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3 years ago

Need only group b 1st question

Physics

1 answer:

galina1969 [7]3 years ago

7 0

Answer:

49 m/s

Explanation:

This is the question from the document:

1. A bag dropped from a helicopter falls with an acceleration of 9.8m/s2. What is its velocity after 5s?

We can find the answer by using the equation a = (v-u) / t

Acceleration a = 9.81m/s2

initial velocity u = 0 m/s (the bag is only dropped, so there's no initial velocity already existing)

time t= 5s

final velocity v = unknown

9.8 = (v -0) / 5

v - 0 = 9.8x5

v = 49 m/s

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When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

When the magnetic flux through a single loop of wire increases by 30 Tm^2 , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of 2.5 ohms , (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

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3 years ago

Half Time At 3:00 the hour hand and the minute hand of a clock point in directions that are 90.0

jenyasd209 [6]

Answer:

3 : 08 : 10.9

Explanation:

assuming a 12 hour clock

angular velocity of the hour hand is 2π/(3600(12)) rad/s

angular velocity of the minute hand is 2π/3600 rad/s

difference is 2π/3600 - 2π/(3600(12)) = 11(2π/(3600(12)) rad/s

45° = π/4 radians

This angle is covered in a time of

π/4 rad / 22π / (3600(12)) = 900(12) / 22 = 490.909090... s

or about 8 minutes 10.9 s

ANSWER 3:08:10.9

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Answer:

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Explanation:

APEX verified

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g100num [7]

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Explanation:

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leva [86]

Answer:

3,1,4,2

Explanation:

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