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2 years ago

15

A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F

= 40.0 N that is directed at an angle of 37.0∘ below the horizontal, and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton’s laws to calculate the normal force that the floor exerts on the chair.

1 answer:

Tanya [424]2 years ago

3 0

Answer: Normal force, N = 141.64 Newton

Explanation:

All the forces acting on the system and described in free body diagram are:

1) gravitational pull in downward direction

2) Normal force in upward direction

3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:

i) F Cos 37° along the horizontal plane in forward direction and

ii) F Sin 37° along the vertical plane in downward direction

Applying the Newton's second law, net forces in the vertical plane are:

Net force, f = N - (mg + F Sin 37°)

As there is no acceleration in the vertical plane hence, net force f = 0.

So,

N - (mg + F Sin 37°) = 0

Adding (mg + F Sin 37°) both the sides in above equation, we get

N = mg + F Sin 37°

N = 12 $\times$ 9.8 + 40 $\times$ 0.601 because (Sin 37° = 0.601)

N = 117.6 + 24.04

N = 141.64 Newton

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In coordinates with the origin at the barn door, the cow walks from x 0 to x 6.9 m as you apply a force with x component Fx 320.

Stella [2.4K]

Answer:

-209.42J

Explanation:

Here is the complete question.

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?

Solution

The work done by a force W = ∫Fdx since our force is variable.

Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.

So, the workdone by the force on the cow is

W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx

= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx

= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹

= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]

= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J

= -209.415 J ≅ -209.42J

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2 years ago

A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0

GuDViN [60]

Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

$E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}$

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

$F=qE$

Put the value into the formula

$F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}$

We need to calculate the total charge on the rod

$Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}$

Put the value into the formula

$Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}$

$Q=47.8\times10^{-9}\ C$

$Q=47.8\ nC$

Hence, The total charge on the rod is 47.8 nC.

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2 years ago

A stranded soldier shoots a signal flare into the air to attract the attention of a nearby plane. The flare has an initial verti

Ipatiy [6.2K]

Answer:

Explanation:

h = ut - 16 t² = ut - 1/2 x32 t² = ut - 1/2 g t² , g = acceleration = - 32 ft / s²

1) v² = u² - 2 g h , v = 0 so

h = u² / 2g = 1500² / 2 x 32 = 35156.25 ft

2) v = u - gt

t = u / g = 1500 / 32 = 46.875 s

3) It will hit the ground after 2 x 46.875 = 93.75 s

4 ) time to reach 30000 ft height t is given by

h = ut - 16 t²

30000 = 1500t - 16t²

16t²-1500t + 30000 = 0

t = 28.92 s and 64.82 s

Time required to travel 50000 by plane

= 50000/880 = 56.82 . There is no match of timing so plane will not hit it.

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2 years ago

What do energy advisors mean by the phrase "the greenest kilowatt is the one you never use?"

d1i1m1o1n [39]

Answer:

"Energy deficiency, no coal-burning, no-cost mining pollution" is the correct answer.

Explanation:

“The greenest kilowatt-hour seems to be the one this really doesn't should use,” explained Joe Stepenovitch, co-owner as well as COO of something like the electricity IQ Group. Whether a kilowatt becomes generated is far less essential instead of not needing to do something with it.
It, therefore, reduces operational costs, appeals to progressives and green-conscious consumers, prepares the business for impending emissions reductions policy caps, as well as coincides with you including an imminent future focused on renewable energy sources.

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1 year ago

A seaplane flies horizontally over the ocean at 50 meters/second. It releases a buoy, which lands after 21 seconds. What's the v

pantera1 [17]

The motion of the buoy consists of two independent motions on the horizontal and vertical axis.

On the horizontal axis, the motion of the buoy is a uniform motion with constant speed $v=50 m/s$ . On the vertical axis, the motion of the buoy is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ . The vertical position of the buoy at time t is given by
$y(t)=h- \frac{1}{2}gt^2$
where h is the initial heigth of the buoy when it is released from the plane. At the time t=21 s, the buoy reaches the ground, so y(21 s)=0. If we substitute these two numbers inside the equation, we can find the value of h, the vertical displacement from the plane to the ocean:
$0=h- \frac{1}{2}gt^2$
$h= \frac{1}{2}gt^2= \frac{1}{2}(9.81 m/s^2)(21 s)^2=2163 m$

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2 years ago