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2 years ago

15

Show how multiplying both sides of the acceleration equation a equals F net over in Byam results in the net force equation F net

equals MA

1 answer:

Murrr4er [49]2 years ago

8 0

What are u trying to solve

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** PLEASE PLEASE HELP A student uses the right-hand rule as shown.

guapka [62]

As per right hand thumb rule if we put our thumb of right hand along the current in the wire then curl of fingers will show the direction of magnetic field.

So here as we can see the fingers in the above case is curled such that in front position its showing towards right.

So the direction of magnetic field near the student in front position must be towards Right.

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2 years ago

Read 2 more answers

Which scientist discovered the laws that determine the motion of a spring?

Kruka [31]

Answer: Robert Hooke

Explanation: Hooke's law, law of elasticity discovered by the English scientist Robert Hooke in 1660, which states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load.

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1 year ago

Name 4 element of weather

andriy [413]

Answer:

They are temperature, atmospheric pressure, wind, humidity, precipitation, and cloudiness. Together, these components describe the weather at any given time.

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1 year ago

An object of 4 cm length is placed at a distance of 18 cm in front of a convex mirror of radius of curvature 30 cm. Find the pos

erica [24]

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

$\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm$

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

$\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm$

3 0

2 years ago

A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,

Trava [24]

Given data:

* The mass of the ball is 2 kg.

* The gravitational field strength at the surface of planet X is 5 N/kg.

Solution:

The weight of the ball on the planet X is,

$W=ma$

where m is the mass of ball, a is the gravitational field strength,

Substituting the known values,

$\begin{gathered} W=2\times5 \\ W=10\text{ N} \end{gathered}$

Thus, the weight of the ball on the surface of planet X is 10 N.

3 0

8 months ago