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musickatia [10]
2 years ago
13

What do (3.9 x 10^2)(4.1 x 10^4) equal

Physics
1 answer:
pantera1 [17]2 years ago
8 0

Answer:

Explanation:(3.9x100)(4.1x10,000)                

390x41,000

15,990,000

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A toy cart at the end of a string 0.70 m long moves in a circle on a table. The cart has a mass of 2.0 kg and the string has a b
luda_lava [24]
Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a 
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8 0
3 years ago
A student makes the following statement
Aliun [14]

Answer:

The current is not used up. The electrons flow through the entire circuit and this travel is the current. They flow until they are not charged anymore. That is also why the circuit must be closed or else electrons would escape not just light it up for a second then go out.

Explanation:

8 0
3 years ago
The table below shows the distance from the coast of four locations in Florida.
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6 0
3 years ago
If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?
insens350 [35]
Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
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3 0
3 years ago
In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie
Evgesh-ka [11]

Answer:

8.4 V

Explanation:

induced emf, e1 = 5.8 V

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magnetic field, B2 = 0.55 T

induced emf, e2 = ?

As we know that the induced emf is directly proportional to the magnetic field strength.

When the other parameters remains constant then

\frac{e_{1}}{e_{2}}=\frac{B_{1}}{B_{2}}

\frac{5.8}{e_{2}}=\frac{0.38}{0.55}

e2 = 8.4 V

Thus, the induced emf is 8.4 V.

4 0
2 years ago
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