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AnnyKZ [126]
3 years ago
7

Two boxes on a horizontal plane with coefficient of friction µ are connected by a massless string. The left-hand box has mass m

and the right-hand box has mass 2m. A force F is applied to the right-hand mass at an angle θ with respect to the horizontal so that they move to the right at constant acceleration. What is the magnitude of the tensi
Physics
1 answer:
pochemuha3 years ago
5 0

Answer:

T = (3μmg - Fcosθ)/2

Explanation:

Since the boxes move to the right, the net force on box of mass 2m is

Fcosθ - 2μmg + T = ma (1)where Fcosθ = horizontal component of applied force, 2μmg = frictional force and T = tension in string and a = acceleration of box

The net force on the box with mass m is

μmg - T = ma (2) where μmg = frictional force, T = tension and a = acceleration of boxes.

Equating (1) and (2) we have

Fcosθ - 2μmg + T = μmg - T

collecting like terms

Fcosθ - 2μmg - μmg = -T - T

Fcosθ - 3μmg = -2T

(3μmg - Fcosθ)/2 = T

T = (3μmg - Fcosθ)/2

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Question 10 of 34
labwork [276]

Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4  \ block /min

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

#SPJ1

3 0
2 years ago
I will give the Branliest!!! In which body of water can you see your reflected image better, a pond or an ocean? Why? (Please ex
GREYUIT [131]

Answer:

I would say a pond

Explanation:

A pond is more still than an ocean, therefore you could see your reflection better

4 0
3 years ago
Read 2 more answers
A car is moving in the positive direction along a straight highway and accelerates at a constant rate while going from point A t
rusak2 [61]

Answer:

The time where the avergae speed equals the instaneous speed is T/2

Explanation:

The velocity of the car is:

v(t) = v0 + at

Where v0 is the initial speed and a is the constant acceleration.

Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

v_{ave} = \frac{1}{T}\int\limits^T_0 {v(t)} \, dt

v_ave = v0+a(T/2)

We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>

v(T/2)=v_ave

Now we want to know where the car should be, the osition of the car is:

x(t) = x_A + v_0 t + \frac{1}{2}at^2

Where x_A is the position of point A. Therefore, the car will be at:

<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>

8 0
3 years ago
A vertical cylindrical tank 10 ft in diameter, has an inflow line of 0.3 ft inside diameter and an outflow line of 0.4 ft inside
neonofarm [45]

Answer:

\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}, level is rising.

Explanation:

Since liquid water is a incompresible fluid, density can be eliminated of the equation of Mass Conservation, which is simplified as follows:

\dot V_{in} - \dot V_{out} = \frac{dV_{tank}}{dt}

\frac{\pi}{4}\cdot D_{in}^2 \cdot v_{in}-\frac{\pi}{4}\cdot D_{out}^2 \cdot v_{out}= \frac{\pi}{4}\cdot D_{tank}^{2} \cdot \frac{dh}{dt} \\D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out} = D_{tank}^{2} \cdot \frac{dh}{dt} \\\frac{dh}{dt}  = \frac{D_{in}^2 \cdot v_{in} - D_{out}^2 \cdot v_{out}}{D_{tank}^{2}}

By replacing all known variables:

\frac{dh}{dt} = \frac{(0.3 ft)^{2}\cdot (5 \frac{ft}{s} ) - (0.4 ft)^{2} \cdot (2 \frac{ft}{s} )}{(10 ft)^{2}}\\\frac{dh}{dt} = 1.3 \times 10^{-3} \frac{ft}{s}

The positive sign of the rate of change of the tank level indicates a rising behaviour.

6 0
3 years ago
The wavelength of light that has a frequency of 1.20 × 1013 s-1 is ________ m.
klio [65]
The relationship between frequency and wavelength for an electromagnetic wave is
c=f \lambda
where
f is the frequency
\lambda is the wavelength
c=3 \cdot 10^8 m/s is the speed of light.

For the light in our problem, the frequency is f=1.20 \cdot 10^{13} s^{-1}, so its wavelength is (re-arranging the previous formula)
\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{1.20 \cdot 10^{13} s^{-1}}=  2.5 \cdot 10^{-5}m
8 0
3 years ago
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