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2 years ago

8

A square sheet of rubber has sides that are 20 cm long. If you stretch it so that the area of the rubber is four times larger, h

ow
long is each of the sides now?
Each side = ____cm

2 answers:

MA_775_DIABLO [31]2 years ago

8 0

Answer:

40

Explanation:

The area is 400cm^2 but the area on each side is 40cm.

oee [108]2 years ago

7 0

Answer:

400

Explanation:

Trust me its correct

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A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

$\theta=274rev$

Explanation:

From the question we are told that:

Angular velocity $\omega=510rpm$

Mass $m=40.kg$

Diameter d $75=>0.75m$

Off Time $t=40.0s$

Oscillation at Power off $N=210$

Generally the equation for Angular displacement is mathematically given by

$\theta_{\infty}=\frac{w+w_0}{t}t$

$w=\frac{2*\theta_{\infty}}{t}-w_0$

$w=\frac{28210}{40*(\frac{1}{60})}-510$

$w=120rpm$

Generally the equation for Time to come to rest is mathematically given by

$t=(\frac{\omega_0}{\omega_0-\omega})t$

$t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})$

$t=0.87min$

Therefore Angular displacement is

$\theta =(\frac{120+510}{2})0.87$

$\theta=274rev$

6 0

2 years ago

Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium

ss7ja [257]

Answer:

The number of atoms are $1.86\times10^{8}$ .

Explanation:

Given that,

Diameter $D = 1.40\times10^{2}\ pm$

$D=1.40\times10^{2}\times10^{-12}\ m$

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

$Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}$

$Number\of\atoms =1.86\times10^{8}$

Hence, The number of atoms are $1.86\times10^{8}$ .

8 0

2 years ago

A 300 kg car initially travels with a velocity of 20 m/s to the right. A net force F acts on the car for 5 s, which causes the v

hram777 [196]

Answer:

<em>600N.</em>

Explanation:

From the question, we are to calculate the net force acting on the car.

According to Newton's second law of motion:

F = ma

m is the mass of the car

a is the acceleration = change in velocity/Time

a = v-u/t

F = m(v-u)/t

v is the final velocity = 30m/s

u is the initial velocity = 20m/s

t is the time = 5secs

m = 300kg

Get the net force:

Recall that: F = m(v-u)/t

F = 300(30-20)/5

F = 60(30-20)

F = 60(10)

<em>F = 600N</em>

<em>Hence the net force acting on the car is 600N.</em>

<em></em>

<em></em>

3 0

3 years ago

A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0

2 years ago

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele

kirza4 [7]

Answer:

$W = 462.5 keV$

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

$F = qE$

$F = (1.6 \times 10^{-19})(1.85 \times 10^6)$

$F = 2.96 \times 10^{-13} N$

now the distance moved by the electron is given as

$d = 0.25 m$

so we have

$W = F.d$

$W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)$

$W = 7.4 \times 10^{-14} J$

now we have to convert it into keV units

so we have

$1 keV = 1.6 \times 10^{-16} J$

$W = 462.5 keV$

5 0

2 years ago