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m_a_m_a [10]
2 years ago
3

A skateboarder is moving to the right with a velocity of 8 \,\dfrac{\text m}{\text s}8 s m ​ 8, space, start fraction, m, divide

d by, s, end fraction. After a steady gust of wind that lasts 5 \,\text{s}5s5, space, s, the skateboarder is moving to the right with a velocity of 5\,\dfrac{\text m}{\text s}5 s m ​ 5, space, start fraction, m, divided by, s, end fraction.
Physics
1 answer:
malfutka [58]2 years ago
4 0

Answer:

-0.6 m/s^2 (to the left)

Explanation:

The question is missing in the text. The complete text of the problem is:

<em>"A skateboarder is moving to the right with a velocity of 8 m/s. After a steady gust of wind that lasts 5 s, the skateboarder is moving to the right with a velocity of 5 m/s. Assuming the acceleration is constant, what is the acceleration of the skateboarder during the 5 s time period?"</em>

<em />

Solution:

The acceleration of the skateboarder is given by:

a=\frac{v-u}{t}

where

a is the acceleration

v is the final velocity after a time t

u is the initial velocity

t is the time interval

In this problem, taking the right as positive direction, we have:

u = +8 m/s (to the right)

v = +5 m/s (to the right)

t = 5 s

Substituting into the equation, we find

a=\frac{+5-(+8)}{5}=-0.6 m/s^2

And the negative sign means the acceleration is towards the left.

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Which of the following is not a factor in whether a reaction will spontaneously occur? A. Entropy change of the system B. Enthal
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D

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pressure change have nothing to do with the spontaneity.

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2 years ago
A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow
leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

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a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}

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7 0
2 years ago
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s
Bezzdna [24]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?

0.57 m

0.64 m  

0.80 m  

1.25 m

Data:

E_{pe}\:(elastic\:potential\:energy) = 5184\:J

K\:(constant) = 16200\:N/m

x\:(displacement) =\:?

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

E_{pe} = \dfrac{k*x^2}{2}

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:  

E_{pe} = \dfrac{k*x^2}{2}

5184 = \dfrac{16200*x^2}{2}

5184*2 = 16200*x^2

10368 = 16200\:x^2

16200\:x^2 = 10368

x^{2} = \dfrac{10368}{16200}

x^{2} = 0.64

x = \sqrt{0.64}

\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark

Answer:  

The displacement of the spring = 0.8 m (or 0.80 m)

_________________________________________

I Hope this helps, greetings ... Dexteright02! =)

8 0
2 years ago
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A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
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Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

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let v is the image distance

use, 1/u + 1/v = 1/f

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1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

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= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

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v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
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