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3 years ago

How are distance and displacement different?

1 answer:

7 0

Distance is a scalar as it provides magnitude without direction
Ex. 5 m

Displacement is a vector as it has magnitude and direction
Ex. 5 m [N]

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What is the kinetic energy of an object with a mass of 10 kilograms traveling at a speed of 10 meters per second?

bagirrra123 [75]

Answer:

500 J

Explanation:

Kinetic Energy = (mass x velocity x velocity) ÷ 2

We know that mass = 10 kg and velocity = 10 m/s.

KE = (m × v × v) ÷ 2

KE = (10 kg × 10 m/s × 10 m/s) ÷ 2

KE = 500 J

The kinetic energy of the object is 500 J (joules). Hope this helps, thank you !!

5 0

3 years ago

A virtual image produced by a lens is always

Sever21 [200]

C. located in front of the lens

3 0

3 years ago

Read 2 more answers

Hello!

Mnenie [13.5K]

Answer:

$T_0=80695.17162...$

Explanation:

Given equation:

$\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124$

To solve the given equation:

$\textsf{Apply log rules}: \quad e^{\ln (x)}=x$

$\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}$

Multiply both sides by T₀:

$\implies T_0-100=T_0e^{-0.00124}$

Add 100 to both sides:

$\implies T_0=T_0e^{-0.00124}+100$

Subtract $T_0e^{-0.00124}$ from both sides:

$\implies T_0-T_0e^{-0.00124}=100$

Factor out the common term T₀:

$\implies T_0(1-e^{-0.00124})=100$

Divide both sides by $(1-e^{-0.00124})$

$\implies T_0=\dfrac{100}{1-e^{-0.00124}}$

Carry out the calculation:

$\implies T_0=\dfrac{100}{1-0.99876...}$

$\implies T_0=\dfrac{100}{0.001239231...}$

$\implies T_0=80695.17162...$

6 0

3 years ago

A note on a piano vibrates 262 time per second. What is the period of the wave

Alexxandr [17]

Answer: it is period 93

Explanation:

7 0

4 years ago

"The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers a

My name is Ann [436]

Answer:

Angle of incidence that entered material b= 63.1°

Angle of incidence between a and b = 55.9°

Explanation: Using the formular:

n1sintheta1= n2sintheta2

The light ray which enters material B will be

1.4Sin72.8° = 1.5Sin theta

1.3373= 1.5Sintheta

sintheta = 1.3373/1.5

Sin^-1 0.8916 = Theta

63.1 = theta

When the ray hits interface with material a

1.5Sin63.1 = 1.3 Sin theta

1.3374 = 1.3Sin theta

Sintheta= 1.3374/1.3

Sin theta = 1.0877

There will be total reflection off the boundary b c because sin theta exceeded 1 in value.

The equation should be

1.4sin63.1 = 1.4 sin theta

Sin theta=72.8°

When the ray hits air-c boundary:

1.4sin72.8=1.00sin theta

Sin theta=1.3374/1 =1.3374

There is total reflection.

In material a,the ray will:

1.3sin72.8° = 1.00sin theta

There will be total reflection when the ray hits a-b boundary.

1.3sin72.8= 1.5sintheta

Sin theta= 1.2419/ 1.5

Sin theta =0.8279

Theta= Sin^-10.8279= 55.88°

When ray hits c-air boundary

1.4sin63.1= 1.00sintheta

1.2485= sin theta = Toal reflection.

Therefore when the ray of light pass through the layers of material a, b and c the boundary with air on top and bottom will be total reflection.

7 0

4 years ago