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3241004551 [841]

3 years ago

15

Freida stands on a hill with a slope of 12 degrees. If her mass is 65 kg, what is the magnitude of the normal force acting on he

r?
A. 623 N
B. 637 N
C. 132 N
D. 601 N

2 answers:

aliina [53]3 years ago

7 0

Answer:

b

Explanation:

Svetradugi [14.3K]3 years ago

7 0

Answer:

b

Explanation:

I did my math and made sur got it correct

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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

3 years ago

Question 4 (2 points)

Igoryamba

The answer is a) Teres Major Muscle

7 0

2 years ago

Read 2 more answers

A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the

Goshia [24]

To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.

The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as

$P = \frac{E}{t}$

Here,

E = Energy

t = time

As the energy is equal to the potential Energy we have tat

$P = \frac{mgh}{t}$

The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:

$P = \frac{700*8}{10}$

$P = 560W$

Therefore the correct answer is A.

8 0

4 years ago

A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

labwork [276]

Answer:

a

$\theta _2 = 13^o$

b

$\theta _1 =32.94^o$

c

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

So

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

4 0

3 years ago

In a cell what substance is analogous to a factory manger and where would it be found?

charle [14.2K]

The substance is DNA and it would be found in the nucleus.

5 0

3 years ago