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alex41 [277]
3 years ago
12

Please help!

Physics
1 answer:
tankabanditka [31]3 years ago
6 0

Answer:

35 cm

Explanation:

There's a -0.50N force at 20cm.

There's another -0.50N force at 50cm.

You apply a force of F at a position x.

Sum of the forces:

∑F = ma

F − 0.5 N − 0.5 N = 0

F = 1.0 N

Sum of the moments (or torques):

∑τ = Iα

(-0.50 N) (20 cm) + (-0.50 N) (50 cm) + (1.0 N) (x) = 0

x = 35 cm

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valina [46]

The operating coefficient or performance coefficient of a heat pump is the ratio between the heating or cooling provided and the electricity consumed. The higher coefficients are equivalent to lower operating costs. The coefficient can be greater than 1, because it is a percentage of the output: losses, other than the thermal efficiency ratio: input energy. Mathematically can be written as,

\text{Coefficient of performance} = \frac{\text{Heat extracted}}{\text{Electrical energy supplied}}

COP = \frac{Q}{E}\rightarrow Q =COP\times E

Replacing,

Q = 2.1\times 1.82kJ

Q = 3.822kJ

Therefore the heat is 3.822kJ

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3 years ago
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

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3 years ago
You can comfortably hold your fingers close beside a candle flame, but not very close above the flame. why? challenge (optional)
Inessa05 [86]
The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.
6 0
3 years ago
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What influences the strength of an electric field?
Slav-nsk [51]

Answer:

Explanation:

The charge alters that space, causing any other charged object that enters the space to be affected by this field. The strength of the electric field is dependent upon how charged the object creating the field is and upon the distance of separation from the charged object.

4 0
3 years ago
HELP IM TIMED. ILL MARK BRAINLIEST PLSSSSSS.
Alex17521 [72]

Answer:

According to Newton's Second Law of Motion :    

Where,

F = Force Applied

m = Mass of the object

a = Acceleration

Now, we will use this law to solve this question.

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<u>m = 2.9 kg</u>

5 0
3 years ago
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