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3 years ago

A girl is sitting in a sled sliding horizontally along some snow (there is friction present).

Physics

1 answer:

prisoha [69]3 years ago

6 0

Answer:

Ff = 159 [N]

Explanation:

Since in the problem give us two values of friction coefficient one static and the other dynamic, it is important to keep in mind that in order to solve this problem we must know the conditions in which the sled is located. We can read that the sled is in motion, in this way we use the dynamic coefficient of friction.

The friction force is calculated as the product of normal force, by the appropriate coefficient of friction value.

Ff = N * μ

Ff = 636 * 0.25

Ff = 159 [N]

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the fireman wishes to direct the flow of water from his hose to the fire at b. determine two possible angles u1 and u2 at which

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The two possible angles obtained by using the qudratic equation are;

θ $_{1}$ = 15.10° and θ2 = 73.51°

Given, speed of water = $v_{A}$ = 50ft/s

For the motion along x direction, time period can be calculated as follows:

$s_{x} = (v_{A}) _x_{} } t$

35 = (50 × cosθ) t

t = 0.64 / cosθ

For the motion in y direction, an equation can be obtained as follows:

$s_{y} = (v_{A})_{y} t +\frac{1}{2} (a_{y} )t^{2}$

$s_{y} = (-v_{A}$ $sin$ θ) $} t +\frac{1}{2} (a_{y} )t^{2}$

Plugging in the values we get:

$-20 = (-50_$ $sin$ θ) $} t +\frac{1}{2} (-32.2} )t^{2}$

-20 = -32tanθ - 10.304 $sec^{2}$ θ

Upon solving the above quadratic equation, we get,

tanθ = 0.27 , -3.38

Therefore,

tanθ $_{1}$ = 0.27

θ $_{1}$ = 15.10°

and, tanθ $_{2}$ = -3.38

θ $_{2}$ = 73.51

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1 year ago

Suppose you take a short piece of wire that is not attached to anything and move it up and down in a magnetic field. Explain whe

Otrada [13]

Answer:

No.

Explanation:

According to Faraday's law, the induced emf in the circuit is given by :

$e=\dfrac{d\phi}{dt}$ , it is proportional to the rate of change of magnetic flux.

In this case, a short piece of wire that is not attached to anything and move it up and down in a magnetic field. It means that the circuit is not completed here. It is an open circuit. For the induction of current, a circuit must be completed.
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3 years ago

Read 2 more answers

A 4kg block and a 2kg block can move on horizontal frictionless surface. The blocks are accelerated by a +12-N force that pushes

Stolb23 [73]

Answer:

a) -4 N

b) +4 N

Explanation:

Draw a free body diagram for each block.

For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.

For the small block, there is 1 force, F pushing to the right.

There are also weight and normal forces in the vertical direction, but we can ignore those.

Sum of forces on the large block in the x direction:

∑F = ma

12 − F = 4a

Sum of forces on the small block in the x direction:

∑F = ma

F = 2a

2F = 4a

Substitute:

12 − F = 2F

12 = 3F

F = 4

The small block pushes on the large block 4 N to the left (-4 N).

The large block pushes on the small block 4 N to the right (+4 N).

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3 years ago

An object moves on a trajectory given by Bold r left parenthesis t right parenthesis equals left angle 10 cosine 6 t comma 10 si

spayn [35]

Answer:

Explanation:

(r) = <10 cos 6t, 10 Sin 6t>

The distance traveled by the object is the magnitude of vector r.

The magnitude of vector r is given by

$r = \sqrt{(10 Cos 6t)^{2}+(10 Sin 6t)^{2}}$

$r =10 \sqrt{(Cos^{2} 6t)+(Sin^{2} 6t)$

r = 10

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3 years ago

Suppose you were asked to find the torque about point p due to the normal force n in terms of given quantities. which method of

igomit [66]

Answer:

T=Lnsin $\alpha$

Please check the attached

Explanation:

The torque can simply be calculated by multiplying the length of the rod by the perpendicular force n as shown in the attached figure.

Note that sin90=1

T=Lsin $\alpha$ (nsin90)

T=Lsin $\alpha$ xn

T=Lnsin $\alpha$

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3 years ago