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const2013 [10]
3 years ago
13

The fully loaded bus accelerates uniformly from rest to a speed of 14 m / s. The time taken to reach a speed of 14 m / s is 20 s

.
Physics
1 answer:
tino4ka555 [31]3 years ago
3 0

Answer:

0.7m/s^2

Explanation:

acceleration=(final-initial velocity)/time

x=(14-0)/20

x=14/20

x=0.7

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A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the
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Answer:

3.4 x 10^-4 T

Explanation:

A = 1.5 x 10^-3 m^2

N = 50

R = 180 ohm

q = 9.3 x 106-5 c

Let B be the magnetic field.

Initially the normal of coil is parallel to the magnetic field so the magnetic flux is maximum and then it is rotated by 90 degree, it means the normal of the coil makes an angle 90 degree with the magnetic field so the flux is zero .

Let e be the induced emf and i be the induced current

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B = q / ( A x R)

B = (9.3 x 10^-5) / (1.5 x 10^-3 x 180) = 3.4 x 10^-4 T

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4 years ago
A student sees a newspaper ad for an apartment that has 1330 square feet (ft^2) or floor space. How many square meters of area a
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3 years ago
Two horses on opposite sides of a narrow stream are pulling a barge up the stream. Each hors pulls with a force of 720N. The rop
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For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
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8 0
3 years ago
Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)
stepladder [879]

Answer:

m(t)=20e^{-0.5t}

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope (t_{1/2}) = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

m(t)=m_0e^{-kt}

Where,

m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year

So, the equation is: m(t)=20e^{-kt}

At half-life, the mass is reduced to half of the initial value.

So, at t=t_{1/2},m(t)=\frac{m_0}{2}. Plug in these values and solve for 'k'. This gives,

\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5

Hence, the equation for the mass remaining is given as:

m(t)=20e^{-0.5t}

8 0
4 years ago
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