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3 years ago

Is a force pushing up on an object as gravity is pulling down?

Physics

1 answer:

aliya0001 [1]3 years ago

6 0

Answer:

Logically yes, because Newton's Third law state "When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body."

If force wasn't pushing up then neither gravity is pulling down.

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47. Calculate the final steady temperature obtained when

Rashid [163]

Answer:

the final steady temperature of the mixture is 46.67 ⁰C

Explanation:

Given;

mass of first water,m₁ = 6 g = 0.006 kg

initial temperature, t₁ = 50 ⁰C

mass of the second water, m₂ = 3.0 g = 0.003 kg

initial temperature of the second water, t₂ = 40 ⁰C

specific heat capacity of water, C = 4,200 J/kg.K

Let the final temperature of the mixture = T

Based on Law of conservation of energy, the temperature of the final mixture is calculated as follows;

m₁c(t₁ - T) = m₂c(T - t₂)

0.006 x 4200 x (50 - T) = 0.003 x 4200 x (T - 40)

1,260 - 25.2T = 12.6T - 504

1,260 + 504 = 12.6T + 25.2T

1764 = 37.8T

T = 1764/37.8

T = 46.67 ⁰C

Therefore, the final steady temperature of the mixture is 46.67 ⁰C

4 0

2 years ago

A 20 kg box is being pulled across a floor by a horizontal rope. The tension in the rope is 99 Newtons. The coefficient of frict

fgiga [73]

Answer:

friction = 49 N, a = 2.5 m/s/s

Explanation:

big brain

5 0

3 years ago

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In this experiment, the

tigry1 [53]

Result of the other variable

6 0

2 years ago

A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N,

Serggg [28]

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

$\sum F_x = ma_x\\$

$Fm - Ff = ma_x\\$ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

$Ff = \mu R\\Ff = \mu mg\\$

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

$Fm - Ff = ma_x\\Fm - \mu mg = ma\\$

Substitute the given parameters

$Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a = 1.736m/s^2\\$

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

7 0

2 years ago

Which statements describe the characteristics of a magnet? Select four options.

PolarNik [594]

Answer:

its a, c, d, and f

Explanation:

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2 years ago

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