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Svetach [21]
3 years ago
13

Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.

Physics
1 answer:
ValentinkaMS [17]3 years ago
4 0

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

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A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an an
frozen [14]

Answer:

a) <em>2.278 x 10^-5 volts</em>

b) <em>1.139 x 10^-6 Ampere</em>

c) <em>2.59 x 10^-11 W</em>

Explanation:

The radius of the wire r = 2 mm = 0.002 m

the number of turns N = 200 turns

direction of the magnetic field ∅ = 25°

magnetic field strength B = 0.02 T

varying time = 2 sec

The cross sectional area of the wire = \pi r^{2}

==> A = 3.142 x 0.002^{2} = 1.257 x 10^-5 m^2

Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°

==> Φ = 2.278 x 10^-7 Wb

The induced EMF is given as

E = NdΦ/dt

where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7

E = 200 x 1.139 x 10^-7 = <em>2.278 x 10^-5 volts</em>

<em></em>

<em></em>

b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as

I = E/R

where R is the resistor

I = (2.278 x 10^-5)/20 = <em>1.139 x 10^-6 Ampere</em>

<em></em>

<em></em>

<em> </em>c) power delivered to the resistor is given as

P = IE

P = (1.139 x 10^-6) x (2.278 x 10^-5) = <em>2.59 x 10^-11 W</em>

4 0
3 years ago
Calculate the work done when a 100-W lightbulb is lit for 30 seconds.
Mila [183]

Answer:

3000 J

Explanation:

8 0
3 years ago
In a typical rear-end collision, the victim's ________ is/are accelerated faster and harder than the torso
Georgia [21]

The victim's head is accelerated faster and harder than the torso when the victom is involved in a typical rear-end collision.

The traffic accident where a vehicle crashes into another vehicle that is directly in front of it is called a rear-end collision.

 

One of the most common accident in the United States is the rear-end collision, and in a lot of cases, rear-end collisions are prompted by drivers who are inattentive, unfavorable conditions of the road, and poor following distance.

 

<span>An enough room in front of your car so you can stop when the car in front of you stops suddenly is one basic driving rule. The person isn’t driving safely if he / she is behind you and couldn’t stop.</span>

8 0
3 years ago
Read 2 more answers
Name three characteristics of scientists that are important to their work but are also found in nonscientists.
Alina [70]

Answer:

The three characteristics of a good scientist are his curiosity, creativity and problem-solving skills.

Explanation:

3 0
2 years ago
You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

3 0
3 years ago
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