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3 years ago

A force of 40N is applied to a 28 g mass, what is the acceleration? (round to the hundredths place)

Physics

1 answer:

Leno4ka [110]3 years ago

7 0

Answer:

1428.6m/s²

Explanation:

Given parameters:

Force applied on the body = 40N

Mass of the body = 28g

1000g = 1kg

28g will therefore be 0.028kg

Unknown:

Acceleration = ?

Solution:

To solve this problem, we use the expression derived from Newton's second law of motion.

Force = mass x acceleration

Insert the parameters and solve;

40 = 0.028 x acceleration

Acceleration = $\frac{40}{0.028}$ = 1428.6m/s²

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What happens to the force attraction of the distance two objects is increased?

Fed [463]

Answer:

Explanation:

The attraction weakens. Two objects that are farther apart are not drawn together as strongly as if they were close together.

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3 years ago

A transform boundary occurs where two tectonic plates _____.?

gladu [14]

Where they slide over each other.

Transform boundaries are formed or occur when two plates slide past each other in a sideways motion. They do not tear or crunch into each other (but the rock in between them may be ground up) and therefore none of the spectacular features are seen such as occur in divergent and convergent boundaries.

In a transform boundary, neither plate is added to at the boundary nor destroyed. They are marked in some places by features like stream beds that have been split in half and the two halves moved in opposite directions.

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3 years ago

Read 2 more answers

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 25 m lower vert

Musya8 [376]

Answer:

x = 5.79 m

Explanation:

given,

mass of the car = 39000 Kg

spring constant = 5.7 x 10⁵ N/m

acceleration due to gravity = 9.8 m/s²

height of the track = 25 m

length of spring compressed = ?

using conservation of energy

potential energy is converted into spring energy

$m g h = \dfrac{1}{2}kx^2$

$x =\sqrt{\dfrac{2 m g h}{k}}$

$x =\sqrt{\dfrac{2\times 39000 \times 9.8 \times 25}{5.7 \times 10^{5}}}$

$x =\sqrt{33.5263}$

x = 5.79 m

the spring is compressed to x = 5.79 m to stop the car.

3 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0

2 years ago

A horizontal force of 200 N is applied to move a 55 kg television set across a 10 m level surface. What is the work done by the

vivado [14]

Answer:

The work done is "2000 J".

Explanation:

The given values are:

Force,

F = 200 N

Mass,

m = 55 kg

Displacement,

d = 10 m

Now,

The work done will be:

⇒ $Work \ done= Force\times displacement$

On substituting the given values, we get

⇒ $=200\times 10$

⇒ $=2000 \ J$

3 0

3 years ago