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3 years ago

How much would a 25 kg suitcase weigh on the surface of…?

Physics

1 answer:

photoshop1234 [79]3 years ago

8 0

Answer:

A. 95N

B. ?

C.225N

D.?

Explanation:

For Mars and Pluto im not so sure about those too but for A and C I am positive those are correct.

Sorry I could not help you all the way, please dont be mad ;(

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Question is in the picture!

raketka [301]

Any change in velocity is called an acceleration. The rock is experiencing the gravitational force from the earth, accelerating it towards the ground.

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4 years ago

How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 1.6 % increase

morpeh [17]

Answer:

The value is $v_o = 5.488 \ m/s$

Explanation:

From the question we are told that

The emitted frequency increased by $\Delta f = 1.6 \% = 0.016 \$

Let assume that the initial value of the emitted frequency is

$f = 100\% = 1$

Hence new frequency will be $f_n = 1 + 0.016 = 1.016$

Generally from Doppler shift equation we have that

$f_1 = [\frac{ v \pm v_o}{v \pm + v_s } ] f$

Here v is the speed of sound with value $v = 343 \ m/s$

$v_s$ is the velocity of the sound source which is $v_s = 0 \ m/s$ because it started from rest

$v_o$ is the observer velocity So

Generally given that the observer id moving towards the source, the Doppler frequency becomes

$f_1 = [\frac{ v + v_o}{v + 0 } ] f$

=> $1.016 = [\frac{ 343 + v_o}{343 } ] * 1$

=> $v_o = 5.488 \ m/s$

8 0

3 years ago

Two boxes (m1 = 43.0 kg and m2 = 39.5 kg) are connected by a light string that passes over a light, frictionless pulley. One box

wlad13 [49]

The net force on the system:
F = m₂g - m₁gsin(∅)
F = 39.5 x 9.81 - 43 x 9.81 x sin(30)
F = 176.58 N

Now, we use F = ma to find the acceleration on each mass.
F = m₁a₁
a₁ = 176.58 / 43
a₁ = 4.11 m/s²

F = m₂a₂
a₂ = 176.58 / 39.5
a₂ = 4.47 m/s²

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3 years ago

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3 years ago

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The gravitational force between two objects is ______ proportional to the products of the masses and _______ proportional to the

Rudiy27

directly, inversely ... newton's grav eqn

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3 years ago

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