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gogolik [260]
3 years ago
14

What needs to be done to balance this

Physics
1 answer:
navik [9.2K]3 years ago
4 0

Answer:

I'm not sure, but "Add coefficient 2 to O2...". You didn't really write out the equation properly.

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point A is at the bottom of a rough plane which is inclined at angle tita to the horizontal . A body of mass M is projected from
ziro4ka [17]

<h2>The work done , when body moves along the plane </h2>

Explanation:

A body is projected from bottom of the inclined plane . When body is going up the plane .

The downward force = m g sinθ is developed due to its weight

As body is moving upwards , the force of friction will act downwards

The force of friction = μ R

here μ is the coefficient of friction ans R is the normal reaction

Thus force of friction f = μ mg cosθ

Let the acceleration upwards is a

The upward force required = m a

Thus m a = mg sinθ + μ mg cosθ

or acceleration a = g ( sinθ + μ cosθ )

The work done in moving upwards  W = F S

Thus W =  mg ( sinθ + μ cosθ ) S

here S is the displacement on the plane

When body moves down , the force of friction acts upwards

Thus m a = m g ( sinθ - μ cosθ )

The work done W = m g ( sinθ - μ cosθ ) S

As the body is projected with velocity u

which can be calculated by the relation v² - u² = - 2 a X

Here v = 0 at the highest point

Thus u = \sqrt{2ax}

here a = g ( sinθ + μ cosθ )

Similarly , when it moves down , the initial velocity u = 0

Thus v² - 0 = 2 a x

or  v = \sqrt{2ax}

here a = g ( sinθ - μ cosθ )

3 0
3 years ago
Why are microwaves beneficial to use for transmitting information
Darya [45]

Microwaves. Microwave radiation can be used to transmit signals such as mobile phone calls. ... Certain microwave radiation wavelengths pass through the Earth's atmosphere and can be used to transmit information to and from satellites in orbit

5 0
4 years ago
A nonconducting solid sphere of radius 8.40 cm has a uniform volume charge density. The magnitude of the electric field at 16.8
mina [271]

Answer:

The sphere's volume charge density is 2.58 μC/m³.

Explanation:

Given that,

Radius of sphere R= 8.40 cm

Electric field E= 2.04\times10^{3}\ N/C

Distance r= 16.8 cm

We need to calculate the sphere's volume charge density

Using Gauss's law

\int{\vec{E}\cdot\vec{da}}=\dfrac{Q_{enc}}{\epsilon_{0}}

E\times 4\pi r^2=\dfrac{1}{\epsilon_{0}}\times\dfrac{4}{3}\piR^3\rho

E=\dfrac{\rho R^3}{3\epsilon_{0}r^2}

\rho=\dfrac{3\times E\times\epsilon_{0}r^2}{R^3}

Put the value into the formula

\rho=\dfrac{3\times2.04\times10^{3}\times8.85\times10^{-12}\times(16.8\times10^{-2})^2}{(8.40\times10^{-2})^3}

\rho=2.58\times10^{-6}\ C/m^3

\rho=2.58\ \mu C/m^3

Hence, The sphere's volume charge density is 2.58 μC/m³.

5 0
4 years ago
12*3a car is stopped at a traffic light. it then travels along a straight road so that its distance from the light is given by w
Digiron [165]
Missing parts in the text of the exercise:
- The distance from the traffic light is given by x(t) = bt^2 -ct^3, where b=2.4 m/s^2 and c=0.13 m/s^3

Solution:

part a) <span>calculate the average velocity of the car for the time interval t=0 to t=8.0 s
- The average velocity is given by the ratio between the distance covered in the time interval:
</span>v_{ave} =  \frac{x(8.0 s)-x(0 s)}{8.0 s-0 s}
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
x(8.0 s)=b(8 s)^2-c(8 s)^3 = (2.4 m/s^2)(8 s)^2 -(0.13 m/s^3)(8 s)^3=
=87.04 m
Therefore, the average velocity is
v_{ave}= \frac{x(8.0 s)-x(0)}{8.0s-0}= \frac{87.04 m}{8.0s}= 10.88 m/s

part b) <span>calculate the instantaneous velocity of the car at t=0, t=4.0 s and t=8.0 s
- The instantaneous velocity can be found by performing the derivation of x(t):
</span>v(t) = \frac{dx(t)}{dt}=2bt-3ct^2
<span>So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: </span>v(4.0 s)=2(2.4 m/s^2)(4.0 s)-3(0.13 m/s^2)(4.0s)^2=12.96 m/s
- t=8 s: v(8.0 s)=2(2.4 m/s^2)(8.0 s)-3(0.13 m/s^2)(8.0s)^2=13.44 m/s

part c) <span>how long after starting from rest is the car again at rest?
- To solve this part we must find the value of t for which v(t)=0, so:
</span>2bt-3ct^2=0
t(2b-3ct)=0
<span>The first solution is t=0 s, which corresponds to the beginning of the motion, so we are not interested in this value. The second solution is
</span>t= \frac{2b}{3c}= \frac{2\cdot 2.4 m/s^2}{3 \cdot 0.13 m/s^3}=12.31 s
<span>and this is the time at which the car is at rest again.</span>
8 0
3 years ago
A cart starting from rest rolls down a frictionless 1.0 m high hill. It travels a distance 2.0 m along the rough bottom surface
Rzqust [24]

Answer:

Option B is correct.

Explanation:

Given data

Height of the hill = AB = 1 m

Distance traveled  along the rough bottom surface = AC = 2 m

Now from the ΔABC

\sin \theta = \frac{AB}{AC}

\sin \theta = \frac{1}{2}

\theta = 30 °

We know that  the coefficient of kinetic friction is

\mu = \tan \theta

\mu = \tan 30

\mu = 0.5

This is the value of the coefficient of kinetic friction

Thus option B is correct.

8 0
3 years ago
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