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dsp73
3 years ago
9

A 5 newton force and a 7 newton force act concurrently on a point. As the angle between the forces is increased from 0 to 180 th

e magnitude of the resultant of the two forces changes from
Physics
1 answer:
Reika [66]3 years ago
4 0

Answer:

The magnitude of the resultant decreases from A+B to A-B

Explanation:

The magnitude of the resultant of two vectors is given by

R=\sqrt{A^2 +B^2 +2AB cos \theta}

where

A is the magnitude of the first vector

B is the magnitude of the second vector

\theta is the angle between the directions of the two vectors

In the formula, A and B are constant, so the behaviour depends only on the function cos \theta. The value of cos \theta are:

- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is

R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B

- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B


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A cassette recorder uses a plug-in transformer to convert 120 V to 12.0 V, with a maximum current output of 200 mA. What is the
goldfiish [28.3K]

Answer:

20 mA

Explanation:

We have given that the transformer convert 120 V to 12 V means it is a step down transformer

The turn ratio of the transformer is given as a=\frac{N_1}{N_2}=\frac{V_1}{V_2}=\frac{120}{12}=10 here V_! is primary voltage and V_2 is secondary voltage

As the voltage decreases from primary to secondary so current will increase from primary to secondary

We have given output current that is secondary current so primary that is input current will be less

So input current =\frac{200}{10}=20mA

4 0
3 years ago
Question 8 of 10 When is there the least amount of heat transfer within a liquid? O A. When the substance changes into a gas OB.
AlexFokin [52]

Answer:

C. When the temperature of the liquid is the same throughout

Explanation:

4 0
2 years ago
The engine of a locomotive exerts a constant force of 8.1*10^5 N to accelerate a train to 68 km/h. Determine the time (in min) t
Bumek [7]

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of 8.1\times 10^5 N

Mass of train=1.9\times 10^7

Final speed (v)=68 km/h \approx 18.88 m/s

F=ma

so acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}

a=0.042631 m/s^2

and acceleration is

a=\frac{v-u}{t}

0.042631=\frac{18.88-0}{t}

t=443.089 \approx 443.1 s

8 0
3 years ago
Read 2 more answers
3. A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of
Alex Ar [27]

Answer:

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555

Again using Snell's law for blue light as :

n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283

The angular separation between the refracted red and refracted blue beams while they are in the glass is 42.555 - 42.283 = 0.272 degrees.

7 0
3 years ago
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
Read 2 more answers
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