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Harlamova29_29 [7]
2 years ago
13

A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f

riction *
Fg sin 36 degrees
Fg cos 36 degrees
Fg tan 36 degrees
Fg x N
Physics
1 answer:
sweet-ann [11.9K]2 years ago
5 0

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.

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PtichkaEL [24]

Answer:

\tau=3.3*10^{-6}s

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

C*\frac{dV}{dt}+\frac{V}{R}=0

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

\frac{dV}{V}=-\frac{1}{RC}dt

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt

Evaluating the integrals:

ln(\frac{V}{v})=e^{\frac{-t}{RC} }

natural logarithm to both sides in order to isolate V:

V(t)=ve^{-\frac{t}{RC} }

Where the term RC is called time constant and is given by:

\tau=R*C=10*(0.330*10^{-6})=3.3*10^{-6}s

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3 years ago
If the wavelength is 10 m and the frequency is 5 Hz, what is the wave speed?
timama [110]
V = f x wavelength
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4 0
2 years ago
Two 8.0 Ω lightbulbs are connected in a 12 V parallel circuit. What is the power of both glowing bulbs?
kati45 [8]

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3 years ago
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio
Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is   = 5.676*10^{-3} \ mm

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft  × 4  sorings

K = 100 lb/ft

Amplitude of the microscope  \frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon  \frac{\omega}{W_n})^2}]

where;

\epsilon = 0

W_n = \sqrt { \frac{k}{m}}

= \sqrt { \frac{100*32.2}{200}}

= 4.0124

replacing them into the above equation and making X the subject of the formula:

X = Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}

X = 0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}

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8 0
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You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type
alexdok [17]

Answer:

30 N \cdot m

Explanation:

The torque applied by a force can be calculated as

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In this problem, we have

F = 15 N

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Substituting into the equation, we find

\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m

7 0
3 years ago
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