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3 years ago

HELP PLEASECan someone please explain to me why the answer would be 2.45m/s?

Physics

2 answers:

trapecia [35]3 years ago

8 0

Solution with explanation is given below in attachment.

marissa [1.9K]3 years ago

4 0

It's basically asking for the acceleration. When an object is in free fall, the acceleration is just gravity. so in this case ∆v = 2.45 because it is the gravity on that planet.

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create a poem that incorporates those ten words. Feel free to make it as silly as you like! MINIMUM of 6 lines with a MINIMUM of

Luba_88 [7]

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3 years ago

An elevator carrying a person of mass m is moving upward and slowing down. How does the magnitude f of the force exerted on the

V125BC [204]

The force f of the elevator on the man keeps reducing as the elevator keeps going up while the gravitational force mg keeps increasing moving upwards.

<h3>What is an elevator?</h3>

An elevator is an electrical device that lifts people up and down a tall building or structure.

for the elevator to go up, f > mg.

for the elevator to come down mg > f.

Analysis

since the force on the man is f = ma

where a is the acceleration of the elevator, then it means when a increases, f will increase and when it decreases, f would decrease. slowing down means a, is decreasing going up and this reduces the force as the elevator keeps going up.

on the other hand, gravity acts faster on bodies that are slower in motion so since g, increases going up, mg would also increase.

Learn more about forces in an elevator : brainly.com/question/13526583

#SPJ1

6 0

1 year ago

Net force of 8.0 N acts on an 18 kg body for one minute. Determine the impulse due to the force.

boyakko [2]

Answer:

p = FΔt = 8.0 N(60 s) = 480 N•s

Explanation:

not asked for, but in that time a frictionless 18 kg mass on a horizontal surface will have change velocity by 480/18 = 26.7 m/s.

An impulse results in a change of momentum.

3 0

3 years ago

A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan

sergejj [24]

Here mass of the iron pan is given as 1 kg

now let say its specific heat capacity is given as "s"

also its temperature rise is given from 20 degree C to 250 degree C

so heat required to change its temperature will be given as

$Q = ms \Delta T$

$Q = 1*s*(250 - 20)$

$Q = 1*s*230$

now if we give same amount of heat to another pan of greater specific heat

so let say the specific heat of another pan is s'

now the increase in temperature of another pan will be given as

$Q = ms'\Delta T$

$1*s*230 = 1* s' * \Delta T$

now we have

$\Delta T = (\frac{s}{s'})*230$

now as we know that s' is more than s so the ratio of s and s' will be less than 1

And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature

So correct answer is

<u>A) The second pan would reach a lower temperature.</u>

3 0

2 years ago

Read 2 more answers

A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

5 0

3 years ago