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Anettt [7]
3 years ago
5

What is a element that tends to be shiny, easy shaped, and a good conductor of heat?

Physics
1 answer:
ozzi3 years ago
8 0
The answer to your question is Metal
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If two people are running at the same speed in the same direction. One person is one meter ahead of the other. The person in fro
Gre4nikov [31]

Answer:yes

Explanation:

5 0
3 years ago
A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia
aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the <u>initial point</u> is 19820 J

2) The potential energy at <u>point A</u> is 19620 J

3) The kinetic energy at <u>point B</u> is 10010 J

4) The potential energy at <u>point C</u> is zero

5) The kinetic energy at <u>point C</u> is 19820 J

6) The velocity of the roller coaster at <u>point C</u> is 19.91 m/s

1) The total energy for the roller coaster at the <u>initial point</u> can be found as follows:

E_{t} = KE_{i} + PE_{i}

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The<em> total energy</em> is:

E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J

Hence, the total energy for the roller coaster at the <u>initial point</u> is 19820 J.

   

2) The <em>potential energy</em> at point A is:

PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J

Then, the potential energy at <u>point A</u> is 19620 J.

3) The <em>kinetic energy</em> at point B is the following:

KE_{A} + PE_{A} = KE_{B} + PE_{B}

KE_{B} = KE_{A} + PE_{A} - PE_{B}

Since

KE_{A} + PE_{A} = KE_{i} + PE_{i}

we have:

KE_{B} = KE_{i} + PE_{i} - PE_{B} =  19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J

Hence, the kinetic energy at <u>point B</u> is 10010 J.

4) The <em>potential energy</em> at <u>point C</u> is zero because h = 0 meters.

PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J

5) The <em>kinetic energy</em> of the roller coaster at point C is:

KE_{i} + PE_{i} = KE_{C} + PE_{C}            

KE_{C} = KE_{i} + PE_{i} = 19820 J      

Therefore, the kinetic energy at <u>point C</u> is 19820 J.

6) The <em>velocity</em> of the roller coaster at point C is given by:

KE_{C} = \frac{1}{2}mv_{C}^{2}

v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s

Hence, the velocity of the roller coaster at <u>point C</u> is 19.91 m/s.

Read more here:

brainly.com/question/21288807?referrer=searchResults

I hope it helps you!

3 0
3 years ago
In a nuclear station if the temperature of the superheated water starts to decrease what should the engineer in charge of the re
Sophie [7]

Answer:

They sh0uld g0 t0 the reactor and then, see what the issue is...

Explanation:

then, see if they can fix the problem, im sorry if its wr0ng.

8 0
2 years ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 29.5 km/h and the μs
Yakvenalex [24]

Answer:

Radius=15.773 m

Explanation:

Given data

v=29.5 km/h=8.2 m/s

μs=0.435

To find

Radius R

Solution

The acceleration is a centripetal acceleration  which is experienced by the bicycle given by

a=v^{2}/R

This acceleration is only due to static force which given as

f=ma\\f=m(v^{2}/R )

The maximum value of the static force is given as

Fs_{max}=u_{s}F_{N}

where

FN is normal force equal to mass*gravity

Therefore when the car is on the verge of sliding

f=fs_{max}\\ m(v^{2}/R )=u_{s}mg

Therefore the minimum radius should be found by the bicycle move without  sliding

So

v^{2}/R=u_{s}g\\  R=v^{2}/u_{s}g\\R=(8.2)^{2}/(0.435*9.8)\\R=15.773m

8 0
3 years ago
How to solve 30(a)<br><br> Give solution asap.
expeople1 [14]

Answer:

They will not meet

h-hX=1.2*g*t²

hX=v0*t-(1/2*g*t²)

Explanation:

fall h=1/2*g*t²

elevation time if v0=20 m/s  te=v0/g=20 m/s /9.81 m/s²=2.0387s

hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m

free fall

t=2.0387s yields hX=1/2*g*t²=20.387 m

h-hX=200m - 20.387 m=179,613 m.

so, the second body has not enough initianoal speed to reach a meeting point

5 0
3 years ago
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