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Zina [86]
3 years ago
5

Using the rules for the significant figures what do you get when you add 24.545 and 307.3

Physics
1 answer:
Tamiku [17]3 years ago
4 0
1
2 4. 5 4 5
+3 0 7. 3 0 0
——————
3 3 1 8 4 5
line up the decimal points and add.
hope this helps!
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A car’s tire rotates 5.25 times in 3 seconds. What is the tangential velocity of the tire?
Lena [83]

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.

<h3>Angular velocity of the tire</h3>

The angular velocity of the tire is the rate of change of angular displacement of the tire with time.

The magnitude of the angular velocity of the tire is calculated as follows;

ω = 2πN

where;

  • N is the number of revolutions per second

ω = 2π x (5.25 / 3)

ω =  11 rad/s

<h3>Tangential velocity of the tire</h3>

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.

The magnitude of the tangential velocity is caculated as follows;

v = ωr

where;

  • r is the radius of the car's tire

v = 11r m/s

Learn more about tangential velocity here: brainly.com/question/25780931

4 0
2 years ago
A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is
Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ  = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

  = √(T/A×ρ)

  = √[2.96×10^4/(4.49×10^-3×7860)]

  = 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

7 0
3 years ago
Use the exact values you enter to make later calculations. You repeat the same experiment that you did in the lab with the force
frutty [35]

Answer:

sorry I don't know but I hope you get a answer soon

7 0
3 years ago
How much heat is released as a 5.89 kg block of aluminum cools from 462 °C to 315 °C. The specific heat capacity of aluminum is
s2008m [1.1K]
Mass, m = 5890g
Change in temperature, θ = Final_temperature - Initial_temperature
= 315 - 462°C
= -147°C

Specific heat capacity of aluminum, c = 0.900 J/(g*K) 
=mcθ
=5890g x 0.900 J/(g*K) x -147°C
=-779,247j

Answer would be C. 
5 0
3 years ago
Read 2 more answers
A cannonball is launched from the ground at an angle of 30 degrees above the horizontal and a speed of 30 m/s. Ideally (no air r
ICE Princess25 [194]
B is the answer please mark brainliest
6 0
3 years ago
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