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Natali5045456 [20]

3 years ago

5

The energy of the ball has been reduced by joules after bouncing back. This reduction happens because potential energy transform

s to energy.

1 answer:

luda_lava [24]3 years ago

5 0

<span>This reduction happens because potential energy transforms to "thermal energy" and "sound energy"

Hope this helps!</span>

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A 100 kg box sits on an incline held at rest by a very thin rope attached to another mass hanging as shown. The force of frictio

MAVERICK [17]

If friction is acting along the plane upwards

then in this case we will have

For equilibrium of 100 kg box on inclined plane we have

$mgsin\theta = F_f + T$

also for other side of hanging mass we have

$T = Mg = 50(9.8) = 490 N$

now we have

$100(9.8)sin\theta = 100 + 490$

$980sin\theta = 590$

$sin\theta = 0.602$

$\theta = 37 degree$

In other case we can assume that friction will act along the plane downwards

so now in that case we will have

$mgsin\theta + F_f = T$

also we have

$T = Mg = 50(9.8) N$

now we have

$100(9.8)sin\theta + 100 = 50(9.8)$

$980sin\theta + 100 = 490$

$980 sin\theta = 490 - 100$

$sin\theta = 0.397$

$\theta = 23.45 degree$

<em>So the range of angle will be 23.45 degree to 37 degree</em>

6 0

3 years ago

A Jaguar XK8 convertible has an eight-cylinder engine. At the beginning of its compression stroke, one of the cylinders contains

Colt1911 [192]

If only I was smart then I could help you :/ no but like for real im madddd dumb sorry :(

4 0

4 years ago

A girl is sitting in a sled sliding horizontally along some snow (there is friction present).

Over [174]

Answer:

159 N

Explanation:

The force of friction, Fr is a product of coefficient of feiction and the normal force. Therefore, Fr=uN where N is the normal force and u is coefficient of friction. Here, we have two coefficients of friction but since it is sliding, then we use coefficient of kinetic energy. Substituting 0.25 for u and 636 N for N then

Fr=0.25*636=159 N

Therefore, the force of friction is equivalent to 159 N

4 0

3 years ago

How many bonds can antimony form?

bija089 [108]

Answer:

An antimony can form 3 bonds

Let me know if this helps

Thanks!

:)

Explanation:

4 0

3 years ago

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago