>action</em><em>) on an object B, then object B exerts an equal and opposite force (called </em><em>reaction</em><em>) on object A"</em>

In this problem, we can identify the larger mass sphere as object A and the smaller mass sphere as object B. This law tells us that the two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the larger mass sphere on the smaller mass sphere is equal to the force exerted by the smaller mass sphere to the larger mass sphere.

2)

The magnitude of the gravitational force between the two spheres is given by

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

where

G is the gravitational constant

m_1, m_2 are the masses of the two spheres

r is the separation between the two spheres

In this problem, we are asked to find what happens when the distance between the spheres is halved, therefore when the new distance is

r'=\frac{r}{2}

Substituting into the equation, we find

F'=G\frac{m_1 m_2}{r'^2}=G\frac{m_1 m_2}{(r/2)^2}=4(\frac{Gm_1 m_2}{r^2})=4F

So, the force between the two spheres will quadruple.

3)

We can give an estimate for the gravitational force exerted by your notebook on you.

As we said, the magnitude of the gravitational force is

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

Where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

Let's estimate the following:

m_1 = 60 kg is your mass

m_2 = 2 kg is the mass of the notebook

r=1 m, assuming the notebook is at 1 metre from you

Substituting,

F=(6.67\cdot 10^{-11})\frac{(60)(2)}{1^2}=8.0\cdot 10^{-9} N

We see that this force has an extremely small value: therefore, it is almost negligible in daily life, where other much stronger forces act on you.

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

8 0
3 years ago
Jenny pushes a 40 N crate down the hall 2m. How much work did she do?
Shalnov [3]
Work done = force x distance = 40 x 2 = 80 Joules.
8 0
3 years ago
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