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aalyn [17]
3 years ago
11

A coil with 55 loops has its flux change from 0 Wb to 0.0266 Wb in a certain amount of time, generating 5.22 V of EMF. How much

time elapsed?
Physics
1 answer:
Eduardwww [97]3 years ago
4 0

Answer:

0.280 s

Explanation:

I set it up as 5.22=(55)(0.0266)/x and then solved for x to be 2.80.

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GIVING BRAINLIEST PLEASE HELP ME!!!
kumpel [21]

Answer:

I'm Pretty sure the answer your looking for is C

4 0
3 years ago
A bike accelerates uniformly(from rest to a speed of 7 m/s over a distance of 40 m. Determine
marshall27 [118]

Answer:

0.61 m/s^2

Explanation:

The bike's acceleration can be found by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity of the bike

u is the initial velocity

a is the acceleration

s is the distance covered

For the bike in the problem,

u = 0

v = 7 m/s

d = 40 m

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{7^2-0}{2(40)}=0.61 m/s^2

3 0
3 years ago
1) Displacement is to velocity as ____________ is to acceleration. A) direction B) speed C) time D) velocity
vivado [14]

Answer: C) time

Explanation:

7 0
3 years ago
A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails
harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

y=h+ut+\frac{1}{2}gt^2

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

v^2-u^2 = 2ad

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

a=\frac{v-u}{t}

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0
3 years ago
A motorcycle is travelling at a constant velocity of 30ms. The motor is in high gear and emits a hum of 700Hz. The speed of soun
timurjin [86]

Answer:

a) T=1.43\times 10^{-3}\ s

b) d=0.0429\ m

c) \lambda=0.4857\ m

d) f_o=767.7\ Hz

Explanation:

Given:

  • velocity of the sound from the source, S=340\ m.s^{-1}
  • original frequency of sound from the source, f_s=700\ Hz
  • speed of the source, v_s=30\ m.s^{-1}

(a)

We know time period is inverse of frequency:

Mathematically:

T=\frac{1}{f}

T=\frac{1}{700}

T=1.43\times 10^{-3}\ s

(b)

Distance travelled by the motorcycle during one period of sound oscillation:

d=v_s.T

d=30\times 1.43\times 10^{-3}

d=0.0429\ m

(c)

The distance travelled by the sound during the period of one oscillation is its wavelength.

\lambda=\frac{S}{f}

\lambda=\frac{340}{700}

\lambda=0.4857\ m

(d)

observer frequency with respect to a stationary observer:

<u>According to the Doppler's effect:</u>

\frac{f_o}{f_s}= \frac{S+v_o}{S-v_s} ...........................(1)

where:

f_o\ \&\ v_o are the observed frequency and the velocity of observer respectively.

Here, observer is stationary.

\therefore v_o=0\ m.s^{-1}

Now, putting values in eq. (1)

\frac{f_o}{700}= \frac{340+0}{340-30}

f_o=767.7\ Hz

5 0
3 years ago
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