A train moves at constant velocity of 90km/h. How far will it go in 0.25h.

A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.

Viefleur [7K]

Answer:

Explanation:

a )

change in the gravitational potential energy of the bear-Earth system during the slide = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J - F x 11 = 756.9 J

F x 11 = 4851 J - 756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N

4 0

2 years ago

A 67 kg soccer player uses 5100 kJ of energy during a 2.0 h match. What is the

Oksanka [162]

The average power produced by the soccer player is 710 Watts.

Given the data in the question;

Mass of the soccer player; $m = 67kg$

Energy used by the soccer player; $E = 5100KJ = 5100000J$

Time; $t = 2.0h = 7200s$

Power; $P =\ ?$

Power is simply the amount of energy converted or transferred per unit time. It is expressed as:

$Power = \frac{Energy\ converted }{time}$

We substitute our given values into the equation

$Power = \frac{5100000J}{7200s}\\\\Power = 708.33J/s \\\\Power = 710J/s \ \ \ \ \ [ 2\ Significant\ Figures]\\\\Power = 710W$

Therefore, the average power produced by the soccer player is 710 Watts.

Learn more: brainly.com/question/20953664

8 0

2 years ago

Which of the following is an example of chemical change?

Maru [420]

Answer:

c

Explanation:

all the others r physical

8 0

2 years ago

Read 2 more answers

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Pachacha [2.7K]

Explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

$u -v= d$

$u=71+26=97\ cm$

We need to calculate the focal length

Using formula of lens

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

put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-26}+\dfrac{1}{97}$

$\dfrac{1}{f}=-\dfrac{71}{2522}$

$f=-35.52\ cm$

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

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

Put the value in to the formula

$\dfrac{1}{-29}=\dfrac{1}{v}-\dfrac{1}{95}$

$\dfrac{1}{v}=\dfrac{1}{-29}-\dfrac{1}{95}$

$\dfrac{1}{v}=-\dfrac{124}{2755}$

$v=-22.21\ cm$

We need to calculate the magnification

Using formula of magnification

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

$m=\dfrac{22.21}{95}$

$m=0.233$

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.

4 0

3 years ago

How many protons are in Calcium? (* Hint: The Atomic number of Calcium is 20).

ryzh [129]

Answer:

The answer to your question is: 20

Explanation:

Atomic number is the number of proton and atom has. Each element has a specific number of protons, if the number of protons change, then this is a new element.

Mass number is the number of protons and neutrons and atom has.

Mass number = protons + neutrons

Data

Number of protons = ?

Atomic number = 20

Then,

atomic number = number of protons = 20

3 0

2 years ago