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2 years ago

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Physics

1 answer:

pishuonlain [190]2 years ago

5 0

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A capacitor with C = 6.00 μF is fully charged by connecting it to a battery that has emf 50.0 V. The capacitor is disconnected f

Arte-miy333 [17]

Answer:

1.99×10^-4coulombs

Explanation:

The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor(C) is the proportionality constant. Mathematically, Q = CV

If V is the voltage across the resistor, V = IR (according to ohm's law) where I is the current in the resistor and R is the resistance.

We need to calculate the voltage on the resistor first when 0.18A current is passed through it.

V = 0.18 × 185

V = 33.3Volts

The charge Q on the resistor will be;

Q = CV

Were C = 6.00 μF, V = 33.3

Q= 6×10^-6 ×33.3

Q = 0.0001998

Q= 1.99×10^-4Coulombs

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2 years ago

Which equation is most likely used to determine the acceleration from a velocity vs:time graph?

tresset_1 [31]

Acceleration, a = (v - u)/t

where v is the final velocity, u is the initial velocity, and t is the time.

This formula on a velocity time graph represents the slope of the graph.

7 0

2 years ago

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An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

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2 years ago

A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow

leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

$F=ma$

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

$a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}$

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

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2 years ago

A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force

boyakko [2]

Answer:

Explanation:

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1 year ago

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