Answer:
The length of the rod is 0.6 m
Solution:
Mass of the copper rod, m = 100 g = 0.1 kg
Angular speed at lowest position, [tex]\omega_{L} = 7 rad/s[/tex]
Now,
By using the law of conservation of energy, the overall mechanical energy of the system taken about the center of mass remain conserve.
Thus at the initial position of the rod, i.e., horizontal:
[tex]\frac{1}{2}mv_{c}^{2} + \frac{1}{2}I\omega^{2} + mg\frac{L}{2} =0 + mg\frac{L}{2}[/tex]
[tex]\frac{1}{2}mv_{c}^{2} + \frac{1}{2}I\omega^{2} = 0[/tex]
(Since, [tex]v_{c} = 0[/tex] and [tex]\omega = 0[/tex] at horizontal position).
where
Tranlational Kinetic energy about center of mass = [tex]\frac{1}{2}mv_{c}^{2}[/tex]
Rotational K.E about the center of mass, [tex]\frac{1}{2}I\omega^{2}[/tex]
Potential energy about Center of Mass = [tex]mg\frac{L}{2}[/tex]
Now, applying the law of conservation at the lowest point of the rod:
[tex]\frac{1}{2}mv'_{c}^{2} + \frac{1}{2}I\omega'^{2} + 0 =\frac{1}{2}m(\frac{\omega' L}{2})^{2} + \frac{1}{2}I\omega'^{2}[/tex]
[tex]\frac{1}{2}mv'_{c}^{2} + \frac{1}{2}I\omega'^{2} =\frac{1}{2}m(\frac{\omega' L}{2})^{2} + \frac{1}{2}\frac{mL^{12}}{12}\omega'^{2} = \frac{mL^{2}\omega'^{2}}{6}[/tex]
where
Moment of inertia about the center of mass at the lowest position is [tex]\frac{mL^{2}}{12}[/tex]
[tex]v'_{c} = \frac{\omega L}{2}[/tex]
Thus
From these, potential energy about center of mass = [tex]\frac{mL^{2}\omega'^{2}}{6}[/tex]
[tex]mg\frac{L}{2} = \frac{mL^{2}\omega'^{2}}{6}[/tex]
At the lowest point, [tex]\omega' = 7 rad/s[/tex]
Thus
[tex]mg\frac{L}{2} = \frac{mL^{2}\times 7^{2}}{6}[/tex]
[tex]g = \frac{49L}{3}[/tex]
g = 9.8[tex]m/s^{2}[/tex]
[tex]9.8 = \frac{49L}{3}[/tex]
L = 0.6 m
The total length of the cord is L = 7.00 m, the mass of the cord is m = 7.00 g, the mass of the hanging object is M = 2.50 kg, and the pulley is a fixed a distance d = 4.00 m from the wall. You pluck the cord between the wall and the pulley and it starts to vibrate. What is the fundamental frequency (in Hz) of its vibration?
Answer:
frequency = 19.56 Hz
Explanation:
given data
length L = 7 m
mass m = 7 g
mass M = 2.50 kg
distance d = 4 m
to find out
fundamental frequency
solution
we know here frequency formula is
frequency = [tex]\frac{v}{2d}[/tex] ...........1
so here d is given = 4
and v = [tex]\sqrt{\frac{T}{\mu} }[/tex] ..........2
tension T = Mg = 2.50 × 9.8 = 24.5 N
and μ = [tex]\frac{m}{l}[/tex] = [tex]\frac{7*10^{-3} }{7}[/tex] = [tex]10^{-3}[/tex] kg/m
so from equation 2
v = [tex]\sqrt{\frac{24.5}{10^{-3}} }[/tex]
v = 156.52
and from equation 1
frequency = [tex]\frac{v}{2d}[/tex]
frequency = [tex]\frac{156.52}{2(4)}[/tex]
frequency = 19.56 Hz
Final answer:
The fundamental frequency of the vibrating cord is approximately 14.18 Hz.
Explanation:
To determine the fundamental frequency of the vibrating cord, we can use the equation for the fundamental frequency of a vibrating string:
f1 = 1/2L * sqrt(T / μ)
Where f1 is the fundamental frequency, L is the total length of the cord, T is the tension in the cord, and μ is the linear density of the cord.
Plugging in the given values, we have: f1 = 1/2 * 7.00 * sqrt(90 / 0.007)
Simplifying this equation gives us the fundamental frequency of the cord: f1 ≈ 14.18 Hz
A basketball referee tosses the ball straight up for the starting tip-off. At what velocity (in m/s) must a basket ball player leave the grownd to rist 1.24 m above the floor in an attempt to get the ball?
Answer:4.93 m/s
Explanation:
Given
height to reach is (h )1.24 m
here Let initial velocity is u
using equation of motion
[tex]v^2-u^2=2ah[/tex]
here Final Velocity v=0
a=acceleration due to gravity
[tex]0-u^2=2\left ( -g\right )h[/tex]
[tex]u=\sqrt{2gh}[/tex]
[tex]u=\sqrt{2\times 9.81\times 1.24}[/tex]
[tex]u=\sqrt{24.328}[/tex]
u=4.93 m/s
During constant linear acceleration, what changes uniformly? A. acceleration B. distance C. displacement D. velocity E. all of these
Answer:D-velocity
Explanation:
Given
Constant linear acceleration i.e. acceleration is constant
a=constant
and we know [tex]\frac{\mathrm{d}V}{\mathrm{d} t}=a[/tex]
Thus velocity is changing uniformly to give constant acceleration
While displacement and distance are function of time thus they are not changing uniformly.
During constant linear acceleration, the following quantity changes uniformly: velocity. Therefore option D is correct.
Constant linear acceleration refers to a situation where an object's acceleration remains constant over a specific period of time. In this case, the rate of change of velocity is uniform, meaning the velocity increases or decreases by the same amount in equal time intervals.
Velocity is the rate of change of displacement and is directly affected by acceleration. During constant linear acceleration, the velocity changes uniformly. It either increases or decreases at a constant rate, depending on the direction of acceleration.
Therefore,
During constant linear acceleration, the following quantity changes uniformly: velocity.
Know more about linear acceleration:
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A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in a circular orbit about the first at a distance of 201 meters as a result of their mutual gravitational attraction. What is the speed of the second asteroid?
Answer:
1.81 x 10^-4 m/s
Explanation:
M = 98700 kg
m = 780 kg
d = 201 m
Let the speed of second asteroid is v.
The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.
[tex]\frac{GMm}{d^{2}}=\frac{mv^2}{d}[/tex]
[tex]v=\sqrt{\frac{GM}{d}}[/tex]
Where, G be the universal gravitational constant.
G = 6.67 x 10^-11 Nm^2/kg^2
[tex]v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}[/tex]
v = 1.81 x 10^-4 m/s
According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. A 150-lbm swimmer is floating in a nearby pool; 95% of his or her body's volume is in the water while 5% of his or her body's volume is above water. Determine the density of the swimmer's body. The density of water is 0.036lbm/in^3.
Answer:
Density of the swimmer = [tex]0.0342\ lbm/in^3[/tex].
Explanation:
Assuming,
[tex]\rho[/tex] = density of the swimmer.[tex]\rho_w[/tex] = density of the water.[tex]m[/tex] = mass of the swimmer.[tex]m_w[/tex] = mass of the water displaced by the swimmer.[tex]V_w[/tex] = volume of the displaced water.[tex]V[/tex] = volume of the swimmer.Given:
[tex]m=150\ lbm.[/tex][tex]\rho_w = 0.036\ lbm/in^3.[/tex]The density of an object is defined as the mass of the object per unit volume.
Therefore,
[tex]\rho =\dfrac{m}{V}\ \Rightarrow m = \rho V\ \ .........\ (1).[/tex]
Since only 95% of the body of the swimmer is inside the water, therefore,
[tex]V_w = 95\%\ \text{of}\ V=\dfrac{95}{100}\times V = 0.95V.[/tex]
According to Archimedes' principle,
[tex]m=m_w\\[/tex]
Using (1),
[tex]\rho V=\rho_w V_w\\\rho V = 0.036\ lbm/in^3\times 0.95 V\\\rho=0.036\times 0.95\ lbm/in^3=0.0342\ lbm/in^3.[/tex]
The kilowatt-hour is a unit of speed
power
force
momentum
energy
Answer:
Answer:
energy
Explanation:
The capacity to do work is called energy.
There are several types of energy. Some of the types of energy are given below:
1. Kinetic energy
2. Potential energy
3. Solar energy
4. thermal energy
5. Sound energy
6. Nuclear energy, etc.
The SI unit of energy is Joule.
Some other units of energy are
erg, eV, etc.
1 J = 10^7 erg
1 eV = 1.6 x 10^-19 J
The commercial unit of energy is kilowatt hour.
1 Kilo watt hour = 3.6 x 10^6 J
Explanation:
Consider two metallic rods mounted on insulated supports. One is neutral, the other positively charged. You bring the two rods close to each, but without contact, and briefly ground the the neutral rod by touching it with your hand. What would be resulting charge (if any) on initially neutral rod?
Hi!
Letw call A to the initially neutral rod, and B to the positively charged. When they are close to each other, the positive charges in B attract the negative charges in A, and repelle the positive ones. If you ground A, negative charges from ground (your body, in this case), flow to A attracted by the positive charges in B, and positive charges in A flow to ground, so finally A results negatively charged
Answer:
Negative charges
Explanation:
The procedure described above is known in physics as charging by electrostatic induction. If we desire to impart negative charges to a hitherto neutral rod, we bring a positively charged rod near it without allowing the two insulated rods to touch each other. If the neutral rod is earthed, negative charges remain on the rod.
Alternative sources of energy include : geothermal
hydrogen
wind
solar cells All of the above
Answer:
All of the above
Explanation:
The correct answer is option E (All of the above)
All the options are alternative source of energy.
The option given are not the traditional way (energy production from coal) of extracting energy as the loss of natural resources does not take place in these source of energy.
energy extracted from wind, solar light , hydrogen ,etc are the source of energy the alternative source of production of energy because do not exploit the natural resources, reduce the carbon emission and energy produced by them is clean energy.
A diver springs upward from a board that is 3.90 m above the water. At the instant she contacts the water her speed is 11.8 m/s and her body makes an angle of 78.2° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.
Answer:
V = 7.91 m/s(magnitude) @ 72.3°(direction)
Explanation:
We will have to do a parabolic movement analysis in order to determine the initial velocity. For this we will mainly use the following kinetic formulae:
[tex]1) X_{f}=X_{0}+V_{0}t+\frac{at^{2}}{2}\\\\2) V_{f}=V_{0}+at[/tex]
We will work backwards from the initial information: a final velocity of 11.5 m/s at a 78.2° angle. From this we can extract the final horizontal and vertical velocity:
[tex]V_{fx}=11.8\frac{m}{s}*Cos(78.2)=2.41\frac{m}{s}\\\\V_{fy}=11.8\frac{m}{s}*Sin(78.2)=11.55\frac{m}{s}[/tex]
We will use [tex]V_{fy}[/tex] with a negative value (since the diver is moving in a downwards direction). That way we can find the time elapsed between the highest point of the trayectory to instants before she contacts the water. At the maximum height of the movement, the velocity in the vertical direction is zero. Therefore we can rewrite equation 2) and solve for t:
[tex]-11.55\frac{m}{s}=0\frac{m}{s} - 9.8\frac{m}{s^{2}}t\\\\t= \frac{-11.55\frac{m}{s}}{- 9.8\frac{m}{s^{2}}}=1.17s[/tex]
This t that we found represents the time from the maximum height to the surface of the water. Now we can use equartion 1) to determine the maximum height. We must remember a starting point of 3.9m above the water:
[tex]0 = h+3.9m+0\frac{m}{s}*t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\ \\h=\frac{9.8\frac{m}{s^{2}}1.17s^{2}}{2}-3.9m=2.9m\\\\[/tex]
Since we now know what is the maximum height we can determine the time that it took for the diver to reach that point using equation 1) and 2):
[tex]2.9m =V_{0y}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\0\frac{m}{s}= V_{0y}t-9.8\frac{m}{s^{2}}t[/tex]
We will solve for [tex]V_{ox}[/tex] in equation 2) and replace it in equation 1):
[tex]V_{0y} = 9.8\frac{m}{s^{2}}t\\\\2.9m =(9.8\frac{m}{s^{2}}t)t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}[/tex]
We solve for t (time from diving board to maximum height):
[tex]2.9m =(9.8\frac{m}{s^{2}}t)t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\2.9m=\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\5.8m=9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{ \frac{5.8m}{9.8\frac{m}{s^{2}}}}=0.77s[/tex]
Finally we can replace our t value in [tex]V_{0y}t=9.8\frac{m}{s^{2}}t[/tex] to obtain:
[tex]V_{0y}t=9.8\frac{m}{s^{2}}t=9.8\frac{m}{s^{2}}*0.77s=7.54\frac{m}{s}[/tex]
With this we have the information we need in order to determine the initial velocity of the diver. The magnitude will be calculated from the initial horizontal and vertical velocities (consider final and initial horizontal velocity to be equal in PARABOLIC MOVEMENTS):
[tex]V_{0}=\sqrt{V_{0x}^{2}+V_{0y}^{2}}=\sqrt{(2.41\frac{m}{s})^{2}+(7.54\frac{m}{s})^{2}}=7.91\frac{m}{s}[/tex]
And the direction is determined by an arctangent:
[tex]tan\theta=\frac{7.54\frac{m}{s}}{2.41\frac{m}{s}}\\\\\theta=arctan(\frac{7.54\frac{m}{s}}{2.41\frac{m}{s}})=72.3[/tex]
Final answer:
To find the diver's initial velocity, use equations involving vertical and horizontal components considering her final speed and displacement. The magnitude is 15.93 m/s, and the direction is horizontal.
Explanation:
To determine the diver's initial velocity, we first find the vertical component of her velocity using the equation vf=vi+at. Here, vf = 11.8 m/s, vi is the vertical component of the initial velocity, a = -9.81 m/s^2, and t = time in the air. Solving for vi, we get vi = 15.93 m/s. Then, to find the horizontal component of the initial velocity, we use the equation dx=vit+1/2at^2. Since the diver contacts the water at the highest point, her vertical displacement is 3.9m, and we can find the horizontal component of the initial velocity to be 0 m/s. Therefore, her initial velocity magnitude is 15.93 m/s, and her initial velocity direction is horizontal.
Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible for 0.20 s as it moves a distance of 1.19 m from the bottom to the top of the window. How long does it take before the ball reappears?
Firstly , let's consider the motion of the ball as it travelled up past the window. Given that the distance covered by the ball is 1.19 m in 0.20 s, we can calculate the initial velocity of the ball using the straightforward kinematic equation for velocity, which is v = d/t. Here, 'd' is the distance travelled by the ball, and 't' is the time taken to cover that distance.
Plugging the values into the equation:
vi = d/t
vi = 1.19 m / 0.20 s
vi = 5.95 m/s
Therefore, the initial velocity of the ball is approximately 5.95 m/s.
Next, we want to determine how long it would take for the ball to reach its peak and then reappear. The acceleration due to gravity (g) is 9.81 m/s^2 and acts downwards. Meaning that, as the ball ascends, it slows down until it reaches its peak, where its velocity becomes zero.
To find the time at peak, we can use the equation of motion v = vi + at, where 'v' is the final velocity, 'vi' is the initial velocity, 'a' is the acceleration, and 't' is time. At the peak, the final velocity 'v' becomes 0, thus:
t_peak = vi / g
t_peak = 5.95 m/s / 9.81 m/s²
t_peak = 0.61 s
Therefore, the time it takes to reach the peak is approximately 0.61 seconds.
As we're dealing with symmetrical motion (upwards and downwards motion are equal), the time for the ball to descend from its peak would be identical to the time it took to ascend, which means:
Time for the ball to reappear, t_reappear = 2 * t_peak
t_reappear = 2 * 0.61 s
t_reappear = 1.22 s
Hence, it should take approximately 1.22 seconds before the ball reappears.
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Why is it easier to use a potentiometer in a circuit rather than two separte resistors in series with one another?
Answer:
Because by potentiometer we can drop voltage according to our requirement but in set of two series resistors we can not do this.
Explanation:
As we know that the potentiometer can act as a variable resistor but resistors have constant value.
In any circuit if we insert resistors they will drop a constant value if we insert a potentiometer we will drop the value of voltage by according to our will.
Therefore, by the above discussion it can be say that the potentiometer is better than two resesters in series.
What is the approximate uncertainty in the area of a circle of radius 4.5×10^4 cm? Express your answer using one significant figure.
To find the uncertainty in the area of a circle with a radius containing two significant figures, use the formula A = πr² and round the result to two significant figures, taking into account the rules of rounding and the implications for uncertainty.
Explanation:The question pertains to the calculation of the uncertainty in the area of a circle when provided with the radius which has a certain number of significant figures. Given the radius (r) of a circle as 4.5×104 cm, which contains two significant figures, the area (A) can be calculated using the formula A = πr². However, since π is a constant with an infinite number of significant digits and the radius has only two, the answer must be limited to two significant figures to reflect the precision of the given measurement. Consequently, the area will be calculated and then rounded to two significant figures to give an approximate value of the uncertainty.
Assuming the calculated area is A, we must consider the rounding rules to determine the uncertainty. When we round to two significant figures, this may imply an uncertainty, as the final digit in the rounded number can vary up or down by one. This is an important concept to understand, as rounding can lead to a loss of information and increase the relative uncertainty of the final answer.
The first accurate measurements of the properties of high-pressure gases were made by E. H. Amagat in France between 1869 and 1893. Before developing the dead-weight gauge, he worked in a mineshaft and used a mercury manometer for measurements of pressure to more than 400 bar. Estimate the height of the manometer required.
Answer:
Height, h = 300.27 meters
Explanation:
Given that,
Pressure of the gas, [tex]P=400\ bar=4\times 10^7\ Pa[/tex]
We need to find the height of the manometer required. The pressure at a height is given by :
[tex]P=\rho gh[/tex]
Where
[tex]\rho[/tex] is the density of mercury, [tex]\rho=13593\ kg/m^3[/tex]
h is the height of the manometer required.
[tex]h=\dfrac{P}{\rho g}[/tex]
[tex]h=\dfrac{4\times 10^7}{13593\times 9.8}[/tex]
h = 300.27 meters
So, the height of the manometer required is 300.27 meters. Hence, this is the required solution.
A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tall and located 3.60 m from the lens. (a) How far is the CCD sensor from the lens when the person is in focus?
(b) How tall is the person's image on the CCD sensor?
Answer:
a) 35.44 mm
b) 17.67 mm
Explanation:
u = Object distance = 3.6 m
v = Image distance
f = Focal length = 35 mm
[tex]h_u[/tex]= Object height = 1.8 m
a) Lens Equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm[/tex]
The CCD sensor is 35.34 mm from the lens
b) Magnification
[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}[/tex]
[tex]m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm[/tex]
The person appears 17.67 mm tall on the sensor
What is the magnitude of the electric field of a proton at a distance of 50 micrometers? _____________ (in units of N/C)
Answer:
Electric field at distance of 50 micrometer due to a proton is 0.576 N/C
Explanation:
We have given distance where we have to find the electric field [tex]r=50\mu m=50\times 10^{-6}m[/tex]
We know that charge on proton [tex]q=1.6\times 10^{-16}C[/tex]
Electric field due to point charge is given by [tex]E=\frac{Kq}{r^2}[/tex], here K is a constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]
So electric field [tex]E=\frac{9\times 10^9\times 1.6\times 10^{-16}}{(50\times 10^{-6})^2}=0.576N/C[/tex]
Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4.0 cm/yr. Assuming this to be a constant rate, how many years will pass before the radius of the Moon's orbit increases by 3.84x 10^7 m (10%)?
Answer:
967500000 years
Explanation:
The Speed at which the radius of the orbit of the Moon is increasing is 4 cm/yr
Converting to m
1 m = 100 cm
[tex]1\ cm=\frac{1}{100}\ m[/tex]
[tex]4\ cm\y=\frac{4}{100}=0.04\ m/yr[/tex]
The distance by which the radius increases is 3.84×10⁷ m
Time = Distance / Speed
[tex]\text{Time}=\frac{3.87\times 10^7}{0.04}\\\Rightarrow \text{Time}=967500000\ yr[/tex]
967500000 years will pass before the radius of the orbit increases by 10%.
Final answer:
The radius of the Moon's orbit increases by approximately 4 cm/year. To calculate the number of years it will take for the radius to increase by 3.84 × 10^6 m, we can use the formula Time = Change in Distance / Rate of Increase. The answer is approximately 9.6 × 10^7 years.
Explanation:
The radius of the Moon's orbit is increasing at a rate of approximately 4 cm/year. To find out how many years will pass before the radius of the Moon's orbit increases by 3.84 × 10^6 m, we can use the formula:
Time = Change in Distance / Rate of Increase
Substituting the given values, we get:
Time = (3.84 × 10^6 m) / (4 cm/year)
Now, we need to convert the units so that they are consistent. 3.84 × 10^6 m is equivalent to 3.84 × 10^8 cm. Substituting this value into the equation, we get:
Time = (3.84 × 10^8 cm) / (4 cm/year)
Canceling out the units, we find that:
Time = 9.6 × 10^7 years.
We attach two blocks of masses m1 = 7 kg and m2 = 7 kg to either end of a spring of spring constant k = 1 N/m and set them into oscillation. Calculate the angular frequency ω of the oscillation.
Answer:
The angular frequency [tex]\omega[/tex] of the oscillation is [tex]0.58s^{-1}[/tex]
Explanation:
For this particular situation, the angular frequency of the system is given by
[tex]\omega=\sqrt{\frac{m_1+m_2}{m_1m_2}k}=\sqrt{\frac{7 kg+5 kg }{7kg *5 kg}1\frac{N}{m}}=\sqrt{\frac{3}{35s^2}}\approx 0.58s^{-1}[/tex]
A 7.80-nc charge is located 1.81 m from a 4.30-nc point charge. (a) Find the magnitude of the electrostatic force that one charge exerts on the other. N (b) Is the force attractive or repulsive? O attractive O repulsive
Explanation:
Given that,
Charge 1, [tex]q_1=7.8\ nC=7.8\times 10^{-9}\ C[/tex]
Charge 2, [tex]q_2=4.3\ nC=4.3\times 10^{-9}\ C[/tex]
Distance between charges, r = 1.81 m
(a) The electrostatic force that one charge exerts on the another is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{7.8\times 10^{-9}\times 4.3\times 10^{-9}}{(1.81)^2}[/tex]
[tex]F=9.21\times 10^{-8}\ N[/tex]
(b) As both charges are positively charged. So, the force of attraction between them is repulsive.
A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball? (Assume that the player gets the ball at his maximum height.)
Answer:4.95 m/s
Explanation:
Given
Ball rises to a height of 1.25 m
Let u be the velocity of player while leaving ground .
Its final velocity will be zero as he reaches a height of 1.25 m
thus
[tex]u^2=2gh[/tex]
[tex]u=\sqrt{2\times 9.81\times 1.25}[/tex]
[tex]u=\sqrt{24.525}[/tex]
u=4.952 m/s
Final answer:
To reach a height of 1.25 meters, a basketball player must leave the ground with an initial upward velocity of 4.95 m/s, calculated using the kinematic equation for motion under constant acceleration due to gravity.
Explanation:
To find the required initial velocity, we use the kinematic equation which relates distance (d), initial velocity ([tex]v_i[/tex] ), final velocity ([tex]v_f[/tex]), acceleration (a), and time (t). We can ignore time since we are not asked for it and we know that the final velocity at the peak of the jump is 0 m/s because the ball will momentarily stop moving upwards before it starts falling down. We will use the acceleration due to gravity, which is approximately 9.81 m/s².
The equation is:
[tex]v_f^2 = v_i^2 + 2ad[/tex]
Setting [tex]v_f[/tex] to 0 m/s (since the ball stops at the peak), the equation simplifies to:
[tex]v_i[/tex] = √(2ad)
Plugging in the values we get:
[tex]v_i[/tex] = √(2 * 9.81 m/s² * 1.25 m) = √(24.525 m²/s²)
[tex]v_i[/tex] = 4.95 m/s
Therefore, the basketball player must leave the ground with an initial velocity of 4.95 m/s to reach a height of 1.25 meters.
A boat moves ar 10.0 m/s relative to
thewater. If the boat is in a river where the current is 2.00 m/s,
howlong does it take the boat to make a complete round trip of 1000
mupsteam followed by a 1000- m trip downstream?
Answer: 208.3 s
Explanation:
Hi!
You need to calculate the velocity of the boat relative to the shore, in each trip.
Relative velocities are transformed according to:
[tex]V_{b,s} = V_{b,w} + V_{w,s}\\ V_{b,s} = \text{velocity of boat relative to shore}\\V_{b,w} = \text{velocity of boat relative to water} \\V_{b,w} = \text{velocity of water relative to shore}\\[/tex]
Let's take the upstram direction as positive. Water flows downstream, so it's velocity relative to shore is negative , -2 m/s
In the upstream trip, velocity of boat relative to water is positive: 10m/s. But in the downstream trip it is negatoive: -10m/s
In upstream trip we have:
[tex]V_{b,s} = (10 - 2)\frac{m}{s} = 8 ms[/tex]
In dowstream we have:
[tex]V_{b,s} = (-10 - 2)\frac{m}{s} = -12 ms[/tex]
In both cases the distance travelled is 1000m. Then the time it takes the round trip is:
[tex]T = T_{up} + T_{down} = \frac{1000}{8}s + \frac{1000}{12}s = 208.3 s[/tex]
A 60.0 kg astronaut is freely floating in space and pushes on a freely floating 120.0 kg spacecraft with a force of 30.0 N for 1.50 s. (a) Compare the forces exerted on the astronaut and the spacecraft, and (b) compare the acceleration of the astronaut to the acceleration of the spacecraft
Answer:
Explanation:
mass of astronaut, m = 60 kg
mass of space craft, M = 120 kg
t = 1.5 s
Force on space craft = 30 N
(a) According to Newton's third law
Force on spacecraft by the astronaut = Force on astronaut by the space craft
Force on astronaut by the space craft = 30 N
(b) According to Newton's second law
Force = mass x acceleration
Let a be the acceleration of the astronaut
30 = 60 x a
a = 0.5 m/s^2
Let A be the acceleration of the spacecraft
30 = 120 x A
a = 0.25 m/s^2
Consider the scenario where a person jumps off from the edge of a 1 m high platform and lands on the ground Suppose his initial jumping speed was 3 m/s. For how long was this person in the air?
Answer:
For 0.24 sec the person was in the air.
Explanation:
Given that,
Height = 1 m
Initial velocity = 3 m/s
We need to calculate the time
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Where, u = initial velocity
s = height
Put the value into the formula
[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]4.9t^2+3t-1=0[/tex]
[tex]t = 0.24\ sec[/tex]
On neglecting the negative value of time
Hence, For 0.24 sec the person was in the air.
The pilot of an airplane carrying a package of mail to a remote outpost wishes to release the package at the right moment to hit the recovery location A. What angle θ with the horizontal should the pilot’s line of sight to the target make at the instant of release? The airplane is flying horizontally at an altitude of 86 m with a velocity of 283 km/h.
Answer:
The angle is [tex]\theta\approx 14.61[/tex] degrees.
Explanation:
Se the attached drawing if you need a visual aid for the explanation. Let [tex]\theta[/tex] be the angle of elevation of the plante which in itself is the same drop angle that the pilot measures. Let [tex]d[/tex] be the horizontal distance from the target and [tex]h[/tex] the height of the plane. We know that the package is dropped without any initial vertical speed, that means that it has a y-position equation of the form:
[tex]y(t)=-\frac{1}{2}gt^2+h[/tex]
If we set [tex]y(t)=0[/tex] we are setting the condition that the package is in the ground. We can then solve for t and get the flight time of the package.
[tex]0=-\frac{1}{2}gt^2+h\implies t_f=\sqrt{\frac{2h}{g}}[/tex].
If the flight time is -[tex]t_f[/tex] then the distance b can be found in meters by taking into account that the horizontal speed of the plane is [tex]v=283\, Km/h=78.61 \, m/s[/tex].
[tex]d=v\cdot t_f=78.61\cdot \sqrt{\frac{2h}{g}}[/tex]
The angle is thus
[tex]\theta=\arctan{\frac{h}{v\cdot t_f}}=\arctan{\frac{h}{v\cdot \sqrt{\frac{2\cdot h}{g}}}\approx 14.61 [/tex] degrees.
The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on each body, such that the net force (gravitational plus electrical) on each is zero. What is the magnitude of the charge placed on each body?
Answer:
The magnitude of charge on each is [tex]5.707\times 10^{13} C[/tex]
Solution:
As per the question:
Mass of Earth, [tex]M_{E} = 5.98\times 10^{24} kg[/tex]
Mass of Moon, [tex]M_{M} = 7.35\times 10^{22} kg[/tex]
Now,
The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:
[tex]F_{G} = \frac{GM_{E}M_{M}}{d^{2}}[/tex] (1)
Now,
If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:
[tex]F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex] (2)
Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:
Equating eqn (1) and (2):
[tex]\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}[/tex]
[tex](6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}[/tex]
[tex]\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q[/tex]
Q = [tex]\pm 5.707\times 10^{13} C[/tex]
When a car is on an inclined bank of angle θ and rounding a curve with no friction, what is the centripetal force equal to? a. The weight of the car b. N cos (θ) c. N sin (θ) d. Zero
Answer:
[tex]F_{net}=N\ sin\theta[/tex]
Explanation:
Let a car of m is on an incline bank of angle θ and it is rounding a curve with no friction. We need to find the centripetal force acting on it.
The attached free body diagram shows the car on the banked turn. It is clear that,
In vertical direction,
[tex]N\ cos\theta=mg[/tex]
In horizontal direction,
[tex]F_{net}=F_{centripetal}[/tex]
[tex]F_{net}=N\ sin\theta[/tex]
So, the centripetal force is equal to [tex]N\ sin\theta[/tex]. Hence, the correct option is (c).
The position vector of an object of mass 0.5 kg is given by r 2.0t2i+ 3.0tj m where t represents time. Which ONE of the following statements is FALSE when a time of t 2.0 s has elapsed. a) The object has moved a distance of 10 m. b) The momentum of the object is 41+1.5/ kg.m.s c) The kinetic energy of the object is 18 J d) The force on the object acts perpendicular to the direction of the unit vectorj e) The magnitude of the force acting on the object is 4 N
Answer:
Statement E is false.
Explanation:
Statement a says that the object would have moved a distance of 10 m.
At t = 0
x0 = 0
y0 = 0
At t= 2
x2 = 2 * 2^2 = 8
y2 = 3 * 2 = 6
[tex]d2 = \sqrt{x2^2 + y2^2}[/tex]
[tex]d2 = \sqrt{8^2 + 6^2} = 10 m[/tex]
Statement A is true.
Statement B says that the momentum of the object would be 4*i+1.5*j
To test this we need to know the speed.
The speed is the first derivative by time:
v = 4*t*i + 3*j
At t = 2 the speed is
v2 = 8*i + 3*j
The momentum is
P = m * v
P2 = 0.5 * (8*i + 3*j) = (4*i+1.5*j)
This statement is true.
Statement C says that the kinetic energy of the object would be 18 J.
The magnitude of the speed is
[tex]v = \sqrt{8^2 + 3^2} = 8.54 m/s[/tex]
The kinetic energy is
Ek = 1/2 * m * v^2
Ek = 1/2 * 0.5 * 8.54^2 = 18 J
Statement C is true
Statement D says that the force acts perpenducular to the direction of the unit vector j.
The acceleration is the second derivative respect of time
a = 4*i + 0*j
Since there is no acceleration in the direction of j we can conclude that the force is perpendicular to the direction of j.
Statement D is true.
Statement E says that the force acting on the object is of 4 N
f = m * a
f = 0.5 * 4 = 2 N
Statement E is false.
(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the capacitor been connected to a 6.00-V battery, how much energy would have been stored?________ μJ
Answer:
(a) [tex]E_{ c} = 112.5 \mu J[/tex]
(b) [tex]E'_{ c} = 18 \mu J[/tex]
Solution:
According to the question:
Capacitance, C = [tex]1.00\mu F = 1.00\times 10^{- 6} F[/tex]
Voltage of the battery, [tex]V_{b} = 15.0 V[/tex]
(a)The Energy stored in the Capacitor is given by:
[tex]E_{c} = \frac{1}{2}CV_{b}^{2}[/tex]
[tex]E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}[/tex]
[tex]E_{ c} = 112.5 \mu J[/tex]
(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:
[tex]E'_{c} = \frac{1}{2}CV'_{b}^{2}[/tex]
[tex]E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}[/tex]
[tex]E'_{ c} = 18 \mu J[/tex]
(a) The energy stored in the capacitor is [tex]1.125 \times 10^{-4} \ J[/tex]
(b) The energy stored in the capacitor is [tex]1.8 \times 10^{-5} \ J[/tex]
The given parameters;
charge of the capacitor, q = 1 -μFvoltage across the capacitor, V = 15The energy stored in the capacitor is calculated as follows;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 15^2\\\\E = 1.125 \times 10^{-4} \ J[/tex]
When the battery voltage changes to 6 V, the energy stored in the capacitor is calculated as follows;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 6^2\\\\E = 1.8 \times 10^{-5} \ J\\\\[/tex]
Learn more here:https://brainly.com/question/24943210
A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer reading of the car in km/hr?
Answer:
17.64 km/h
Explanation:
mass of car, m = 1000 kg
Kinetic energy of car, K = 1.2 x 10^4 J
Let the speed of car is v.
Use the formula for kinetic energy.
[tex]K = \frac{1}{2}mv^{2}[/tex]
By substituting the values
[tex]1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}[/tex]
v = 4.9 m/s
Now convert metre per second into km / h
We know that
1 km = 1000 m
1 h = 3600 second
So, [tex]v = \left (\frac{4.9}{1000} \right )\times \left ( \frac{3600}{1} \right )[/tex]
v = 17.64 km/h
Thus, the reading of speedometer is 17.64 km/h.
To find the speedometer reading, use the kinetic energy and mass to calculate the speed in meters per second and then convert it to kilometers per hour. The calculated speed is approximately 88 km/hr.
Explanation:To determine the speedometer reading of the car in km/hr based on the kinetic energy given, we use the kinetic energy formula KE = ½ mv². We rearrange this formula to solve for speed (v): v = √(2KE/m).
Plugging in the provided kinetic energy of 1.2 x 10´ J and the car's mass of 1000 kg, we get:
v = √(2 × 1.2 x 10´ J / 1000 kg)
v = √(24 x 10´ J/kg)
v ≈ 24.49 m/s
To convert from m/s to km/hr, we multiply by 3.6:
v = 24.49 m/s × 3.6 km/hr per m/s ≈ 88.16 km/hr
Therefore, the speedometer would read approximately 88 km/hr.
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the electron be moving when it is 3.00 cm from the proton?
Answer:
electron moving at 91.82 m/s
Explanation:
given data
distance d1 = 6 cm
distance d2 = 3 cm
to find out
how fast will the electron be moving
solution
we know potential energy formula that is
potential energy = [tex]k\frac{q1q2}{r}[/tex]
here k is 9× [tex]10^{9}[/tex] N-m²/C²
m mass of electron is 9.11 × [tex]10^{-31}[/tex] kg
and q = 1.60 × [tex]10^{-19}[/tex] C
we consider here initial potential energy = U1
U1 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.06^2}[/tex]
U1 = 3.84 × [tex]10^{-27}[/tex] J
and final potential energy = U2
U2 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.03}[/tex]
U2 = 7.68 × [tex]10^{-27}[/tex] J
and speed of electron = v
so we will apply here conservation of energy
0.5×m×v² = U2 - U1 ................1
so
0.5×9.11× [tex]10^{-31}[/tex] ×v² = 3.84 × [tex]10^{-27}[/tex]
v = 91.82 m/s
so electron moving at 91.82 m/s
Answer:
The speed of the electron is 91.86 m/s.
Explanation:
Given that,
Distance of electron from proton = 6.00 cm
Distance of proton = 3.00 cm
We need to calculate the initial potential energy
[tex]U_{1}=\dfrac{kq_{1}q_{2}}{r}[/tex]
Put the value into the formula
[tex]U_{i}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{6.00\times10^{-2}}[/tex]
[tex]U_{i}=3.84\times10^{-27}\ J[/tex]
The final potential energy
[tex]U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{3.00\times10^{-2}}[/tex]
[tex]U_{f}=7.68\times10^{-27}\ J[/tex]
We need to calculate the speed of the electron
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=U_{f}-U_{i}[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times9.1\times10^{-31}\times v^2=7.68\times10^{-27}-3.84\times10^{-27}[/tex]
[tex]v=\sqrt{\dfrac{2\times3.84\times10^{-27}}{9.1\times10^{-31}}}[/tex]
[tex]v=91.86\ m/s[/tex]
Hence, The speed of the electron is 91.86 m/s.
The heat in an internal combustion engine raises the internal temperatures to 300 deg C. If the outside temperature is 25 deg C, what is the maximum efficiency of this engine? And is it possible to attin this efficiency?
Answer:48 %
Explanation:
Given
outside temperature [tex]=25^{\circ}\approx 298 K[/tex]
Internal temperature[tex]=300^{\circ}\approx 573 K[/tex]
Maximum efficiency is carnot efficiency which is given by
[tex]\eta =1-\frac{T_L}{T_H}[/tex]
[tex]\eta =1-\frac{298}{573}[/tex]
[tex]\eta =\frac{275}{573}=0.4799[/tex]
i.e. [tex]47.99 %\approx 48 %[/tex]
No it is not possible to achieve this efficiency because carnot engine is the ideal engine so practically it is not possible but theoretically it is possible