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const2013 [10]
4 years ago
8

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 10.5 cm2 is rotated during the time interval 3.10

×10−2 s from a position in which its plane is perpendicular to earth's magnetic field to one in which its plane is parallel to the field. the magnitude of earth's magnetic field at the lab location is 5.30×10−5 t .
Physics
1 answer:
olchik [2.2K]4 years ago
7 0
I guess the problem is asking for the induced emf in the coil.

Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
\epsilon = -N \frac{\Delta \Phi}{\Delta t}
where
N is the number of turns in the coil
\Delta \Phi is the variation of magnetic flux through the coil
\Delta t is the time interval

The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
\Phi_i = BA=(5.30 \cdot 10^{-5}T)(10.5 \cdot 10^{-4} m^2)=5.57 \cdot 10^{-8} Wb
At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
\Phi_f = 0

Therefore, we can calculate now the induced emf by using the first formula:
\epsilon = -N  \frac{\Delta \Phi}{\Delta t}=- (250)  \frac{5.57 \cdot 10^{-8} Wb - 0}{3.10 \cdot 10^{-2} s} = -4.5 \cdot 10^{-4} V
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A 110 kg football player running with a velocity of 5.0 m/s hits another stationary football
KengaRu [80]

Answer:

The final velocity of the second player is 6.1 m/s.

Explanation:

The final velocity of the second player can be calculated by conservation of linear momentum (p):

p_{i} = p_{f}  

m_{a}v_{a_{i}} + m_{b}v_{b_{i}} = m_{a}v_{a_{f}} + m_{b}v_{b_{f}}  (1)

Where:

m_{a}: is the mass of the first football player = 110 kg

m_{b}: is the mass of the second football player = 90 kg

v_{a_{i}}: is the initial velocity of the first football player = 5.0 m/s

v_{b_{i}}: is the initial velocity of the second football player = 0 (he is at rest)

v_{a_{f}}: is the final velocity of the first football player = 0 (he stops after the impact)

v_{b_{f}}: is the final velocity of the second football player =?

By solving equation (1) for v_{b_{f}} we have:

110 kg*5.0 m/s + 0 = 0 + 90 kg*v_{b_{f}}

v_{b_{f}} = \frac{110 kg*5.0 m/s}{90 kg} = 6.1 m/s

Therefore, the final velocity of the second player is 6.1 m/s.

I hope it helps you!

8 0
3 years ago
If a machine has an efficiency of 94% and you apply 574J of work, how much work do you get out of the machine
Mariana [72]
<h3>Answer:</h3>

539.56 Joules

<h3>Explanation:</h3>
  • Efficiency of a machine is the ratio of work output to work input expressed as a percentage.
  • Efficiency = (work output/work input) × 100%
  • Efficiency of a machine is not 100% because so energy is lost due to friction of the moving parts and also as heat.

In this case;

Efficiency = 94%

Work input = 574 Joules

Therefore, Assuming work output is x

94% = (x/574 J) × 100%

0.94 = (x/574 J)

<h3>x = 539.56 J</h3>

Thus, you get work of 539.56 J from the machine

7 0
3 years ago
The acceleration due to gravity on earth is 9.8m/s2. What is the weight of a 75 kg person on earth?
MissTica
Weight= 9.8*75=735N
OptionD
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3 years ago
A damped harmonic oscillator consists of a mass on a spring, with a small damping force that is proportional to the speed of the
exis [7]

Answer:

2.19 N/m

Explanation:

A damped harmonic oscillator is formed by a mass in the spring, and it does a harmonic simple movement. The period of it is the time that it does one cycle, and it can be calculated by:

T = 2π√(m/K)

Where T is the period, m is the mass (in kg), and K is the damping constant. So:

2.4 = 2π√(0.320/K)

√(0.320/K) = 2.4/2π

√(0.320/K) = 0.38197

(√(0.320/K))² = (0.38197)²

0.320/K = 0.1459

K = 2.19 N/m

4 0
4 years ago
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kati45 [8]
C. distance

sometimes the equation can be:

w = F x D x cos(x)

where cos(x) is the angle between where the force is exerted and the object's displacement
4 0
3 years ago
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