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Vesnalui [34]
3 years ago
13

A box is being pulled to the right. What is the magnitude of the Kinect frictional force?

Physics
2 answers:
Anna35 [415]3 years ago
3 0
The answer to this question is A - 25 N
natima [27]3 years ago
3 0

The so-called "force" of friction always acts exactly opposite to the direction in which an object is moving.  So for this question, since the box is moving to the right, the "force" of friction is acting toward the left.

There's not enough information given in the question to determine its magnitude.

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Right Hand Rule 1 requires you to put the thumb of your right hand in the direction of current and your curled fingers will indi
Jet001 [13]
Answer: false
explanation: magnetic field does that
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3 years ago
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A power company charges its customers for electricity based upon which one of the following?a. powerb. energyc. power per unit o
Vlad [161]

B. Energy

A power company charges its customers for electricity based upon B. Energy.

<h3>Explanation:</h3>

Kilo-watt Hours (kWh) is the unit that measures the electricity consumption of customers. Since Power is defined as the rate at which electrical energy is transferred by an electrical circuit per unit time,

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

If energy is transmitted at a constant rate over a period of time, the total energy in kilowatt hours is the product of power in kilowatts(kW) and time in hours (h)

1\ kWh = 3.6\times 10^6\ Joules

7 0
3 years ago
Air (14.5 lb) undergoes a polytropic process in a closed system from p1 = 80 lbf/in2, υ1 = 4 ft3/lb to a final state where p2 =
Yanka [14]
The energy transfer in terms of work has the equation:

W = mΔ(PV)

To be consistent with units, let's convert them first as follows:

P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm

W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf

In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>


4 0
3 years ago
An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete
dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a)=2 m/s^2

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

v=0+2\times 6=12 m/s

After this object will start moving under gravity

height reached in first 6 s

s=ut+\frac{at^2}{2}

s=0+\frac{2\times 6^2}{2}

s=36 m

After fuel run out distance traveled in upward direction is

v^2-u^2=2as_0

here v=0

u=12 m/s

a=9.8 m/s^2

0-12^2=2(-9.8)(s)

s_0=\frac{144}{2\times 9.8}=7.34 m

s+s_0=36+7.34=43.34 m

7 0
3 years ago
Muscles that are fatigued ____. A. Are injured , B. Should always be rested , C. Can sometimes be exercised further D. Should be
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Answer:

<u>B. Should be rested</u>

Explanation:

That is the only reasonable answer.

5 0
3 years ago
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