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4 years ago

Find the length of an arc with a radius of 6.0m swept across 2.5 radians.

Physics

1 answer:

soldier1979 [14.2K]4 years ago

7 0

Hello!

In terms of an arc length, radians is the measurement of an arc length in terms of the original radius. This arc length is 2.5 radians, so we multiply it by our original radius.

2.5(6)=15

Therefore, the arc length is 15 meters.

I hope this helps!

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Select the correct answer. Which of the following causes air pollution inside the house? A. Smoking cigarettes B. Growing housep

Troyanec [42]

Answer: Smoking cigarettes

Explanation: The other options don't cause pollution to form in the air besides smoking, from the particles it creates causing harm and damage to your lungs and fill the air with smoke particles.

3 0

2 years ago

Read 2 more answers

Both high-side pressures and low-side pressures are low with the engine running and the selector set to the air-conditioning pos

elena-14-01-66 [18.8K]

Answer:

Technician A only

Explanation:

Both high-side pressures and low-side pressures are low with the engine running and the selector set to the air-conditioning position. Technician A says that the system is undercharged. Technician B says the cooling fan could be inoperative. Which technician is correct?

usually . An overcharged system will result in lower than normal low side pressures

An undercharged system will not enable the compressor to create pressure. As a result of the low amount of refrigerant, the cooling ability is reduced. When we say undercharged, we mean the refrigerant in the system is low, so the both the high side pressures and low side pressures will be low.

8 0

3 years ago

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

3 years ago

330 g of water at 55°c are poured into an 855 g aluminum container with an initial temperature of 10°

Olenka [21]

The final temperature of the system is 32.5°
we know, H = mcT
where, H = Heat content of the body
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry

The net heat remains constant i.e.
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x $\frac{900}{4200}$ x (T - 10)

⇒ 14,850 - 330T = 183.21T - 1832

⇒ - 513.21 T = - 16682

or T = 32.5°

3 0

3 years ago

A hi-lo lifts an 25 N skid to top of a pallet rack. The pallet rack is 3.6 meters tall. The hi-lo takes 12 seconds to get the sk

mojhsa [17]

Answer:

7.5Watts

Explanation:

Given parameters:

Force of lift = 25N

Height = 3.6m

Time = 12s

Unknown:

Power output = ?

Solution:

Power is the rate at which work is done ;

Power = $\frac{force x height }{time}$

Power = $\frac{25 x 3.6}{12}$ = 7.5Watts

8 0

3 years ago