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3 years ago

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ONLINE CALCULATOR .A force of 187 pounds makes an angle of 73 degrees 36 ' with a second force. The resultant of the two forces

makes an angle of 29 degrees 1 ' to the first force. Find the magnitudes of the second force and of the resultant.

1 answer:

saul85 [17]3 years ago

3 0

Answer:

The magnitudes of the second force is $Z = 129.9 N$

The magnitudes of the resultant force is $R = 256.047 N$

Explanation:

From the question we are told that

The force is $F = 187 \ lb$

The angle made with second force $\theta_o = 73 ^o 36' = 73 + \frac{36}{60} = 73.6^o$

The angle between the resultant force and the first force $\theta _1 = 29 ^o 1 ' = 29 + \frac{1}{60} = 29.0167^o$

For us to solve problem we are going to assume that

The magnitude of the second force is Z N

The magnitude of the resultant force is R N

According to Sine rule

$\frac{F}{sin (\theta _o - \theta_1 } = \frac{Z}{\theta _1}$

Substituting values

$\frac{187}{sin(73.3 - 29.01667)} =\frac{Z}{sin (29.01667)}$

$267.82 =\frac{Z}{0.4851}$

$Z = 129.9 N$

According to cosine rule

$R = \sqrt{F ^2 + Z^2 + 2(F) (Z) cos (\theta _o) }$

Substituting values

$R = \sqrt{187^2 + 129.9 ^2 + 2 (187 ) (129.9) cos (73.6)}$

$R = 256.047 N$

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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

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Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

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$\Delta t$ is the time interval

The change in magnetic flux is given by

$\Delta \phi = \phi _{f} - \phi _{i}$

Where $\phi _{f}$ is the final magnetic flux

and $\phi _{i}$ is the initial magnetic flux

Magnetic flux is given by the formula

$\phi = BAcos(\theta)$

Where $B$ is the magnetic field

$A$ is the area

and $\theta$ is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, $\theta = 0^{o}$

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

$A = \pi r^{2}$

∴ $A = \pi (0.15)^{2}$

$A = 0.0225\pi$

∴ $\phi_{i} = (1.5)(0.0225\pi)cos(0^{o} )$

$\phi_{i} = (1.5)(0.0225\pi)$

( NOTE: $cos (0^{o}) = 1$ )

$\phi_{i} = 0.03375\pi$ Wb

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The field pointed upwards, that is $\theta = 90^{o}$ . Since $cos (90^{o}) = 0$

Then

$\phi_{f} = 0$

Hence,

$\Delta \phi = 0- 0.03375\pi$

$\Delta \phi = - 0.03375\pi$

From the question

$\Delta t = 2.8 ms = 2.8 \times 10^{-3} s$

Here, $N$ = 1

Hence,

$Emf = -N\frac{\Delta \phi}{\Delta t}$ becomes

$Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }$

$Emf = 37.9 V$

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

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3 years ago

A large crate is pushed across the floor with an effort of 45 Newtons. The box is pushed a distance of 3.5 meters. How much work

I am Lyosha [343]

Answer:

$\boxed {\boxed {\sf 157.5 \ Joules }}$

Explanation:

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Substitute the values into the formula.

$W= 45 \ N * 3.5 \ m$

Multiply.

$W= 157.5 \ N*m$

1 Newton meter is equal to 1 Joule
Our answer of 157.5 N*m equals 157.5 J

$W= 157.5 \ J$

<u>157.5 Joules </u> of work are done on the crate.

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