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3 years ago

A 2.1W iPod is used for 30 minutes. How much energy does it use? (3

Physics

2 answers:

dusya [7]3 years ago

8 0

Answer:

Energy used, E = 3780 Joules

Explanation:

It is given that,

Power, P = 2.1 W

Time, t = 30 minutes = 1800 seconds

We need to find the energy used by the iPod. The product of power and time is called energy used i.e.

E = P × t

$E=2.1\ W\times 1800\ s$

E = 3780 Joules.

So, the energy used by the iPod is 3780 Joules. Hence, this is the required solution.

shutvik [7]3 years ago

3 0

It us 200 energy if it the right awnser

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You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.6

AveGali [126]

Answer:

0.01

Explanation:

Given the data:

10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90

True value = 9.81

Mean value :

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Sample size, n = 9

(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9

= 88.69 / 9

= 9.854

Standard deviation (σ) :

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[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9

Sqrt(0.113824 / 9)

Sqrt(0.0126471)

σ = 0.1124593

Standard Error = σ / sqrt(n)

Standard Error = 0.1124593 / 9

Standard Error = 0.0124954

Standard Error = 0.01 ( 1 significant digit)

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3 years ago

according to information received from radio telescopes, where is our sun positioned in the milky way? a. at the exact center of

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Within one of the spiral arms

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3 years ago

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One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t

Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2 $\frac{m}{s^{2} }$

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

$K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=F$ Δx= $(0.400m)\\$

W_{ext}=20.4 Nm

$U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2} \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}$

To determine, the acceleration we solve the following equation for a:

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4 years ago

How much Power does a truck use to complete 4,100 J of Work in 30<br> seconds? *

alexandr1967 [171]

Answer:

136.67 watts

Explanation:

P=W/t

P=4100/30

P= 136.67 watts

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3 years ago