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postnew [5]
3 years ago
9

Two stones, one of mass m and the other of mass 2m, are thrown directly upward with the same velocity at the same time from grou

nd level and feel no air resistance. Which statement about these stones is true?
a. The heavier stone will go twice as high as the lighter one because it initially had twice as much kinetic energy.
b. Both stones will reach the same height because they initially had the same amount of kinetic energy.
c. At their highest point, both stones will have the same gravitational potential energy because they reach the same height.
d. At Its highest point, the heavier stone will have twice as much gravitational potential energy as the lighter one because it is twice as heavy.
e. The lighter stone will reach its maximum height sooner than the heavier one.
Physics
1 answer:
brilliants [131]3 years ago
8 0

Answer:

c.  At their highest point, both stones will have the same gravitational potential energy because they reach the same height.

Explanation:

The above statement is true going by the fact that, the gravitational potential energy of both stones is a function of its mass (weight) as well as the height at which it reach. <em>Despite the fact that, both stones has different weights, it would reach different heights. At their various maximum height, the will have same potential energy.</em>

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What environmental factors affect sound?
Mariana [72]

Answer:

Here are 5:

Distance from source to receiver

Wind speed and direction

Wind gradients

Temperature gradients

Atmospheric attenuation

and there are many more...

Hope that was helpful.Thank you!!!

7 0
3 years ago
Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

5 0
3 years ago
The law of conservation of energy states...
iris [78.8K]
Can neither be created or destroyed.
3 0
3 years ago
Read 2 more answers
on a high way a car is driven 80 km the first 1 he of travel, 50km during the next 0.5 he, and 40 km in the final 0.5 hr. What i
Leno4ka [110]

Average speed = (total distance covered) / (total time to cover the distance) .

Total distance = (80 + 50 + 40) = 170 km

Total time = (1 + 0.5 + 0.5) = 2 hours

Average speed = (170 km) / (2 hrs) = 85 km/hr .
 
4 0
3 years ago
THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN
timurjin [86]

Answer:

f_{o} = 391.67 Hz

Explanation:

The sound of lowest frequency which is produced by a vibrating sting is called its fundamental frequency (f_{o}).

The For a vibrating string, the fundamental frequency (f_{o}) can be determined by:

f_{o} = \frac{v}{2L}

Where v is the speed of waves of the string, and L is the length of the string.

L = 42.0 cm = 0.42 m

v = 329 m/s

f_{o} = \frac{329}{2*0.42}

   = \frac{329}{0.84}

f_{o} = 391.6667 Hz

The fundamental frequency of the string is 391.67 Hz.

3 0
3 years ago
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