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3 years ago

I have to submit my homework soon and I am not familiar with these concepts. can anybody do then for me?

Physics

1 answer:

AlladinOne [14]3 years ago

6 0

#4). The concepts are: A). acceleration is always in the direction of the force, and B). friction always acts in the direction opposite to motion. (that's B)

#5). The concepts are: A). the NET force is the sum of all the individual forces acting (on this car, it's 1600N forward). and B). Force = (mass) x (acceleration). So Acceleration = (force) / (mass). For this car, that's (1600N forward) / (800 kg) .

#6). The concept is: As long as you don't exceed the "proportionality limit" of a spring, its extension is proportional to the load on it. That means that the change in extension is always proportional to the change in the load. So now, look at the table: As long as the load is 10N or less, the spring stretched 3cm longer for every 2N more of load. But if the load is somewhere between 10N and 12N, that relationship disappears. Something changes between 10N and 12N of load. The spring's "elastic limit" is somewhere in that slot.

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A force of 500 N is used to slide a box across a smooth surface; the box moved 5 m in 1.2 seconds. What power is used?

Katen [24]

F = 500N
S = 5m
T = 1.2s
----------------------------------
P = ?

We can use formula P = F * V, just because it was a smooth slide we can assume that average speed was V = S/T = 5 / 1.2 = 50/12

So the final answer would be:

P = F * V = 500 * 50/12

Are you sure those are correct numbers? The answer don't look nice :D

8 0

3 years ago

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According to Einstein's special theory of relativity, in which frames of reference will all laws of physics apply?

kenny6666 [7]

Answer: Within any frame of reference that is accelerating

Special relativity was proposed on 1905 by Einstein, who developed his theory based on the following two postulates:

1. The laws of physics are the same in all inertial systems. There is no preferential system.

2. The speed of light in vacuum has the same value for all inertial systems.

Focusing on the first postulate, it can be affirmed that any measurement on a body is made with reference to the system in which it is being measured.

Now, taking into account that an inertial reference system is the one that complies with the principle of inertia:

"For a body to have acceleration, an external force must act on it."

The correct answer is

Within any frame of reference that is accelerating

3 0

3 years ago

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-g With what tension must a rope with length 2.50 m and mass 0.120 kg be stretched for transverse waves of frequency 40.0 Hz to

Alex17521 [72]

Answer:

43.2 N

Explanation:

$\lambda$ = Wavelength = 0.75 m

f = Frequency = 40 Hz

m = Mass of string = 0.12 kg

L = Length of string = 2.5 m

$\mu$ = Linear density = $\dfrac{m}{L}$

Velocity of wave is given by

$v=\lambda f\\\Rightarrow v=0.75\times 40\\\Rightarrow v=30\ m/s$

The tension in string is given by

$T=v^2\mu\\\Rightarrow T=30^2\times \dfrac{0.12}{2.5}\\\Rightarrow T=43.2\ N$

The tension in the string is 43.2 N

8 0

3 years ago

Students have a small steel ball and a large steel ball, and they have a short ramp and a tall ramp.

lozanna [386]

I would say it’s “B” for the reason that the student did not experiment with a short ramp and large ball as well as a large ramp and a short ball. They have to do those experiments as well before drawing their conclusion.

3 0

1 year ago

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Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.

kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees

ELECTRIC FORCE (F)

F = $\frac{KQq}{d^{2} }$

Where K = 9 x $10^{9}$ N $m^{2}$ / $C^{2}$

The distance between $q_{1}$ and $q_{3}$ can be calculated by using Pythagoras theorem.

d = $\sqrt{33^{2} + 33^{2} }$

d = 46.7 cm = 0.467 m

For force $F_{1}$ , substitute all the parameters into the formula above

$F_{1}$ = (9 x $10^{9}$ x 3 x 1)/ $0.467^{2}$

$F_{1}$ = 2.7 x $10^{10}$ /0.218

$F_{1}$ = 1.24 x $10^{11}$ N

For force $F_{4}$ , substitute all the parameters into the formula above

$F_{4}$ = (9 x $10^{9}$ x 3 x 4)/ $0.33^{2}$

$F_{4}$ = 1.08 x $10^{11}$ /0.1089

$F_{4}$ = 9.92 x $10^{11}$ N

For force $F_{2}$ , substitute all the parameters into the formula above

$F_{2}$ = (9 x $10^{9}$ x 3 x 2)/ $0.33^{2}$

$F_{2}$ = 5.4 x $10^{10}$ /0.1089

$F_{2}$ = 4.96 x $10^{11}$ N

Summation of forces on Y component will be

$F_{y}$ = $F_{4}$ - $F_{1}$ Sin 45

$F_{y}$ = 9.92 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{y}$ = 9.04 x $10^{11}$ N

Summation of forces on X component will be

$F_{x}$ = $F_{2}$ - $F_{1}$ Cos 45

$F_{x}$ = 4.96 x $10^{11}$ - 1.24 x $10^{11}$ Sin 45

$F_{x}$ = 4.08 x $10^{11}$ N

Net Force = $\sqrt{F_{x} ^{2} + F_{y} ^{2} } }$

Net force = $\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }$

Net force = 9.9 x $10^{11}$ N

The direction will be

Tan ∅ = $F_{y}$ / $F_{x}$

Tan ∅ = 9.04 x $10^{11}$ / 4.08 x $10^{11}$

Tan ∅ = 2.216

∅ = $Tan^{-1}$ (2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x $10^{11}$ N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0

2 years ago

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