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2 years ago

how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance

Physics

1 answer:

Sergio039 [100]2 years ago

8 0

Answer:

W = F * s

Work done equals applied force * distance traveled

Apparent weight = M g (1 - sin θ) since some of applied force will lighten sled

μ = coefficient of kinetic friction

F cos θ = force applied to motion of sled

s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

Note that F = μ M g if applied force is in the horizontal direction

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An 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its

nikdorinn [45]

Answer:

The average induced emf in the loop is 0.20 V

Explanation:

Given:

Radius of loop $r = \frac{d}{2} = 9.25 \times 10^{-2}$ m

Magnetic field $B = 1.5$ T

Change in time $\Delta t = 0.20$ sec

According to the faraday's law,

Induced emf is given by

$\epsilon = -\frac{\Delta \phi}{\Delta t}$

Where $\phi =$ magnetic flux

$\phi = BA\cos0$ ( here $\theta = 0$ )

Where $A = \pi r^{2}$

We neglect minus sign because it's shows lenz law

$\epsilon = \frac{B \pi r^{2} }{\Delta t}$

$\epsilon = \frac{1.5 \times 3.14 \times (9.25 \times 10^{-2} )^{2} }{0.20}$

$\epsilon = 0.20$ V

Therefore, the average induced emf in the loop is 0.20 V

4 0

3 years ago

Add these two velocity vectors to find the magnitude of their resultant vector.

hammer [34]

The magnitude of their resultant vector is 4.6 meters/seconds

Since we are to add the velocity vectors in order to find the magnitude of their resultant vector.

Hence:

Resultant vector magnitude=5.8 meters/seconds + (1.2 meters/seconds)

Resultant vector magnitude=5.8 meters/seconds-1.2 meters/seconds

Resultant vector magnitude 4.6 meters/seconds

Inconclusion The magnitude of their resultant vector is 4.6 meters/seconds

Learn more here:

brainly.com/question/11134601

6 0

3 years ago

Molds survive best in A) warm, most B) dry, cold C) bright, sunny places

lana66690 [7]

A..........................................

3 0

3 years ago

Read 2 more answers

A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the

olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0

3 years ago

What velocity must a car with a mass of 1280 kg have in order to have the same momentum as a 2230?

sveta [45]

Momentum = (mv).
(2110 x 24) = 50,640kg/m/sec. truck momentum. 
Velocity required for car of 1330kg to equal = (50,640/1330), = 38m/sec

8 0

3 years ago