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lukranit [14]
3 years ago
11

An object of weight W is in freefall close to the surface of Earth. The magnitude of the force that the object exerts on Earth i

s _____.
(A) greater than W
(B) less than W
(C) equal to W
(D) zero
(E) cannot be determined without knowing the relative masses of the object and the earth
Physics
2 answers:
Hoochie [10]3 years ago
4 0
Earth is B)equal to W 
atroni [7]3 years ago
4 0

Answer:

Equal to W

Explanation:

It is given that, an object of weight W is in free fall close to the surface of Earth. Weight of an object is equal to the product of its mass and acceleration. On the surface of Earth, the weight is given by :

W=mg

g=9.8\ m/s^2

The magnitude of the force that the object exerts on Earth is equal to W as per the Newton's third law of motion. The force exerted from one object to another object must be equal to and opposite to the force exerted from object 2 to 1.

So, the correct option is (c) "equal to W".

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Gennadij [26K]

Answer:

48 i believe

Explanation:

3 0
4 years ago
Waves : Transfer of Energy
garri49 [273]

✒ Answer

In the case of still lake and ocean water how are they different in transferring energy from one location to another?

  • Answer:Energy is transferred in waves through the vibration of particles

In what direction will you move a rope to create transverse waves?

  • Answer: in the direction of the black arrow

In what direction will you move a slinky to create longitudinal waves?

  • Answer: parallel to the direction that energy is transported.

7 0
2 years ago
An object weighing 1.840 kg has a volume of 0.0015 m3. What is the density of the object in g/cm3?
olga_2 [115]

Answer:

1.22gcm³

Explanation:

D = mass/ volume

Mass=1.840kg = 1,840g

1000g = 1kg

1.840kg= x(g)

X(g) = 1.840/1000

= 1840g

Volume = 0.0015m³= 1,500cm³

1m³= 1000,000cm³.

0.0015m³= x(cm³)

X(cm³) = 1000,000×0.0015

X(cm³)= 1500.

Since density is mass/volume, now impute your data's

D=m/v

D=1840/1500

1.22g/cm³

5 0
3 years ago
A step index fiber has a numerical aperture of NA = 0.1. The refractive index of its cladding is 1.465. What is the largest core
Vladimir [108]

Answer:

diameter = 9.951 × 10^{-6} m

Explanation:

given data

NA = 0.1

refractive index = 1.465

wavelength = 1.3 μm

to find out

What is the largest core diameter for which the fiber remains single-mode

solution

we know that for single mode v number is

V ≤ 2.405

and v = \frac{2*\pi *r}{ wavelength} NA

here r is radius    

so we can say

\frac{2*\pi *r}{ wavelength} NA    = 2.405

put here value

\frac{2*\pi *r}{1.3*10^{-6}} 0.1    = 2.405

solve it we get r

r = 4.975979 × x^{-6} m

so diameter is = 2  ×  4.975979 × 10^{-6} m

diameter = 9.951 × 10^{-6} m

3 0
3 years ago
You pull a rope to the left with 300 N and a friend pulls the rope to the right with 425 N
eimsori [14]
I’m assuming you’re supposed to calculate the resultant force?

425N (right) -300N (left)
=125 N to the right
6 0
3 years ago
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