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Hunter-Best [27]
3 years ago
15

What is the acceleration of a car at rest that speeds up to 75m/s over 25 seconds?

Physics
1 answer:
Yuki888 [10]3 years ago
6 0
It’s a because I just did it
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Electrons moving back and forth through a circuit would be
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The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.

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4 years ago
A student does 25 J of work on the handle of a pencil sharpener. If the pencil sharpener does 20 J of work on the pencil, what i
allsm [11]

Answer:

80%

Explanation:

Efficiency of machine = work output/work input  ×100 %

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4 years ago
Read 2 more answers
Star a has an absolute magnitude of 8. what is its apparent magnitude at a distance of 100 pc
tresset_1 [31]
From the definition of apparent magnitude, we know that:
m_{1} - m_{2} = -2.5Log( \frac{F_{1}}{F_{2}} )
where:
m = apparent magnitude
F = corresponding flux

We also know that the flux is given by the formula:
F =  \frac{L}{4 \pi  d^{2} }
where:
L = luminosity
d = distance

Therefore:
\frac{F_{1} }{F_{2}} = (\frac{L_{1} }{4 \pi d_{1}^{2} })(\frac{4 \pi d_{2}^{2} }{L_{2} }) \\ =  \frac{L_{1}d_{2}^{2}}{L_{2}d_{1}^{2}}

Now, let's apply these formulae to the same star (therefore, same luminosity), using apparent magnitude and absolute magnitude (which is defined as the apparent magnitude the star would have if it were at a distance of 10pc):
m - M = 2.5 Log ( \frac{d}{10pc})^{2}

Now, let's solve for m:
m = M + 2.5 Log ( \frac{d}{10})^{2}
= <span>8 + 2.5 Log ( \frac{100}{10})^{2}</span>
= 13

Hence, the apparent magnitude of the star would be m = +13
3 0
4 years ago
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