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yuradex [85]
3 years ago
7

Whats the answer to this question show in the picture 2 questions

Physics
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer: i think it would be watts

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A bicycle accelerates from rest to 6 m/s in 2 s. What is the bicycle's acceleration?
Angelina_Jolie [31]

Answer:

Explanation:I don't say you have to mark my ans as brainliest but if you think it has really helped you plz don't forget to thank me...

4 0
3 years ago
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A transformer with 1000 turns on the primary coil is used to decrease the voltage from 2500 V to 120 V for home use. How many tu
Katyanochek1 [597]

Answer:

The number of turns in the secondary coil is 48.

(B) is correct option.

Explanation:

Given that,

Number of turns on primary coil= 1000

Primary voltage = 2500 V

Secondary voltage = 120 v

We need to calculate the turns  in the secondary coil

Using relation between voltage and number of turns in primary and secondary coil

\dfrac{V_{s}}{V_{p}}=\dfrac{N_{s}}{N_{p}}

Put the value into the formula

\dfrac{120}{2500}=\dfrac{N_{s}}{1000}

N_{s}=\dfrac{120}{2500}\times1000

N_{s}=48\ turns

Hence, The number of turns in the secondary coil is 48.

5 0
4 years ago
An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus a
wlad13 [49]
Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75
fcp = 66.666... = 66 \: newtons

Answer: Last option, 66 N.
7 0
3 years ago
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If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude
Mrrafil [7]

The magnitude of the electric field at this location is 1.5\times 10^{10} N/C

<u>Explanation:</u>

Given

Charge\ of\ the\ particle\ Q=4.3\times 10^-18\ C\\Force\ with\ which\ it\ is\ attracted\ F=6.5\times 10^-8\ N

Electric field at this location determined by the force and charge.

E=F/Q

E=\frac{6.5\times 10^-8}{4.3\times10^-18} =1.5\times\ 10^{10}  \ N/C

7 0
4 years ago
A 300.0-kg speedboat is moving across a lake at 35.0 m/s.
Varvara68 [4.7K]
The answer is 300kg times the 35 m/s10,500 kg•m/s
5 0
3 years ago
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