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3 years ago

The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?

Physics

1 answer:

pentagon [3]3 years ago

5 0

Answer:

2π/[28 x (10^-3)]

Explanation:

Angular speed : ω=2π/T

T = 28ms = 28 x (10^-3) s

Angular speed = 2π/[28 x (10^-3)]

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What is the correct equation kinetic enery

Ivahew [28]

Kinetic energy is directly proportional to the mass of the object and to the square of its velocity: K.E. = 1/2 m v2.

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3 years ago

A object travels at constant negative acceleration. What does the graph of the object's velocity as a fun

melomori [17]

Constant = straight line
“Travels at constant negative acc.”
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Solution: B. Straight line w/ neg. slope

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3 years ago

A ball is thrown upwards at an unknown speed. in a time of

alexandr402 [8]

Answer:

a) Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a) We have equation of motion $s=ut+\frac{1}{2} at^2$ , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8 $m/s^2$

Substituting

$6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s$

Initial speed of the ball = 14.45 m/s

b) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

s = 6 m, u = 14.45 m/s, a = -9.8 $m/s^2$

Substituting

$v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s$

So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

u = 14.45 m/s, v =0 , a = -9.8 $m/s^2$

Substituting

$0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m$

Maximum height reached = 10.65 m

8 0

3 years ago

The net flow of energy into and out of the earths system is referred to as energy budget. which type of energy is lost in space

Svet_ta [14]

D not 100% sure though.

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3 years ago

A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.

V125BC [204]

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar, F = 534 N

Co-efficient of kinetic friction, $\mu_k$ = 0.3

Now,

The force against the kinetic friction is given as:

$f = \mu_k N = u_k Mg$

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

$f = 0.3\times108\times9.8$

$f = 317.52N$

Now, the net force on to the metal safe is

$F_{Net}= F-f$

Substituting the values in the equation we get

$F_{Net}= 534N-317.52N$

$F_{Net}= 216.48$

also,

$F_{Net}= M\times$ acceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe= $\frac{F_{Net}}{M}$

acceleration of the safe= $\frac{216.48}{108}$

acceleration of the safe= $2.00 m/s^2$

Hence, the acceleration of the metal safe will be 2.00 m/s²

3 0

3 years ago