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emmasim [6.3K]
3 years ago
6

How much work is being done if a force of 75 Newtons is used to push a box a distance of 100 meters?

Physics
1 answer:
lukranit [14]3 years ago
5 0
Work= Force x Distance
Answer: 7500 Joules
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A boy whirls a stone in a horizontal circle of radius 1.4 m and at height 1.5 m above ground level. The string breaks, and the s
balandron [24]

Answer:233.23 m/s^2

Explanation:

Given

radius of circle=1.4 m

Height of stone above ground=1.5 m

Horizontal distance(R)=10 m

It is given at the time of break stone flies horizontally thus stone to cover a height of 1.5 m in time t before reaching ground

1.5=0+\frac{gt^2}{2}

t=0.55 s

Initial horizontal velocity at the time of break is given by u

R=u\times t

10=u\times 0.55

u=18.07 m/s

Therefore magnitude of centripetal acceleration is given by

a_c=\frac{u^2}{r}=\frac{18.07^2}{1.4}=233.23 m/s^2

6 0
3 years ago
What is NOT a principle of genetics?
timurjin [86]

The recessive trait will always show up

3 0
3 years ago
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Make the following conversion. 120 mL = _____ cm³ 12.0 1200 120 1.20
omeli [17]
120 is your answer. 120 mL = 120 cm^3
7 0
3 years ago
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What happens to the current supplied by the battery when you add an identical bulb in parallel to the original bulb?
Marysya12 [62]

Answer:

b. The current stays the same.

Explanation:

In the case given current is supplied by the battery to a bulb . Here, we should know that bulb also apply resistance to the flow of current .

Now, when an identical bulb is connected in parallel to the original bulb .

Therefore, both the resistance( bulb) are in parallel.

We know, when two resistance are in parallel , current through them is same and voltage is divided between them.

Therefore, in this case current stays same in the original bulb.

Hence, this is the required solution.

6 0
3 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the lo
Zepler [3.9K]

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

\phi = BAcos \theta

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb

Hence the magnetic flux Φ through the loop is 0.5849Weber

5 0
2 years ago
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