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Jlenok [28]
2 years ago
8

Settlers are trying to determine the value of g on a distant planet. They throw a wrench downward from a height of 3 m. If the i

nitial speed of the wrench is 2 m/s, and the speed right before impact is 10 m/s, what is the value of g?
Physics
1 answer:
Agata [3.3K]2 years ago
8 0
<h2>Value of g on the distant planet is 16 m/s²</h2>

Explanation:

We have equation of motion v² = u² + 2as

Here initial velocity, u = 2 m/s

Final velocity, v = 10 m/s

Displacement, s = 3 m

We need to find acceleration, a.

Substituting

               v² = u² + 2as

               10² = 2² + 2 x a x 3

                6a =  96

                  a = 16 m/s²

Value of g on the distant planet is 16 m/s²

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The chart shows data for a moving object.
Ipatiy [6.2K]

Answer:

number 2

Explanation:

8 0
3 years ago
If the temperature is held constant during this process and the final pressure is 683 torrtorr , what is the volume of the bulb
Anna [14]

Answer:

Explanation:

Let the volume of the unknown bulb = X L

The volume of the system , after opening valve = (X + 0.72 L )

Use Boyles law gas equation,

P1V1 = P2V2 ( at temperature is constant )

Given:

P1 = 1.2 atm

P2 = 683 torr

Converting mmHg to atm,

1 atm = 760 mmHg(torr)

683 mmHg = 683/760

= 0.8987 atm

1.2X = 0.8987*(X + 0.720)

1.2X = 0.8987X + 0.6471

0.3013X = 0.6471

X = 2.15 L

5 0
3 years ago
The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spr
ch4aika [34]

Answer:3.67 m/s

Explanation:

mass of block(m)=2 kg

Velocity of block=6 m/s

spring constant(k)=2 KN/m

Spring compression x=15 cm

Conserving Energy

energy lost by block =Gain in potential energy in spring

\frac{m(v_1^2-v_2^2)}{2}=\frac{kx^2}{2}

2\left [ 6^2-v_2^2\right ]=2\times 10^3\times \left [ 0.15\right ]^2

v_2=3.67 m/s

7 0
3 years ago
A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

f_s = f(\dfrac{v}{v-(-v_s)})

f_s = f(\dfrac{v}{v+v_s})

v_s = v[\dfrac{f_s}{f_o}-1]

      = 343[\dfrac{508}{490}-1]

      v_s=12.6 m/s

distance the tunning fork has fallen

y_1=\dfrac{v^2}{2a_y}

     =\dfrac{12.6^2}{2\times 9.8}

     =8.1 m

now, time required for the observed will be

t = \dfrac{8.1}{343} = 0.023 s

now, for the distance calculation

y_2 = u\ t + \dfrac{1}{2}at^2

  = 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2

  =0.293 m

total distance

 = 8.1 + 0.293 = 8.392 m

3 0
3 years ago
Satellites can focus on specific latitudes using:
Fittoniya [83]
B- east west orbits
4 0
2 years ago
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