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natulia [17]
3 years ago
5

A box is moved 20 m across a smooth floor by a force making a downward angle with the floor, so that there is effectively a 10 N

force acting parallel to the floor in the direction of motion and a 5 N force acting perpendicular to the floor the work done is:
Physics
1 answer:
Usimov [2.4K]3 years ago
5 0

Answer:

100 J

Explanation:

From the question, The work done by the forces in moving the box is given as

W = FxdcosФ+Fydcosα................... Equation 1

Where W = Work done, Fx = force acting parallel to the floor, d = distance moved by the box, Ф = angle the parallel force makes with the floor, Fy = force acting perpendicular to the floor, α = angle the perpendicular force make with the floor.

Give: Fx = 10 N, d = 20 m, Fy = 5 N, Ф = 0°, α = 90°

Substitute into equation 1

W = 10×10×cos0°+5×20×cos90°

W = 10×10×1+0

W = 100 J.

Note: The work done by the perpendicular force is zero

Hence the work done = 100 J

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The power that a student generates when walking at a steady pace of vw is the same as when the student is riding a bike at vb =
Nadya [2.5K]

Answer:

1/9

Explanation:

Sorry, I don't know if this is right, but here is what I did. We are ignoring potential energy because we assume that the student is walking and biking on level ground. Power = W/T, W = Mechanical Energy, or just Kinetic for this case. So P_{w}=\frac{E_{w} }{T}, and P_{b} =\frac{E_{b} }{T}. Ew = \frac{1}{2} mv_{w}^{2}, and Eb =\frac{1}{2} m(3v_{w})^{2}. Put Ew over Eb. the 1/2's cancel, the m's cancel, and you are left with \frac{v_{w} ^{2} }{9v_{w}^{2} }. Finally, this simplifies after cutting out the vw^2's to 1/9.

6 0
3 years ago
Three resistors, A, B, and C, are connected in parallel and attached to a battery, with the resistance of A being the smallest a
FromTheMoon [43]

Answer:

Resistor A

Explanation:

Resistors A, B and C are connected in parallel to a battery of voltage represented by V.

Since they are in parallel, the same voltage, V, passes across the three of them.

And from Ohm's law, the voltage (V) through a resistor is the product of the current (I) flowing through it and the resistance (R) of the resistor. i.e

V = I x R

=> I = \frac{V}{R}

Therefore, assuming the values of the resistances of resistors A, B and C are A, B and C respectively,

(i) the current, I_{A} through A is

I_{A} = \frac{V}{A}

(ii) the current, I_{B} through B is

I_{B} = \frac{V}{B}

(iii) the current, I_{C} through C is

I_{C} = \frac{V}{C}

From the foregoing, it can be deduced that the current is inversely proportional to the resistance. Therefore, the higher the resistance, the lower the current. Consequential of this, the resistor that carries the highest current is the one with the smallest resistance, which is A

5 0
3 years ago
A flat, rectangular coil consisting of 60 turns measures 23.0 cmcm by 34.0 cmcm . It is in a uniform, 1.40-TT, magnetic field, w
Travka [436]

Answer:

(a)   ΔФ = -0.109W

(b)  emf = 28.43V

(c)   Iin = emf/R

Explanation:

(a) In order to calculate the magnetic flux you use the following formula:

\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)   (1)

B: magnitude of the magnetic field = 1.40T

A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2

Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.

Then, you obtain:

\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W

The change in the magnetic flux is -0.109 W

(b) During the rotation of the coil the emf induced is given by:

emf=-N\frac{\Delta \Phi}{\Delta t}         (2)

N: turns of the coil = 60

ΔФ: change in the magnetic flux = 0.109W

Δt: lapse time of the rotation = 0.230s

You replace the values of the parameters in the equation (2):

emf=-(60)(\frac{-0.109W}{0.230s})=28.43V

The induced emf is 28.43V

(c) The induced current in the coil is given by:

I_{in}=\frac{emf}{R}      (3)

R: resistance of the coil     (it is necessary to have this value)

emf :induced emf  = 28.43V

7 0
4 years ago
two identical, thin rods, each with mass mm and length ll, are joined at right angles to form an l-shaped object. this object is
Allushta [10]

The length of time that an action or state persists. The given L-shaped object made of two rods oscillates at a frequency of 2π√√2L/3g.

The rod weighs m grams.

The length of a single rod is L.

If it is connecting at, as indicated by the query, then the moment of inertia of the combined system follows.

mL²/3+3L²/3=2mL²/3

Here, I denotes the rod's moment of inertia, d denotes the separation of the center of gravity from the fixed point, and T denotes the length of time the L-shaped object oscillates.

Length of pendulam

Lcos45=L*1/√2=L/√2

Caluclate the time period

T=2π√I/mgl

=2π√2*mL²*√2/3*2mgl

oscillates at a frequency = 2π√√2L/3g

Learn more about frequency here:

brainly.com/question/10254661

#SPJ4

3 0
1 year ago
In a two-source circuit, one source acting alone produces 10 ma through a given branch. the other source acting alone produces 8
pashok25 [27]
Refer to the figure below.
R = resistance.

Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R

Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R

Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.

Answer: 2 mA

6 0
3 years ago
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