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3 years ago

When no friction is involved,what can and outside object do to a moving object?

1 answer:

3 0

It will continue in the same direction. I hope this helps my friend. If you would like more details look at Newton's law of motion.

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Sharks also swim in waters off the coast of Nova Scotia. When gray seals and harbor seals see sharks nearby, they must swim away

Vlada [557]

No. The seals are responding to a stimulus.

6 0

3 years ago

You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror

joja [24]

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y = 30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

tan( 90 - θ) = 30.1 / 30.7

90 - θ = tan^-1 (30.1/30.7)

90 - θ = 44.4345

θ = 45.56 º

the angle of incidence θ is 45.56 º

4 0

3 years ago

How many joules of work are done on an object when a force of 10 N pushes it 5 m?

zhenek [66]

Answer:

option C

Explanation:

given,

Force on the object = 10 N

distance of push = 5 m

Work done = ?

we know,

work done is equal to Force into displacement.

W = F . s

W = 10 x 5

W = 50 J

Work done by the object when 10 N force is applied is equal to 50 J

Hence, the correct answer is option C

5 0

3 years ago

What is the acceleration of a Porsche that can go from 15 mi/hr to 75 mi/hr in 4 seconds?

Elodia [21]

Hi there!

Acceleration = change in velocity / change in time = Δv/Δt

Thus:

a = (75 - 15)/4 = 60/4 = 15 mi/hr²

8 0

2 years ago

Read 2 more answers

Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

We can solve this first part of the problem by applying the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0

3 years ago