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Nuetrik [128]
3 years ago
10

assume that in an unpolished floor a lady us pushing a 40 kg box to the left with an applied force of 192 N. find the sum of for

ce
Physics
1 answer:
Vesnalui [34]3 years ago
5 0
The sum of force is 7,680...............
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_____ help by supporting the plant and holding the leaves up to the light.
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The answer is the stem.
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3 years ago
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Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an
Juli2301 [7.4K]

Answer:

<h2>23.33 kg </h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m  = \frac{42}{1.8}   = 23.3333... \\

We have the final answer as

<h3>23.33 kg</h3>

Hope this helps you

4 0
2 years ago
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A 5kg box is being pulled for a force of 20 n and is sliding with an acceleration of 2 m/s. Find the coefficient of friction
RUDIKE [14]

Answer:

Coefficient of friction = 0.5

Explanation:

Given:

Mass of box = 5 kg

Force applied = 20 N

Acceleration = 2 m/s²

Find:

Coefficient of friction

Computation:

Friction force = Mass x Acceleration.

Friction force =  5 x 2

Friction force = 10 N

Coefficient of friction = Friction force / Force applied

Coefficient of friction = 10 / 20

Coefficient of friction = 0.5

3 0
3 years ago
A satellite omass1000 kg moves in a circular orbit of radius 8000 km round the earth,assumed to be a sphere of radius 6400 km. C
lubasha [3.4K]

Answer:

ΔE = 37.8 x 10^9 J

Explanation:

The energy required will increased the potential energy and increase the kinetic energy.

As the altitude change is fairly small compared to the earth radius, we can ASSUME that the average gravity will be a good representative

Gravity acceleration at altitude would be 9.8(6400²/8000²) = 6.272 m/s²

G(avg) = (9.8 + 6.272)/2 = 8.036 m/s²

ΔPE = mG(avg)Δh = 1000(8.036)(8e6 - 6.4e6) = 12.857e9 J

The centripetal force at orbit must be equal to the gravity force

mv²/R = mg'

v²/8.0e6 = 6.272

v² = (6.272(8.0e6)) = 50.2e6 m²/s²

The maximum velocity when resting on earth at the equator is about 460 m/s.

The change in kinetic energy is

ΔKE = ½m(vf² - vi²)(1000)

ΔKE = ½(1000)(50.2e6 - 460²) = 25e9 J

Total energy increase is

25e9 + 12.857e9 = 37.8e9 J

3 0
3 years ago
Add these measurements, using significant digit rules:
tia_tia [17]

Here we have to add the two measurements given in the question

The measurement values are given as 1.0090 cm and 0.02 cm.we have to  add them on the basis of significant figure rules.

As per the addition rule in terms of significant figures

1-First we have to select the number of significant digits after the decimal point of each quantity.

2-Now we have to remember that during the addition ,the resultant of two quantities will follow the quantity having least number of significant figures after the decimal point.

3-Here we are considering the minimum number of significant figures after the decimal points not the minimum number of significant figures in case of multiplication and division

Now we have to add these two quantities as per the above rule-

         1.0090 cm +0.02 cm

         =1.0290 cm

Here the result  will follow 0.02 which has minimum number of significant figures after the decimal points.

Hence we have to round off the number from 9 of 1.0290

As 9 is  greater than 5 ,so he actual result will be 1.03 cm

       

3 0
3 years ago
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