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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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A wire is made from a material having a temperature coefficient of resistivity of 0.0003125 (°C^-1). In an experiment, we mainta
beks73 [17]

Answer:

The power decreases by 36%

Explanation:

Given:

At 20° C

Power, P₀ = 300 W

Potential difference, V = 150 volts

Now, power is given as

P = V²/R

where, R is the resistance

on substituting the values, we get

300 = 150²/R₀

or

R₀ = 75 Ω

Now, the variation of resistance with temperature is given as

R = R₀[1 + α(T - T₀)]

where, α is the temperature coefficient of resistivity = 0.0003125 (°C⁻¹)

now, at

T₀ = 20° C

R₀ = 75 Ω

for

T = 1820° C

we have

R = R₀[1 + α(T - T₀)]

substituting the values

we get

R = 75×[1 + 0.0003125 × (1820 - 20)]

or

R = 117.18 Ω

Now using the formula for power

We have,

P = V²/R

or

P = 150²/117.18 = 192 W

Therefore, the percentage change will be

= \frac{P-P_0}{P_0}\times 100

on substituting the values , we get

= \frac{192-300}{300}\times 100

= -36%

here, negative sign depicts the decrease in power

3 0
3 years ago
Read 2 more answers
Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.
Effectus [21]

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, \lambda=608\ nm=608\times 10^{-9}\ m

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

y=\dfrac{mD\lambda}{a}

where

a = width of the slit

a=\dfrac{mD\lambda}{y}

a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}

a = 0.000167 m

a=1.67\times 10^{-4}\ m

a = 0.167 mm

So, the width of the slit is 0.167 mm. Hence, this is the required solution.

7 0
3 years ago
use the distributive property to express 16+48. please explain if you can or work it out so I can see how you got your answer. t
Rom4ik [11]
(40+10)+(8+6)
Hope I helped you
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Which one of the following weight management plans is the most effective
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Which of the following?
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The sun's _____ is the layer that emits visible light.
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The Photosphere is the right answer
7 0
3 years ago
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