ANSWER
T₂ = 10.19N
EXPLANATION
Given:
• The mass of the ball, m = 1.8kg
First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,
In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,
The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,
We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,
Solve for T₁,
Now, we use the second equation to find the tension in the horizontal string,
Solve for T₂,
Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.
The object's final velocity, given the data is 10.5 rad/s
<h3>What is acceleration? </h3>
This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
- a is the acceleration
- v is the final velocity
- u is the initial velocity
- t is the time
<h3>How to determine the final velocity</h3>
The following data were obtained from the question
- Initial velocity (u) = 1.5 rad/s
- Acceleration (a) = 0.75 rad/s²
- Time (t) = 12 s
- Final velocity (v) = ?
The final velocity can be obtained as follow:
a = (v – u) / t
0.75 = (v – 1.5) / 12
Cross multiply
v – 1.5 = 0.75 × 12
v – 1.5 = 9
Collect like terms
v = 9 + 1.5
v = 10.5 rad/s
Thus, the final velocity of the object is 10.5 rad/s
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Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
(1)
Where:
- Explosive force, measured in newtons.
- Barrel length, measured in meters.
- Mass of the shell, measured in kilograms.
, 
- Initial and final speeds of the shell, measured in meters per second.
If we know that 
, 
, 
and 
, then the explosive force experienced by the shell inside the barrel is:
![F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%281250%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cleft%28750%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D-%5Cleft%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D%5Cright%5D%7D%7B2%5Ccdot%20%2815%5C%2Cm%29%7D)
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
The power of man performing 500 J of work in 8 seconds is 62.5 J/s.
Power can be defined as the pace at which work is completed in a given amount of time.
Horsepower is sometimes used to describe the power of motor vehicles and other machinery.
The pace at which work is done on an item is defined as its power. Power is a temporal quantity.
Which is connected to how quickly a project is completed.
The power formula is shown below.
Power = Energy / Time
Power = E / T
Because the standard metric unit for labour is the Joule and the standard metric unit for time is the second, the standard metric unit for power is a Joule / second, defined as a Watt and abbreviated W.
Here we have given Energy as 500 J and Time as 8 second.
Power = Energy / Time
Power = 500 / 8 Joule / sec
Power = 250 / 4 Joule / sec
Power = 125 / 2 Joule / sec
Power = 62.5 Joule / sec or 62.5 watt
Power came out to be 62.5 J/s when the man performed 500 Joule of work in 8 seconds.
So we can conclude that the power in the Energy transmitted per unit of time, and can be find out by dividing Energy by time. In our case the Power came out to be 62.5 Joule / Second.
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Answer:
The speed of the electron is 1.371 x 10⁶ m/s.
Explanation:
Given;
wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m
the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J
The energy of the incident light is given by;
E = hf
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
f = c / λ
Photo electric effect equation is given by;
E = W₀ + K.E
Where;
K.E is the kinetic energy of the emitted electron
K.E = E - W₀
K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J
K.E = 8.563 x 10⁻¹⁹ J
Kinetic energy of the emitted electron is given by;
K.E = ¹/₂mv²
where;
m is mass of the electron = 9.11 x 10⁻³¹ kg
v is the speed of the electron
Therefore, the speed of the electron is 1.371 x 10⁶ m/s.
