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3 years ago

7

While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction b

etween the box and the floor is 0.43. The pushing force is directed downward at an angle θ below the horizontal. When θ is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find that value of θ.

1 answer:

Citrus2011 [14]3 years ago

6 0

Answer:

$\theta = 66.7 degree$

Explanation:

since force is applied downwards at some angle with the horizontal

so here we will have

$F_n = mg + Fsin\theta$

now we know that the box will not move if applied force is balanced by frictional force on it

so we will have

$Fcos\theta = \mu F_n$

$F cos\theta = \mu (mg + F sin\theta)$

$F(cos\theta - \mu sin\theta) = \mu mg$

$F = \frac{\mu mg}{cos\theta - \mu sin\theta}$

so here we can say

$cos\theta - \mu sin\theta > 0$

$tan\theta = \frac{1}{\mu}$

$\theta = tan^{-1}\frac{1}{\mu}$

$\theta = tan^{-1}(\frac{1}{0.43})$

$\theta = 66.7 degree$

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The system's tension is 616 N and acceleration is 5.6 $m / s^{2}$

<u>Explanation:</u>

From newton’s second law of motion which state that net force acting on a body is product of mass of a body and acceleration of a body which is given as,

$F_{n e t}=m_{t o t} \times a$

Where,

$F_{n e t}$ is net force acting on body

$m_{\mathrm{tot}}$ is mass of body

a is acceleration of body

Given values

Table mass (m) = 30 kg

Hanging mass (m) = 40 kg

$a=\frac{F_{n e t}}{m_{\mathrm{tot}}}=\frac{m \times g}{m_{\mathrm{tot}}}$

Put the value for m = hanging mass = 40 kg and $g=9.8 \mathrm{m} / \mathrm{s}^{2}$ , we get

$a=\frac{40 \times 9.8}{30+40}=\frac{392}{70}=5.6 \mathrm{m} / \mathrm{s}^{2}$

The tension in the ropes, $T=(m \times g)+(m \times a)$

Here, m as hanging mass

T = tension, N or $k g m / s^{2}$

m = mass, kg

g = gravitational force, $9.8 \mathrm{m} / \mathrm{s}^{2}$

a = acceleration, $m / s^{2}$

$T = (40 \times 9.8)+(40 \times 5.6) = 392+224 = 616 N$

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2 years ago

A certain string can withstand a maximum tension of 39. N without breaking. A child ties a 0.43 kg stone to one end and, holding

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Answer:

Bottom of the circle.

Explanation:

At the top of the circle the tension and the weight contribute on being the centripetal force, at the middle of the circle only the tension contributes on being the centripetal force (the weight being perpendicular to it), while <u>at the bottom</u> of the circle the tension contributes on being the centripetal force (as always) <em>but the weight against to it</em>, so here is where the tension must be greater to allow the same centripetal force as the other cases, thus here is where the string will break.

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2 years ago

A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half

bekas [8.4K]

Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

Kinetic energy of the hammer ,K.E.=

$\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J$

Half of the kinetic energy of the hammer is transformed into heat in the nail.

Energy transferred to the nail in one blow =

$\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J$

Total energy transferred after 3 blows,Q = $3\times 18.0625 J=54.1875 J$

Mass of the nail = 15 g = 0.015 kg

Change in temperature = $\Delta T$

Specif heat of the steel = c = 448 J/kg K

$Q=mc\Delta T$

$54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T$

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The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).

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2 years ago

You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m

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Answer:

Note that the emf induced is

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---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)] [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

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3 years ago

2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the

Luda [366]

Answer:

<em>12,25 km/h</em>

<em>≈ 3,4 m/s </em>

Explanation:

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>= 12,25km/h</em>

<em>or</em>

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>1h = 60m×60s = 3600s</em>

<em>= 12250m/3600s</em>

<em>≈ 3,4 m/s </em>

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2 years ago