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2 years ago

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What is the acceleration of a skier that goes from 2.50 m/s to 14.5 m/s while traveling 505 m down a slope?

1 answer:

Alik [6]2 years ago

3 0

Answer:

493 m*m/s

Explanation:

14.5-2.50=12

505 x 12=493

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Cindy exerts a force of 40 newtons and moves a chair 6 meters. Her brother Andy pushes a different chair for 6 meters while exer

sesenic [268]

Work formula:
W = F * d
F 1 = 40 N, d 1 = 6 m;
F 2 = 30 N; d 2 = 6 m.
W ( Cindy ) = 40 * 6 = 240 Nm
W ( Andy ) = 30 * 6 = 180 Nm
The difference of their amounts if work:
240 Nm - 180 Nm = 60 nm

hope it helps!

3 0

2 years ago

What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?

slavikrds [6]

I will post my work, but is that 99 degrees Celsius and 25 degrees Celsius?

All you have to do is plug in the initial temperature for gold where it says Tg and the initial temperature for the water where it says Tw and then plug that in and you will have your answer.

8 0

2 years ago

In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9

slega [8]

Answer:

1.6 secs

Explanation:

In a circus act, an acrobat upwards from the surface of a trampoline

At that same moment another acrobat perched 9.0m above him

A ball is released from rest

While still in motion the acrobat catches the ball

He left the ball with a trampoline of 5.6m/s

Since the ball is falling downwards from a distance then acceleration will be negative

a= -9.8

s= d

s= 1/2at^2

= 1/2 × (-9.8)t^2

= 0.5× (-9.8)t^2

d = -4.9t^2

Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows

9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2

9 - 4.9t^2= 5.6t + (-4.9)t^2

9 - 4.9t^2= 5.6t - 4.9t^2

9= 5.6t

t= 9/5.6

t= 1.6 secs

6 0

2 years ago

A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards

Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0

2 years ago

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir

Semmy [17]

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{0.24}=\dfrac{1}{0.25}+\dfrac{1}{u}$

$\dfrac{1}{u}=\dfrac{1}{0.24}-\dfrac{1}{0.25}$

$\dfrac{1}{u}=\dfrac{1}{6}$

$u=6\ m$

We need to calculate the magnification

Using formula of magnification

$m=-\dfrac{v}{u}$

Put the value into the formula

$m=-\dfrac{0.24}{-6}$

$m=0.04$

We need to calculate the height of the object

Using formula of magnification

$m=\dfrac{h'}{h}$

$h=\dfrac{0.080}{0.04}$

$h=2\ m$

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

6 0

2 years ago

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