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2 years ago

As you brake your bicycle, your velocity changes from 20 m/s east to 10 m/s east in 5 seconds. What’s your acceleration?

Physics

1 answer:

charle [14.2K]2 years ago

8 0

Hello!

$\large\boxed{-2m/s^{2} }$

Find the acceleration using the formula a = (vf - vi) / t where:

vf = final velocity

vi = initial velocity

t = time

Plug in the points above:

vf = +10 m/s

vi = +20 m/s

t = 5 sec

(10 - 20) / 5 = -10 / 5 = -2 m/s²

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What property of light waves does the michelson-morley interferometer directly demonstrate?

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The wave nature of light, due to the experiment having bright and dark bands corresponding to places where you have constructive and destructive interference.

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2 years ago

Options: A.) 10 N B.) 15 N C.) 25 N D.) 35N

Lady_Fox [76]

Given:

F_gravity = 10 N

F_tension = 25 N

Let's find the net centripetal force exterted on the ball.

Apply the formula:

$\sum ^{}_{}F_{\text{net}}=F_1+F_2=F_{centripetal}$

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.

Hence, the tensional force is positive while the gravitational force is negative.

Thus, we have:

$F_{\text{net}}=F_{\text{centripetal}}=F_{tension}-F_{gravity}=25N-10N=15N$

Therefore, the net centripetal force exterted on the ball is 15 N.

ANSWER:

15 N

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4 months ago

Some machines do not multiply the force that is applied to them

Katyanochek1 [597]

Answer:

Machine Efficiency

Explanation:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work). The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent

5 0

2 years ago

The radius of a planet is 2400 km, and the acceleration due to gravity at its surface is 3.6 m/s2.

kiruha [24]

Answer:

$3.1\cdot10^{23}\:\mathrm{kg}$

Explanation:

We can use Newton's Universal Law of Gravitation to solve this problem:

$g_P=G\frac{m}{r^2}$ ., where $g_P$ is acceleration due to gravity at the planet's surface, $G$ is gravitational constant $6.67\cdot 10^{-11}$ , $m$ is the mass of the planet, and $r$ is the radius of the planet.

Since acceleration due to gravity is given as $m/s^2$ , our radius should be meters. Therefore, convert $2400$ kilometers to meters:

$2400\:\mathrm{km}=2,400,000\:\mathrm{m}$ .

Now plugging in our values, we get:

$3.6=6.67\cdot10^{-11}\frac{m}{(2,400,000)^2}$ ,

Solving for $m$ :

$m=\frac{2,400,000^2\cdot3.6}{6.67\cdot 10^{-11}},\\m=\fbox{$3.1\cdot10^{23}\:\mathrm{kg}$}$ .

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1 year ago

HELP... 3.00 amu = _____ Mev. 3.22 x 10-3 2.79 x 103 3.10 x 102

DanielleElmas [232]

The correct answer is:
$2.79 \cdot 10^3 MeV$
Let's see why.

1 amu corresponds to the mass of the proton, which is:
$m_p = 1.66 \cdot 10^{-27} kg$
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
$E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J$
Now we can convert it into electronvolts:
$E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV$

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
$3 amu = 3 \cdot 934 MeV \sim 2790 MeV = 2.79 \cdot 10^3 MeV$

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2 years ago

Read 2 more answers