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3 years ago

A force =(16.8N/s)t is applied to an object of m=45.0kg. Ignoring friction, how far does the object travel from rest in 5.00s?

Physics

1 answer:

kirza4 [7]3 years ago

6 0

Answer:

Explanation:

a = F/m = 16.8t/45 = 0.373*t

v = ∫a*dt = ∫(0.373*t)dt = 0.1865*t²

x = ∫v*dt = 0.1865∫(t²)dt = 0.062*t³

For t = 5.0 sec,

x = 0.062*(5)^3 = 7.75 m

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a vertical polarizing filter is used on the lens of a camera. Which of the following do not strike the lens?

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3 years ago

If 1 meter = 3.28 feet, what is the height of the Washington Monument in meters?

jasenka [17]

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3 years ago

Read 2 more answers

It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo

vovikov84 [41]

Answer:

L = 1.11 x $10^{6}$ m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x $10^{5}$ m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x $10^{-2}$ m

Length = ?

So,

we know that,

U = 1/2 C Δ $v^{2}$

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δ $v^{2}$

52000 J = (0.5) x (C) x ( $610^{2}$ )

C = 0.28 F

And we also know that,

C = $\frac{K*E*A}{d}$

E = 8.85 x $10 ^{-12}$

K = 3.7

A = 0.20 x L

d = 2.6 x $10^{5}$ m

Plugging in the values into the formula, we get:

0.28 = $\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }$

Solving for L, we get:

L = 1.11 x $10^{6}$ m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

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3 years ago

An astronaut who is repairing the outside of her spaceship accidentally pushes away a 92.9 cm long steel rod, which flies off at

DiKsa [7]

Answer:

V = 0.0723 volts = 72.3 milivolts

Explanation:

The emf induced in the rod is the motional emf due to the magnetic field. This motional emf can be calculated by the following formula:

$EMF = V = vBl Sin\theta$

where,

V = Motional EMF = ?

v = speed of rod = 12.5 m/s

B = Magnetic Field = 6.23 mT = 0.00623 T

l = Length of rod = 92.9 cm = 0.929 m

θ = angle between v and B = 90°

Therefore,

$V = (12.5\ m/s)(0.00623\ T)(0.929\ m)Sin\ 90^o\\$

<u>V = 0.0723 volts = 72.3 milivolts</u>

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2 years ago

An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y

Xelga [282]

Answer:

The fraction of the protons would have no electrons $=1.88\times 10^{-10}$

Explanation:

We are given that

Amoeba has total number of protons= $1.00\times 10^{16}$

Net charge, Q=0.300pC

Electrons are fewer than protons= $1.88\times 10^6$

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

= $\frac{Fewer\;electrons}{Total\;protons}$

The fraction of the protons would have no electrons

= $\frac{1.88\times 10^{6}}{1.00\times 10^{16}}$

$=1.88\times 10^{-10}$

Hence, the fraction of the protons would have no electrons $=1.88\times 10^{-10}$

6 0

3 years ago