A) 1.05 N
The power dissipated in the circuit can be written as the product between the pulling force and the speed of the wire:
![P=Fv](https://tex.z-dn.net/?f=P%3DFv)
where
P = 4.20 W is the power
F is the magnitude of the pulling force
v = 4.0 m/s is the speed of the wire
Solving the equation for F, we find
![F=\frac{P}{v}=\frac{4.20 W}{4.0 m/s}=1.05 N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BP%7D%7Bv%7D%3D%5Cfrac%7B4.20%20W%7D%7B4.0%20m%2Fs%7D%3D1.05%20N)
B) 3.03 T
The electromotive force induced in the circuit is:
(1)
where
B is the strength of the magnetic field
v = 4.0 m/s is the speed of the wire
L = 10.0 cm = 0.10 m is the length of the wire
We also know that the power dissipated is
(2)
where
is the resistance of the wire
Subsituting (1) into (2), we get
![P=\frac{B^2 v^2 L^2}{R}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BB%5E2%20v%5E2%20L%5E2%7D%7BR%7D)
And solving it for B, we find the strength of the magnetic field:
![B=\frac{\sqrt{PR}}{vL}=\frac{\sqrt{(4.20 W)(0.350 \Omega)}}{(4.0 m/s)(0.10 m)}=3.03 T](https://tex.z-dn.net/?f=B%3D%5Cfrac%7B%5Csqrt%7BPR%7D%7D%7BvL%7D%3D%5Cfrac%7B%5Csqrt%7B%284.20%20W%29%280.350%20%5COmega%29%7D%7D%7B%284.0%20m%2Fs%29%280.10%20m%29%7D%3D3.03%20T)
Complete Question
A student bikes to school by traveling first dN = 0.900 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.
Similarly, let d⃗ W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.
Express your answer as two numbers separated by a comma. Be careful with your signs.
Answer:
The value is ![dT = ( -0.3, 0.8)](https://tex.z-dn.net/?f=dT%20%20%3D%20%20%28%20-0.3%2C%20%200.8%29)
Explanation:
From the question we are told that
The first displacement is
i.e positive y-axis
The second displacement is
i.e negative x-axis
The final displacement is
i.e negative y-axis
Generally dW in component for is
Generally the total displacement of the student is mathematically represented as
![dT = ( -0.3, (0.90 - 0.10))](https://tex.z-dn.net/?f=dT%20%20%3D%20%20%28%20-0.3%2C%20%20%280.90%20-%200.10%29%29)
![dT = ( -0.3, 0.8)](https://tex.z-dn.net/?f=dT%20%20%3D%20%20%28%20-0.3%2C%20%200.8%29)
Answer:
4.988kW
Explanation:
According to the question, energy E extracted from the ocean breaker is directly proportional to the intensity I. It can be expressed mathematically as E ∝ I
E = kI where k is the constant of proportionality.
From the formula; k = E/I
This shows that increase in energy extracted will lead to increase in its intensity and vice versa.
If the device produces 10.0 kW of power on a day when the breakers are 1.20 m high
E = 10kW and I = 1.20m
k = 10/1.20
k = 8.33kW/m
To know how much energy E that will be produced when they are 0.600 m high, we will use the same formula
k = E/I where;
k = 8.33kW/m
I = 0.600m
E = kI
E = 8.33 × 0.6
E = 4.998kW
The device will produce energy of 4.998kW when they are 0.600m high.
Answer:
Fp = 26.59[N]
Explanation:
This problem can be solved using the principle of work and energy conservation, i.e. the final kinetic energy of a body will be equal to the sum of the forces that do work on the body plus the initial kinetic energy.
We need to identify the initial data:
d = distance = 41.9[m]
Ff = friction force = 44.5 [N]
m = mass = 16.3 [kg]
v1 = 1.9 [m/s]
v2 = 12.6 [m/s]
The kinetic energy at the beginning can be calculated as follows:
![E_{k1}= \frac{1}{2}*m*v_{1}^2 \\E_{k1}= \frac{1}{2}*16.3*(1.9)_{1}^2\\E_{k1}= 29.42[J]](https://tex.z-dn.net/?f=E_%7Bk1%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B1%7D%5E2%20%5C%5CE_%7Bk1%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2A16.3%2A%281.9%29_%7B1%7D%5E2%5C%5CE_%7Bk1%7D%3D%2029.42%5BJ%5D)
And the final kinetic energy.
![E_{k2}= \frac{1}{2}*m*v_{2}^2 \\E_{k2}= \frac{1}{2}*16.3*(12.6)^2\\E_{k2}= 1294[J]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2Am%2Av_%7B2%7D%5E2%20%5C%5CE_%7Bk2%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%2A16.3%2A%2812.6%29%5E2%5C%5CE_%7Bk2%7D%3D%201294%5BJ%5D)
The work is performed by two forces, the friction force and the pushing force, it is important to clarify that these forces are opposite in direction.
The weight of the cart also performs a work in the direction of movement since the plane is tilted down, this component of the weight of the cart must be parallel to the surface of the inclined plane.
![W_{1-2}=-(44.5*41.9)+(16.3*9.81*sin(17.5)*41.9)+(F_{p}*41.9) \\therefore:\\E_{k1}+W_{1-2}=E_{k2}\\29.42+150.16+(F_{p}*41.9)=1294\\F_{p}=1114.42/41.9\\F_{p}=26.59[N]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D-%2844.5%2A41.9%29%2B%2816.3%2A9.81%2Asin%2817.5%29%2A41.9%29%2B%28F_%7Bp%7D%2A41.9%29%20%5C%5Ctherefore%3A%5C%5CE_%7Bk1%7D%2BW_%7B1-2%7D%3DE_%7Bk2%7D%5C%5C29.42%2B150.16%2B%28F_%7Bp%7D%2A41.9%29%3D1294%5C%5CF_%7Bp%7D%3D1114.42%2F41.9%5C%5CF_%7Bp%7D%3D26.59%5BN%5D)
Answer:
Explanation:
the scales of temperature are kelvin,celsius and fahrenheit so to interconvert them use the formula:
K-273/100 = C/100 = F-32/180