What force is supplied by a jet’s engines when 7 x 106 Joules of work is required to move the plane down a 450 meter runway?

What happens to the force between the spheres when you increase the mass of one of the spheres?

andrew11 [14]

The force between the spheres increases when the mass increases in one of the spheres.

<u>Explanation:</u>

Newton law of universal gravity extends gravity beyond the earth's surface. This gravity depends directly on the mass of both objects and is inversely proportional to square of distance between their centers.

$\bold{F=\frac{G \times\left(m_{1} \times m_{2}\right)}{\left(r^{2}\right)}}$

Since gravity is directly proportional to “mass of both interacting objects”, stronger objects with greater gravitational force attract. If the mass of one object increases, gravity between them also increases. For example, if an object's mass of one double, force between them also doubles.

4 0

3 years ago

Read 2 more answers

The electric company charges $0.08/kilowatt-hour. If a 2 kilowatt appliance is run for 3hours, how much will you have to pay to

asambeis [7]

Answer:

$ 0.48

Explanation:

We can calculate this quantity easily using successive products and taking into account the units.

$\frac{0.08}{kw*h}*2[kw]*3[hr]\\ \\=0.48$

The amount is $ 0.48

3 0

3 years ago

According to Bernoulli's equation, the pressure in a fluid will tend to decrease if its velocity increases. Assuming that a wind

Pie

Answer:

The pressure drop predicted by Bernoulli's equation for a wind speed of 5 m/s

= 16.125 Pa

Explanation:

The Bernoulli's equation is essentially a law of conservation of energy.

It describes the change in pressure in relation to the changes in kinetic (velocity changes) and potential (elevation changes) energies.

For this question, we assume that the elevation changes are negligible; so, the Bernoulli's equation is reduced to a pressure change term and a change in kinetic energy term.

We also assume that the initial velocity of wind is 0 m/s.

This calculation is presented in the attached images to this solution.

Using the initial conditions of 0.645 Pa pressure drop and a wind speed of 1 m/s, we first calculate the density of our fluid; air.

The density is obtained to be 1.29 kg/m³.

Then, the second part of the question requires us to calculate the pressure drop for a wind speed of 5 m/s.

We then use the same formula, plugging in all the parameters, to calculate the pressure drop to be 16.125 Pa.

Hope this Helps!!!

7 0

3 years ago

A hollow conductor is positively charged. Asmall uncharged metal ball is lowered by a silk thread through asmall opening in the

dimulka [17.4K]

Answer:

Explanation:

According to the property of a conductor, the entire charge will reside on the outer surface of the conductor, there is no charge on the inner side of the conductor. As the uncharged metal ball touches the inner surface of the conductor, it does not attain any charge as the inner side of the conductor has no charge.

So option (c) is correct.

8 0

3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago