Ask question

Ask question

All categories

vampirchik [111]

3 years ago

8

A computer mouse (.234 kg) and a full coffee mug (.647 g) are sitting 42.0 cm apart on a desk. Find the gravitational force betw

een these objects.

1 answer:

shutvik [7]3 years ago

7 0

Answer:

euphoria

Explanation:

because euphoria

You might be interested in

To completely describe the motion of an object, you need

notsponge [240]

Answer:

D

Explanation:

7 0

3 years ago

An electron with a speed of 0.95c is emitted by a supernova, where cc is the speed of light. What is the magnitude of the moment

krok68 [10]

Answer:

2.59×10¯²² Kgm/s

Explanation:

Data obtained from the question include:

Velocity of electron = 0.95c

Momentum =?

Next, we shall determine the velocity of the electron. This can be obtained as follow:

Velocity of electron = 0.95c

Velocity of Light (c) = 3×10⁸ m/s

Velocity of electron = 0.95c

Velocity of electron = 0.95 × 3×10⁸

Velocity of electron = 2.85×10⁸ m/s

Finally, we shall determine the mometum of the electron.

Momentum is simply defined as the product of mass and velocity. Mathematically, it is expressed as:

Momentum = mass x Velocity

Thus, with the above formula, we calculate the momentum of the electron as follow:

Mass of electron = 9.1×10¯³¹ Kg

Velocity of electron = 2.85×10⁸ m/s

Momentum of electron =?

Momentum = mass x Velocity

Momentum = 9.1×10¯³¹ × 2.85×10⁸

Momentum = 2.59×10¯²² Kgm/s

Therefore, the momentum of the electron is 2.59×10¯²² Kgm/s

3 0

3 years ago

Waves on a pond are an example of which kind of wave?

KATRIN_1 [288]

<span>
</span><span>Waves on a pond are an example of which kind of wave?

</span>B. surface waves

6 0

4 years ago

U1=20 m/s turn it to km/h

notka56 [123]

It is 72 km/h
I hope it helps

7 0

3 years ago

Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm

Burka [1]

Answer:

The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

Explanation:

Given that,

Atmospheric pressure $P_{atm}= 1.013\times10^{5}\ Pa$

drop height h'= 27.1 mm

Density of mercury $\rho= 13.59 g/cm^3$

We need to calculate the height

Using formula of pressure

$p = \rho g h$

Put the value into the formula

$1.013\times10^{5}=13.59\times10^{3}\times9.8\times h$

$h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}$

$h=0.76\ m$

We need to calculate the new height

$h''=h - h'$

$h''=0.76-27.1\times10^{-3}$

$h''=0.76-0.027$

$h''=0.733\ m$

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

$P=\rho g h$

Put the value into the formula

$P= 13.59\times10^{3}\times9.8\times0.733$

$P=0.97622\times10^{5}\ Pa$

Hence, The atmospheric pressure is $0.97622\times10^{5}\ Pa$ .

7 0

3 years ago