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2 years ago

Two students, Jenny and Cho, are investigating motion.

Physics

1 answer:

IgorLugansk [536]2 years ago

8 0

Answer:

1: a measuring instruments the students should use for time is a stopwatch

2: a measuring instruments the students should use for distance is a measuring tape

Explanation:

pls mark brainliest

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Plzz answer this question

Vera_Pavlovna [14]

Mass does not change anywhere in the space .
So the given 30N is equal to the product of mass of the body and the acceleration due to gravity at that place in space.
After finding the mass of the body multiply it by the given acceleration due to gravity.
Mg is the approximate force due to gravity between the mass and the planet.

I hope this may help you.

6 0

4 years ago

Is the time in Tennessee the same as florida

geniusboy [140]

I would say the same thing as the first answer

5 0

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

A 40.0-μFcapacitor is connected across a 60.0 Hz generator. An inductor is then connected in parallel with the capacitor. What i

amm1812

Answer:

The value of the inductance is 175.9 mH.

Explanation:

Given that,

Capacitor $C= 40.0\ \muF$

Frequency = 60.0 Hz

The inductor and capacitor is connected in parallel, the voltage across each of these elements is the same.

We have,

$V_{L}=V_{C}$

Using ohm's law

$I_{rms}\times X_{L}=I_{rms}\times X_{C}$

$X_{L}=X_{C}$

$2\pi f L=\dfrac{1}{2\pi f C}$

$L=\dfrac{1}{4\pi^2\times f^2\times C}$

Put the value into the formula

$L=\dfrac{1}{4\times\pi^2\times60.0^2\times40\times10^{-6}}$

$L=0.1759\ H$

$L=175.9\times10^{-3}\ H$

$L=175.9\ mH$

Hence, The value of the inductance is 175.9 mH.

5 0

4 years ago

I need help to figure out how to solve this problem and solve it!!!

wariber [46]

well it looks like the walk at a constant increasing pace then at a constant pace then increaseing pace then constant pace then they slow down then walk at a constant pace then walk at a constantly increasing pace

plz rate me brainliest

4 0

4 years ago

Read 2 more answers