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3 years ago

A machine does 1000 J of work in 3.5 minutes (210 seconds). What is

Physics

1 answer:

irina [24]3 years ago

4 0

Answer: 4.76 watts

Explanation:

Work = 1000 J

Time = 3.5 minutes (210 seconds)

Since the SI unit of time is seconds, 210 seconds will be used in calculation.

Power output of the machine = ?

Recall that power is the rate of doing work. Thus, power is Workdone divided by time taken.

i.e Power = (work/time)

Power = 1000J/210seconds

Power = 4.76 watts

Thus, the power output of the machine is 4.76 watts

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You and your friend Peter are putting new shingles on a roof pitched at 21∘. You're sitting on the very top of the roof when Pet

Marat540 [252]

Answer:

3.69 m/s

Explanation:

Forces :

mgsin Θ - mumgcosΘ = ma

g x sinΘ - mu x g x cosΘ = a

9.8 x sin 21 - 0.53 x 9.8 x cos 21 = a

a = -1.337 m/s²

so you have final velocity = 0 m/s

initial velocity = ? m/s

Given d = 5.1 m

By kinematics

vf² = vo² + 2ad

0 = vo² + 2 x -1.337*5.1

vo = 3.69 m/s

8 0

3 years ago

In order for gravitational potential energy to be stored, there must be _____.

trasher [3.6K]

Answer: 1

an object positioned at some height in a gravitational field

Explanation:

Gravitational potential energy of an object is the energy stored due to position of the object or position at certain height relative to zero position.

Gravitational potential energy can also be expressed as object position at some height above or below zero position in a gravitational field

I think 1 and 2 make sense. But 1 make more sense than 2

3 0

3 years ago

Can someone please help

ki77a [65]

Answer:

Acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

Given:

initial speed of hammer = 0 $\frac{m}{s}$

time = 1 s

distance = 15 m

To find:

Acceleration due to gravity = ?

Formula used:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

Solution:

Distance covered by hammer is given by,

s = ut + $\frac{1}{2} a t^{2}$

s = distance

u = initial speed of hammer

t = time taken by hammer to reach ground

a = acceleration

u = 0

t = 1 s

s = 15 m

a = g

Thus substituting these value in above equation.

15 = 0 + $\frac{1}{2} g 1^{2}$

g = 15 × 2

g = 30 $\frac{m}{s^{2} }$

Thus, acceleration of that planet is 30 $\frac{m}{s^{2} }$ .

8 0

3 years ago

Convert the following to relative uncertainties <br>a) 2.70 ± 0.05cm<br>b) 12.02 ± 0.08cm

DENIUS [597]

data which is expressed in form of following way

$a = a_o + \Delta a$

here in above expression

$a_o$ = true value

$\Delta a$ = uncertainty in the value

now the relative uncertainty is given as

$\frac{\Delta a}{a_o}$

now by above formula we can say

a) 2.70 ± 0.05cm

here

True value = 2.70

uncertainty = 0.05

Relative uncertainty = $\frac{0.05}{2.70}$ = 0.0185

b) 12.02 ± 0.08cm

here

True value = 12.02

uncertainty = 0.08

Relative uncertainty = $\frac{0.08}{12.02}$ = 0.00665

4 0

3 years ago

A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten

kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

$\mu$ = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

$\omega$ = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

$P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W$

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0

3 years ago