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Stells [14]
2 years ago
6

A 2kg object is moving horizontally with a speed of 4 m/s. How much net force is required to keep the object moving at this spee

d and in this direction
Physics
2 answers:
Elza [17]2 years ago
8 0
We can solve this problem using the force equation.

Force = Mass * Acceleration

2kg * 4m/s = 8 N

The net force required to keep the object moving at this speed and in this direction is 8 N.


solniwko [45]2 years ago
8 0

Answer: F = 0, if you apply a force, the velocity of the object may change.

Explanation: If there is no force of friction or something like that, there is no actual force applied to the object, because it has not any acceleration. This means that the object reached an uniform motion state, this means that the object will move with the constant velocity until a force is applied over it. This is because if there is no force applied, there is no acceleration (by the second Newton's law, we have that F = m*a) and the acceleration is the rate of change of the velocity, so if there is no acceleration, there is no change in the velocity.

This means that the net force required to keep the object moving at 4m/s and un the same direction is zero; F = 0

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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

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2 years ago
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The speed of light in benzene is 2.00×108 m/s. what is the index of refraction of benzene?
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n= \frac{c}{v}
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There have been several proposed atomic models during the last 150 years. Which model best illustrates the Bohr model. This mode
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3 0
2 years ago
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An arrow is shot vertically upward at a rate of 250ft/s. Use the projectile formula h=−16t2+v0t to determine at what time(s), in
SIZIF [17.4K]

Answer:

The arrow is at a height of 500 feet at time t = 2.35 seconds.

Explanation:

It is given that,

An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s

The projectile formula is given by :

h=-16t^2+v_ot

We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,

-16t^2+250t=500

On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds

So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.

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