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2 years ago

How would reversing the wheel’s initial direction of rotation affect the result??

Physics

1 answer:

d1i1m1o1n [39]2 years ago

4 0

It would change the sign on the vector quantities and have no change to the scalar quantities

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A physicist is constructing a solenoid. She has a roll of insulated copper wire and a power supply. She winds a single layer of

Leni [432]

Answer:

P =105.44 W

Explanation:

Given that

D= 10 cm ,L= 60 cm

d= 0.1 cm ,B= 6.4 mT

ρ= 1.7 x 10⁻⁸ Ω · m

The number of turns N

N= L/d

N= 60/0.1 = 600 turns

Length of wire

Lc= πDN

Lc= 3.14 x 0.1 x 600

Lc=188.4 m

The magnetic filed given as

$B=\dfrac{\mu_oNI}{L}$

$I=\dfrac{BL}{\mu_oN}$

Now by putting the values

$I=\dfrac{0.0064\times 0.6}{4\pi \times 10^{-7}\times 600}$

I=5.09 A

The resistance R given as

$R=\dfrac{\rho L_c}{A}$

$A=\dfrac{\pi}{4}\times d^2$

$A=\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2$

$R=\dfrac{1.7\times 10^{-8} \times 188.4}{\dfrac{\pi}{4}\times (0.1\times 10^{-2})^2}$

R=4.07 Ω

Power P

p =I²R

P= 5.09² x 4.07 W

P =105.44 W

5 0

2 years ago

The air in tires can support a car because gases __________.

____ [38]

Because the air inside the tires is kept at high pressure.

In fact, the force applied by the tires upwards to counter-balance the weight of the car (pushing downwards) is
$F=pA$
where p is the pressure of the air inside the tires and A is the area of contact between the tire and the car. Therefore, a higher pressure means a larger force F, and eventually if the pressure p is higher enough the force F will be large enough to counterbalance the weight of the car.

8 0

2 years ago

Which of Newton's laws accounts for the following statement? "A force cannot act alone." first law second law third law

enot [183]

Newton's laws A force cannot act alone is the THIRD LAW!

6 0

2 years ago

Read 2 more answers

A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp

Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

$y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}$

1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

2. Finding time t at position y = 1 cm:

$0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}$

3. Find velocity v at time t from equation 2:

$v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))$

5 0

2 years ago

Read 2 more answers

13. Calculate the total heat energy in Joules needed to convert 20 g of substance X from -10°C to 70°C?

sergeinik [125]

The heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

Explanation:

The heat energy required to convert a substance or to heat up or increase the temperature of a substance can be obtained from the specific heat formula.

As per this formula, the heat energy applied should be equal to the product of mass of the substance with temperature gradient and also with specific heat of the substance. Basically, the heat provided to increase or convert a substance should be more than the specific heat of the substance.

$Q = mc del T$

Since, here the mass of the substance X is given as m = 20g and the temperature change is given from -10°C to 70°C.

Then ΔT = (70-(-10))=70+10=80°C.

As the substance is unknown, the specific heat of that substance can also not be determined. Hence keep it as C.

$Q = 20*C*80$

Q = 1600C J

Thus, the heat required to convert the unknown substance X from one phase to another is 1600 J times the specific heat of that substance.

5 0

2 years ago