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3 years ago

An object is propelled along a straight-line path by a force. If the new force were doubled, is acceleration would

Physics

2 answers:

alexandr1967 [171]3 years ago

8 0

The speed would go up
hope it helped!

Aneli [31]3 years ago

6 0

Still go straight but would obviously go up in speed!!

Hope this helps plz mark as brainlist and 5 star

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La energía cinética es la energía que presentan los cuerpos que se encuentran en movimiento.

myrzilka [38]

Do you speak a little English cuz I can’t help you if a can’t understand you

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2 years ago

Why don't astronauts use a fountain pen or a ball point pen with liquid ink in a spacecraft

mojhsa [17]

It's probably because there is no gravity which ballpoint pens need for inkflow to the tip.

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2 years ago

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

const2013 [10]

Answer:

$388 cm^3$

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

$pV=const.$

which can also be rewritten as

$p_1 V_1 = p_2 V_2$

In our case, we have:

$p_1 = 75.9 kPa$ is the initial pressure

$V_1 = 639 cm^3$ is the initial volume

$p_2 = 125 kPa$ is the final pressure

Solving for V2, we find the final volume:

$v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3$

7 0

3 years ago

A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,

fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

$p=mv$

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

$m'=2m$

therefore, the new momentum is

$p'=m'v=(2m)v=2(mv)=2p$

so, the momentum has also doubled.

7 0

2 years ago

You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear

Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

$L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|$

L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?

$log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}$

$\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}$

$r_2 =15 \times 10^{\dfrac{|20-50|}{20}}$

$r_2 =15 \times 10^{-1.5}$

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0

3 years ago