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Cloud [144]
2 years ago
9

A car’s tire rotates 5.25 times in 3 seconds. What is the tangential velocity of the tire?

Physics
1 answer:
Lena [83]2 years ago
4 0

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

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dybincka [34]

Answer:

1] 8500000 = <u>8.5 × 10⁶</u>

2] .000072 = <u>7.2 × 10⁻⁵</u>

3] 5.3 × 10⁴ = <u>53000</u>

4] 2.8 × 10⁻³ = <u>0.0028</u>

5] Velocity = \frac{distance}{time}

 V = \frac{50}{10}

 <u>V = 5 m/s</u>

6] Acceleration = \frac{V1-V2}{time}

 A = \frac{30-15}{3}

 A = \frac{15}{3}

 <u>A = 3 m/s²</u>

7 0
2 years ago
The secondary coil of a neon sign transformer provides 7500 V at 0.01 A. The primary coil operates on 120 V. What is the input c
nikitadnepr [17]

Answer:

0.625 A

Explanation:

Vs = 7500 V, Is = 0.01 A

Vp = 120 V

Let the primary current be Ip.

As the transformer is ideal, so input power is equal to the output power

Vp x Ip = Vs x Is

120 x Ip = 7500 x 0.01

Ip = 0.625 A

8 0
3 years ago
What's the formula to find out power
Julli [10]

In general, 

                 Power = (energy moved) / (time to move the energy) .

If it's mechanical power, then     

                 Power = (work done) / (time to do the work) .

If it's electrical power, then it can be any one of these:

                 Power  =  (volts)  x  (amperes)

                 Power  =  (volts)²  /  (resistance, ohms)

                 Power  =  (amperes)²  x  (resistance, ohms) .

Whatever kind of energy you're dealing with, power always
turns out to be

                  (amount of energy produced, used, or moved)
divided by
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3 0
3 years ago
(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat
vitfil [10]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i)  the gradient of f =  ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is  ñ∙∇F =  4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+  ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+  (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have  

∇f(x, y) = -1·i -3·j  that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.  

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z)  = ∇F(x, y, z) = = ∂f/∂x i+  ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore  

ñ = ⅟4(2·i +3·j -√3·k)  

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)  

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

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