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rusak2 [61]
3 years ago
10

A car starts from rest and accelerates at a constant rate in a straight line. In the first second the car moves a distance of 2.

0 meters. How fast will the car be moving at the end of the second second
Physics
1 answer:
goblinko [34]3 years ago
5 0

Answer:

The speed of the car at the end of the 2nd second = 8.0 m/s

Explanation:

The equations of motion will be used to solve this problem.

A car starts from rest,

u = initial velocity of the car = 0 m/s

Accelerates at a constant rate in a straight line,

a = constant acceleration of the car = ?

In the first second the car moves a distance of 2.0 meters,

t = 1.0 s

x = distance covered = 2.0 m

x = ut + (1/2)at²

2 = 0 + (1/2)(a)(1²)

a = 4.0 m/s²

How fast will the car be moving at the end of the second second

Now,

a = 4.0 m/s²

u = initial velocity of the car at 0 seconds = 0 m/s

v = final velocity of the car at the end of the 2nd second = ?

t = 2.0 s

v = u + at

v = 0 + (4×2)

v = 8.0 m/s

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 340 turns. The inner solenoid is 25.0 cm long an
LUCKY_DIMON [66]

Answer:

a) 1.34*10^-8 W

b) 1.18*10^-5 H

c) 20mV

Explanation:

a) To find the average magnetic flux trough the inner solenoid you the following formula:

\Phi_B=BA=\mu_oNIA

mu_o: magnetic permeability of vacuum = 4pi*10^-7 T/A

N: turns of the solenoid = 340

I: current of the inner solenoid = 0.100A

A: area of the inner solenoid = pi*r^2

r: radius of the inner solenoid = 2.00cm/2=1.00cm=10^-2m

You calculate the area and then replace the values of N, I, mu_o and A to find the magnetic flux:

A=\pi(10^{-2}m)^2=3.141510^{-4}m^2\\\Phi_B=(4\pi*10^{-7}T/A)(340)(0.100A)(3.1415*10^{-4}m^2)=1.34*10^{-8}W\\

the magnetic flux is 1.34*10^{-8}W

b) the mutual inductance is given by:

M=\mu_o N_1 N_2 \frac{A_2}{l}

N1: turns of the outer solenoid = 22

N2: turns of the inner solenoid

A_2: area of the inner solenoid

l: length of the solenoids = 25.0cm=0.25m

by replacing all these values you obtain:

M=(4\pi*10^{-7}T/A)(340)(22)\frac{3.14*10^{-4}m^2}{0.25m}=1.18*10^{-5}H

the mutual inductance is 1.18*10^{-5}H

c) the emf induced can be computed by using the mutual inductance and the change in the current of the inner solenoid:

\epsilon_1=M\frac{dI_2}{dt}

by replacing you obtain:

\epsilon_1=(1.18*10^{-5}H)(1700A/s)=0.02V=20mV

the emf is 20mV

7 0
4 years ago
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tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer (\dot Q), measured in watts, in the hollow cylinder is:

\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})

Where:

k - Thermal conductivity, measured in watts per meter-Celsius.

L - Length of the cylinder, measured in meters.

D_{i} - Inner diameter, measured in meters.

D_{o} - Outer diameter, measured in meters.

T_{i} - Temperature at inner surface, measured in Celsius.

T_{o} - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)

If we know that \dot Q = 40.8\,W, L = 0.6\,m, T_{i} = 50\,^{\circ}C, T_{o} = 20\,^{\circ}C, D_{i} = 0.01\,m and D_{o} = 0.04\,m, the thermal conductivity of the biomaterial is:

k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)

k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

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Answer:

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A mature thunderstorm will contain both updraft and downdrafts. The given statement is true.

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