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2 years ago

The two ends of a coiled wire are connected to the electrodes of a lightbulb.

Physics

1 answer:

barxatty [35]2 years ago

5 0

Answer:

Explanation:

The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the loop, creating an electric current.

in other words, the magnets motion creates a current in the loop

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In a chemical equation, where do the reactants appear?

Bess [88]

Answer: A chemical equation describes a chemical reaction. Reactants are starting materials and are written on the left-hand side of the equation. Products are the end-result of the reaction and are written on the right-hand side of the equation.

Explanation:

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2 years ago

Figure 1 shows the kinetic energy as a function of time for a 2kg object that is released from rest and falls toward Earth’s sur

garri49 [273]

Answer:

The answer is 1400 J, according to my Physics teacher.

Explanation:

You need to take into account everything that is listed in the question; it's important to remember that the question is asking about the change in gravitational potential energy of the object-object-Earth system from 0s to 10s, not 0s to 20s. :)

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2 years ago

Why must humans limit their exposure to X-rays and gamma rays?

Anastasy [175]

Answer:

The US Environmental Protection Agency (EPA) sets limits for exposure to x-rays and gamma rays in part because it recognizes that this form of radiation can cause cancer.

4 0

2 years ago

Read 2 more answers

Two teams are playing tug-of-war. The team on the right is pulling with 4320 N. The team on the left is pulling with 4380 N. Whi

zmey [24]

60 N to the left because 4380-4320is 60

6 0

2 years ago

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human i

Varvara68 [4.7K]

Answer:

$0.0018833\ \text{m/s}$

Explanation:

$d$ = Distance of Andromeda Galaxy from Earth = $2.54\times 10^7\ \text{ly}$

$t$ = Time taken = $90\ \text{years}$

$c$ = Speed of light = $3\times 10^8\ \text{m/s}$

We have the relation

$t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}$

$c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}$

The required answer is $0.0018833\ \text{m/s}$ .

7 0

2 years ago