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Reptile [31]
3 years ago
14

An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?

Physics
1 answer:
hjlf3 years ago
8 0

Answer:

I= 3.5 amps

Explanation:

Step one:

given data

rating of resistor R= 8 ohms

power P= 100W

Required

The current I

Step two

Yet this power is also given by

P = I^2R

make I subject of the formula we have

I= \sqrt{\frac{P}{R} }

substitute

I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps

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podryga [215]

Answer:

C.) Sled Team C 28 kg moving at 12m/s

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7 0
3 years ago
Students in a chemistry class added 5g of zinc (Zn) to 50g of hydrochloric acid (HCl). A chemical reaction occurred that produce
PilotLPTM [1.2K]

Answer:

10. 36 g ZnCl2

Explanation:

Zn + 2HCl  -> ZnCl2 + H2

0.076 mol Zn

1.37 mol HCl

3 mol H2

Limiting reactant: Zn

1 mol Zn        -> 1 mol ZnCl2

0.076 mol Zn  ->x                         x= 0.076 mol ZnCl2=10.36 g

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3 years ago
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Genrish500 [490]

Answer:

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Explanation:

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3 0
3 years ago
Bob is pulling a 30 kg filing cabinet with a force of 200 N , but the filing cabinet refuses to move. The coefficient of static
Arada [10]

Answer:

<em>Magnitude of the Frictional force is 200 N</em>

Explanation:

The frictional force is the force that tries to oppose relative motion between two surfaces that are contacting. The coefficient of static friction is the coefficient of friction of a body that is not moving.

Newton's third law of motion states that action and reaction forces are equal and opposite. So the frictional force felt on the filing cabinet will be equal to the applied force pulling the cabinet.

Frictional force = Force applied

Force applied  = 200 N

Therefore, the magnitude of the friction force on the filing cabinet is 200 N

5 0
3 years ago
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
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Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
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