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mrs_skeptik [129]
3 years ago
12

You are trying to induce a current and have put together the setup shown. What needs to be added in order for current to flow in

the wire?
Physics
2 answers:
hram777 [196]3 years ago
3 0
The answer is a magnet
weqwewe [10]3 years ago
3 0

Answer:

What needs to be added in order for current to flow in the wire?

"Magnet"

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A 3.00-kg object has a velocity 16.00 i ^ 2 2.00 j ^2 m/s.
trapecia [35]

Answer:

390 J

Explanation:

m = 3 kg

u = 16 i + 2 j

(a) Magnitude of velocity = \sqrt{16^{2}+2^{2}} = 16.1245 m/s

KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J

(b) v = 18 i + 14 j

Magnitude of velocity =  \sqrt{18^{2}+14^{2}} = 22.804 m/s

KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J

According to the work energy theorem

Work done = change in KE = KEf - KEi = 780 - 390 = 390 J

8 0
4 years ago
A population has a birthrate of 10/1000 , a death rate of 9/1000, an immigration rate of 3/1000, and an emigration rate of 7/100
Charra [1.4K]

Birth rate 10/1000 = +population



Death rate 9/1000 = -population



Immigration rate 3/1000 = +population



Emigration rate 7/1000 = -population



10/1000 - 9/1000 + 3/1000 - 7/1000= -3/1000



Growth rate = (-3/1000)(100%)= -0.3%


3 0
4 years ago
There is a 50 g sample of Ra-229. It has a half-life of 4 minutes.
Sloan [31]

Via half-life equation we have:


A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h} }


Where the initial amount is 50 grams, half-life is 4 minutes, and time elapsed is 12 minutes.  By plugging those values in we get:

A_{final}=50(\frac{1}{2})^\frac{12}{4}=50(\frac{1}{2})^{3}=50(\frac{1}{8})=6.25g


There is 6.25 grams left of Ra-229 after 12 minutes.

4 0
4 years ago
how much energy is needed to heat 2kg of cooking oil with a specific heat capacity of 2000j/kg°c from 20°c to 120°c​
mash [69]

Answer:

400kj

Explanation:

h =mass × specific heat capacity × change in temperature

= 2×2000×[120-20]

=2×2000×100

=400000j

=400kj

3 0
3 years ago
two identical waves of amplitude 5cm meet in a large ripple tank what will be the aplitude of the combined wave at a point where
timofeeve [1]

(a) For constructive interference the two waves will have higher amplitude of 10 cm.

(b) For destructive interference the two waves will have zero amplitude.

<h3>Amplitude of the waves for constructive interference </h3>

For constructive interference the two waves will have higher amplitude after constructive interference.

Resulting amplitude = 5 cm + 5cm = 10 cm

<h3>Amplitude of the waves for destructive interference </h3>

For destructive interference the two waves will have zero amplitude after destructive interference.

Resulting amplitude = 5 cm - 5cm = 0

Learn more about amplitude here: brainly.com/question/413740

#SPJ1

4 0
3 years ago
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