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qwelly [4]
3 years ago
8

Carter pushes a bag full of basketball jerseys across the gym floor. The he pushes with a constant force of 21 newtons. If he pu

shes the bag 9 meters in 3 seconds, how much power does he use? (Hint: 1 watt = 1 .) A. 3 watts B. 7 watts C. 27 watts D. 63 watts E. 189 watts
Physics
2 answers:
N76 [4]3 years ago
5 0

Work = (force) x (distance)

Work = (21 Newtons) x (9 meters) = 189 Joules

Power = (work done) / (time to do the work)

Power = (189 Joules) / (3 sec)

Power = (63 watts/sec)

<em>Power = 63 watts  </em>(D)


vovangra [49]3 years ago
4 0

Answer:

DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

Explanation:

DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD    IIIIIIIIIIIIIIISSSSSSSSSSSSSSSSSSS       TTTTTTHHHHHHHHEEEEEEEE AAAAAAANSWER

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A quick USB charger claims its output current is 1.97Amp. We know that the standard USB output voltage is 5V. What is the output
Arte-miy333 [17]

Answer:

Output power of the charger is 9.85 watts.

Explanation:

It is given that,

Output current of the USB charger, I = 1.97 A

The standard USB output voltage, V = 5 V

We need to find the output power of the charger. It can be determined using the following formula as :

P = V × I

P=5\ V\times 1.97\ A

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3 years ago
A ‘thermal tap’ used in a certain apparatus consists of a silica rod which
abruzzese [7]

Correct question is;

A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).

Answer:

ΔT = 268.67K

Explanation:

We are given;

d1 = 8mm

d2 = 1mm

At standard temperature and pressure conditions, the temperature is 273K.

Thus; Initial temperature; T1 = 273K,

Using the combined gas law, we have;

P1×V1/T1 = P2×V2/T2

The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:

V1/T1 = V2/T2

Now, volume of the tube is given by the formula;V = Area × height = Ah

Thus;

V1 = (πd1²/4)h

V2 = (π(d2)²/4)h

Thus;

(πd1²/4)h/T1 = (π(d2)²/4)h/T2

π, h and 4 will cancel out to give;

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T2 = ((d2)² × T1)/d1²

T2 = (1² × T1)/8²

T2 = 273/64

T2 = 4.23K

Therefore, Change in temperature is; ΔT = T2 - T1

ΔT = 273 - 4.23

ΔT = 268.67K

Thus, the temperature decreased to 268.67K

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