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4 years ago

What is magnification

Physics

1 answer:

il63 [147K]4 years ago

5 0

The action or process of magnifying something or being magnified, especially visually. Hope this helped

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HELP PLEASE!!!

user100 [1]

D A little longer than a yard (about 3 ft)

5 0

3 years ago

Find the hiker’s gravitational potential energy if the cliff is 60m high

Furkat [3]

Answer:

Potential energy is $U=mgh$

Explanation:

The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.

Potential energy $U=mgh$

In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.

But we know the values of g and h

$g=9.81m/s^2$

$h=60m$

So, the potential energy

$U=m(9.81m/s^2)(60m)\\\\U=588.6*m$

m is the mass of the hiker, wich is not in the description of the problem.

4 0

3 years ago

What is the acceleration of a 7 kg mass if the force of 70 N is used to move it toward the Earth?

Assoli18 [71]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

m is the mass

f is the force

From the question we have

$a = \frac{70}{7} = 10 \\$

We have the final answer as

Hope this helps you

4 0

3 years ago

A plane flies along a straight line path after taking off, and it ends up 220 km farther east and 100.0 km farther north, relati

love history [14]

Answer:

North east

Explanation:

The distance travelled by a plane from one point to another is the shortest distance between the two points. This distance travelled is usually a straight line path from point 1 to point 2.

Since the plane ends up 220 km farther east and 100.0 km farther north, the direction of flight for the plane is in the North-East direction.

Let x represent the distance travelled by the plane. Hence:

x² = 100² + 220²

x² = 58400

x = 241.66 km

3 0

3 years ago

Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive

dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = $\frac{-kQqi}{r (a+r)}$

c. F = $\frac{-kQqi}{r^{2} }$

Explanation:

a. x components:

$dE = \frac{kdq}{(a+r-x^{2}) } \\$

= $\frac{kQdx}{(a(a+r-x)^2}$

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = $\frac{-kQqi}{r (a+r)}$

c. Given that r>>a, the expression becomes:

F = $\frac{-kQqi}{r^{2} }$

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0

4 years ago