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3 years ago

Name two noble gases

Physics

2 answers:

jekas [21]3 years ago

4 0

Argon and Krypton are two noble gases.

Debora [2.8K]3 years ago

4 0

Here are more than two noble gases. Helium, neon, argon, krypton, and xenon. Hope this helps.

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A ball is thrown upward from an initial height of 1.5m the ball reaches a height of 5m then falls to the ground . What Is the di

jeka94

Answer:

The distance traveled by the ball is 8.5 m

Explanation:

Initial height of the ball, h₁ = 1.5 m above the ground

final height of the ball, h₂ = 5m

Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m

Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m

Total distance traveled = upward distance + downward distance

Total distance traveled = 3.5 m + 5m = 8.5 m

Therefore, the distance traveled by the ball is 8.5 m

4 0

3 years ago

BRAINLIEST FOR THE FIRST TO ANSWER

Sonja [21]

Answer:

Wrong its B Use a different amount of mass in the cart for five different trials, roll the cart down a ramp with the same slope for each trial, and measure how long it takes the cart to roll one meter each time.

Explanation:

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3 years ago

los murcielagos se orientan en la oscuridad emitiendo ondas de ultrasonido que,al relajarse con los objetos, le proveen informac

Marina CMI [18]

I uploaded the answer to $^{}$ a file hosting. Here's link:

bit. $^{}$ ly/3gVQKw3

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3 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa

Serga [27]

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

$v_r$ = Speed of river

$v_c$ = Speed of canoe

$v_r+v_c=2.8\ m/s$

$v_r-v_c=-1.9\ m/s$

Adding the equations we get

$2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s$

$0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s$

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

6 0

3 years ago