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Y_Kistochka [10]

2 years ago

6

A 64kg wheel with radius 2.4m is rotating at 280rev/min. if you want to stop it in 30sec. (use point mass of moment of inertia)

A. how much work must be done to stop it? B. what is the required power?

1 answer:

AfilCa [17]2 years ago

4 0

Answer:

Alot of work must be done

Im not sure what the required power will be

Explanation:

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Please Help lol. <br> just trying to get up the character count lol

9966 [12]

Answer:

8612

Explanation:

7 0

3 years ago

A circuit consists of a resistance R, and an inductance L in parallel connected, which is in series with a second resistance R2.

Allisa [31]

Answer:

The impedance of the inductor is ω L:

This is in parallel wth R giving a resistance of the parallel combination:

1 / Rp = 1 / ω L + 1 / R

Rp = (ω L * R) / (ω L + R) for the impedance of the parallel combination

Total resistance (impedance) Rt = Rp + R2 = Z

One can use the term impedance to show that the resulting current is not in phase with the resulting voltage

phase angle φ - cos φ = ω L / Rt

V = I Z shows the relation of voltage and urrent

5 0

2 years ago

Can you answer this math homework? Please!

kap26 [50]

$\large \mathfrak{Solution : }$

9. An object which is in circular motion (moving along a circle) is said to be accelerating because it changes it's direction constantly even if it is moving with a constant speed. cuz acceleration is change in either magnitude or direction of an object with respect to time.

therefore, it's still acceleration as change in direction with time.

10. Average speed of an object can be calculated by dividing the total distance covered by an object by time taken to cover that distance.

i.e

$\boxed{speed = \dfrac{distance}{time} }$

it can be re- arranged to find the distance as :

$\boxed{distance = speed \times time}$

$time = \dfrac{distance}{speed}$

11. speed = 20 m/s : conversion into km/h

distance covered : 4 km = 4000 m

$time = \dfrac{distance}{speed}$

$t = \dfrac{4000}{20}$

$t = 200 \: \: sec$

time taken = 200 seconds

12. let's use the first equation of motion to find the acceleration :

$v = u + at$

$50 = 80 + 120a$

$50 - 80 = 120a$

$a = \dfrac{ - 30}{120}$

$a = - \dfrac{1}{4} m/s {}^{2}$

$- 0.25 \: \: m/s {}^{2}$

3 0

3 years ago

Four forces act on bolt A as shown; F1 150N, F2 80N, F3 110N and F4 100N. Determine the magnitude and direction of the resultant

netineya [11]

Complete Question

The complete question(reference (chegg)) is shown on the first uploaded image

Answer:

The magnitude of the resultant force is $F = 199.64 \ N$

The direction of the resultant force is $\theta = 4.1075^o$ from the horizontal plane

Explanation:

Generally when resolving force, if the force (F )is moving toward the angle then the resolve force will be $Fcos(\theta )$ while if the force is moving away from the angle then the resolved force is $Fsin (\theta )$

Now from the diagram let resolve the forces to their horizontal component

So

$\sum F_x = 150 cos(30) + 100cos(15) -80sin (20)$

$\sum F_x = 199.128 \ N$

Now resolving these force into their vertical component can be mathematically evaluated as

$\sum F_{y} = 150 sin(30) - 100sin(15) -110 +80 cos(20)$

$\sum F_{y} = 14.30$

Now the resultant force is mathematically evaluated as

$F = \sqrt{F_x^2 + F_y^2}$

substituting values

$F = \sqrt{199.128^2 + 14.3^2}$

$F = 199.64 \ N$

The direction of the resultant force is evaluated as

$\theta = tan^{-1}[\frac{F_y}{F_x} ]$

substituting values

$\theta = tan^{-1}[\frac{ 14.3}{199.128} ]$

$\theta = 4.1075^o$ from the horizontal plane

5 0

4 years ago

A test of the prototype of a new automobile shows that the minimum distance for a controlled stop from 95 km/h to zero is 55 m.

iris [78.8K]

Answer:

-0.64525g

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 95 km/h

v = Final velocity = 0 km/h

s = Displacement

a = Acceleration

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-95^2}{2\times 0.055}\\\Rightarrow a=-82045.45\ km/h^2$

Converting to m/s²

$a=82045.45=\frac{82045.45\times 1000}{3600\times 3600}=-6.33\ m/s^2$

g = Acceleration due to gravity = 9.81 m/s²

Dividing both the accelerations, we get

$\frac{a}{g}=\frac{-6.33}{9.81}=-0.64525\\\Rightarrow a=-0.64525g$

Hence, acceleration of the car is -0.64525g

8 0

4 years ago