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AURORKA [14]
3 years ago
10

A 140-kg box is being lowered at constant velocity. A rope is attached to the top of the box, and Jimmy is slightly supporting t

he box from the bottom. If Jimmy is applying 50 N of force to the box, how much force is the rope applying?
Physics
1 answer:
Zinaida [17]3 years ago
4 0

Answer:

the anwser is 90

Explanation:

because 140-50=90

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If a board with an area of 3m2 has a 12 N force exerted on it, what is the pressure on the board?
Cerrena [4.2K]
4 N/cm^2 is the answer. the way you get this is the formula for pressure is P=F/A. So you would plug in the numbers which would make the equation P=12/3 which you know equals 4. The units for pressure is N/cm^2 or N/m^2

I hope this helps 
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Which of kepler’s laws explains why the sun has a slightly larger angular diameter in january than in july?
Bond [772]

Kepler’s three law is the answer. Kepler’s 3 is the amount of time it takes to orbit the sun is related to size and distance.  Kepler’s 3 is one of the planetary motion and can be stated as all planets move in elliptical orbits, having the sun sits at one of the foci.

7 0
3 years ago
the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f
vesna_86 [32]

Answer:

6.63\times 10^8\ N/C

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

\dfrac{E}{B}=c\\\\E=Bc

Put all the values,

E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C

So, the magnitude of the electric field is equal to 6.63\times 10^8\ N/C.

7 0
3 years ago
The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the
elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

3 0
3 years ago
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