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3 years ago

PLEASE HELP I NEED IT RIGHT NOW

Physics

2 answers:

morpeh [17]3 years ago

7 0

The answer should be B :)

hodyreva [135]3 years ago

3 0

The choice is B the volume never needs to change. Plus got answer on test and Ace it.

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If a bike is traveling at 12 km/hr for 20 minutes how far will it travel

geniusboy [140]

1) 20 min = 1/3 h
2) distance = speed * time = 12 * 1/3 = 4 km

3 0

3 years ago

A circular conducting loop with a radius of 0.10 m and a small gap filled with a 10.0 ȍresistor is oriented in the xy-plane. If

lidiya [134]

Answer:

The magnitude of the current is 5.45 mA.

Explanation:

Given that,

Resistance = 10.0 ohm

Radius = 0.10 m

Magnetic field = 1.0 T

Angle = 30°

Increase magnetic field = 7.0 T

Time t = 3.0 s

Number of turns = 1

We need to calculate the initial flux

Using formula of flux

$\phi=NB_{1}A\cos\theta$

Put the value into the formula

$\phi=1\times1.0\times\pi\times(0.10)^2\times\cos30^{\circ}$

$\phi=1\times1.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}$

$\phi=0.027\ wb$

We need to calculate the final flux

$\phi=1\times7.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}$

$\phi=0.1904\ wb$

We need to calculate the induced emf

Using formula of emf

$\epsilon=\dfrac{\phi_{f}-\phi_{i}}{t}$

Put the value into the formula

$\epsilon=\dfrac{0.1904-0.027}{3.0}$

$\epsilon=0.0545\ V$

We need to calculate the current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.0545}{10.0}$

$I=5.45\ mA$

Hence, The magnitude of the current is 5.45 mA.

4 0

3 years ago

A student launches a small 0.5 kg rocket with an initial speed of 30 m/s at an angle of 60°. Approximately how much time will th

Elena L [17]

Answer:

$t=5.30s$

Explanation:

The high reached by a proyectile in an uniformly accelerated motion is given by:

$y=v_{0y}t-\frac{gt^2}{2}$

The time that the rocket spends in the air is obtained for y = 0, since this is the time that the rocket travels before touching the ground. Recall that $v_{0y}=v_0sin\theta$ . Solving for t:

$0=(v_0sin\theta) t-\frac{gt^2}{2}\\\frac{gt}{2}=v_0sin\theta\\t=\frac{2v_0sin\theta}{g}\\t=\frac{2(30\frac{m}{s})sin(60^\circ)}{9.8\frac{m}{s^2}}\\t=5.30s$