Which of the following ways of writing 1000w is incorrect?

What property describes the ability of a material to be flattened into thin sheets by hammering? options:

Ivanshal [37]

D. malleability is the ability to bend or form something ? like if something is malleable you can bend it

8 0

2 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:

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

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

2 years ago

How much is 1.75 inches in cm?

saul85 [17]

Answer:

4.44 cm

Explanation:

:)

3 0

3 years ago

Read 2 more answers

An open organ pipe is 1.6m long. If the speed of sound is 343m/s, what are the pipes: a) fundamental , b) 1st overtone , & c

Yakvenalex [24]

Answer:

a) 107.1875 Hz

b) 214.375 Hz

c) 321.5625 Hz

Explanation:

L = length of the open organ pipe = 1.6 m

v = speed of sound = 343 m/s

f = fundamental frequency

fundamental frequency is given as

$f = \frac{v}{2L}$

inserting the values

$f = \frac{343}{2(1.6)}$

$f = 107.1875$ Hz

b)

first overtone is given as

f' = 2f

f' = 2 (107.1875)

f' = 214.375 Hz

c)

first overtone is given as

f'' = 3f

f'' = 3 (107.1875)

f'' = 321.5625 Hz

3 0

3 years ago

A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net w

tatyana61 [14]

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
$v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s$
And so, the kinetic energy of the object is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J$

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
$v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s$
And so the new kinetic energy is
$K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J$

So, the work done on the object is the variation of kinetic energy of the object:
$W=\Delta K=120 J-60 J=60 J$

7 0

2 years ago