Answer:
424088766.068 m
Explanation:
Radius of the circular orbit that the satellite is 2.6 Earth radii (r) = 2.6 R
R = Radius of earth = 6371000 m (mean radius)
In order to find the distance that the satellite travels in 5.89 hours to complete one complete revolution is the circumference of the circular orbit
Circumference of a circle = 2×π×r
⇒Distance travelled in 5.89 hours = 2×π×2.6 R
⇒Distance travelled in 5.89 hours = 2×π×2.6×6371000
⇒Distance travelled in 5.89 hours = 104078451.3393m
Distance travelled in 1 hour = 104078451.3393/5.89 = 17670365.252 m
∴ Distance travelled in 24 hours = 17670365.252×24 = 424088766.068 m
Final answer:
The satellite orbits Earth at approximately 2.60 Earth radii and travels a total distance of about 4.23×10⁸ meters in one day.
Explanation:
The question pertains to calculating the total distance traveled by a satellite orbiting Earth at a given distance over the duration of one day. To solve this, we need to determine the circumference of the satellite's circular orbit and then calculate how many orbits it completes in a day.
To determine the satellite's orbit circumference, we use the formula for the circumference of a circle, C = 2πr, where r represents the orbit radius. Given that the satellite orbits at a distance of about 2.60 Earth radii, we can express this distance in meters by multiplying the Earth's radius, 6.37×10⁶ meters, by 2.60, yielding a radius of approximately 1.6562×10⁷ meters. Therefore, the satellite's orbit circumference is approximately 1.04×10⁸ meters.
Since the satellite completes one orbit in 5.89 hours, we can now calculate how many orbits it completes in a day (24 hours) by dividing 24 by 5.89, which is approximately 4.07 orbits per day. The total distance traveled in one day is the circumference multiplied by the number of orbits, resulting in a distance of approximately 4.23×10⁸ meters per day.
A total of 600 C of charge passes through flashlight bulb in 0.500 hr. What is the average current? Give your answer in A.
The average current passing through a device is given by:
I = Q/Δt
I is the average current
Q is the amount of charge that has passed through the device
Δt is the amount of elapsed time
Given values:
Q = 600C
Δt = 0.500hr = 1800s
Plug in the values and solve for I:
I = 600/1800
I = 0.333A
The average current flowing in the flashlight bulb is 0.333 A
The given parameters:
quantity of the charge, Q = 600 C
time of the current flow, t = 0.5 hour
To find:
the average current, IThe average current is calculated from quantity of charge that flows in the given time period:
Q = It
where:
I is the average current
Q is the quantity of the charge
t is the time period
From the equation above, make the average current the subject of the formula:
[tex]I = \frac{Q}{t} \\\\I = \frac{600 \ C}{0.5 \ hr \times 3600 \ s} \\\\I = \frac{600 \ C}{1800 \ s} \\\\I = 0.333 \ A[/tex]
Thus, the average current is 0.333 A
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A uniform rod of mass 2.55×10−2 kg and length 0.380 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.200 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.80×10−2 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 rev/min . Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. What is the angular speed of the system at the instant when the rings reach the ends of the rod?
What is the angular speed of the rod after the rings leave it?
Answer:
[tex]\omega _{f}=0.3107rad/sec[/tex]
partb) [tex]\omega _{f}=14.93rad/sec[/tex]
Explanation:
Since the system is isolated it's angular momentum shall be conserved
[tex]L_{system}=I_{system}\omega \\\\I_{system}=I_{rod}+2\times m\times r^{2}\\\\I_{system}=\frac{1}{12}ml^{2}+2\times 0.2\times (4.8\times 10^{-2})^{2}\\\\I_{system}=1.22845\times 10^{-3}kgm^{2}\\\\L_{system}=1.22845\times 10^{-3}kgm^{2}\times 3.66rad/sec[/tex]
[tex]\L_{system}=4.496\times 10^{-3}kgm^{2}[/tex]
Now the final angular momentum of the system
[tex]L_{fianl}=I_{final}\omega \\\\I_{final}=I_{rod}+2\times m\times r_{f}^{2}\\\\I_{final}=\frac{1}{12}ml^{2}+2\times 0.2\times (0.19)^{2}\\\\I_{final}=0.01447kgm^{2}\\\\L_{final}=0.01447kgm^{2}\times \omega _{final}rad/sec[/tex]
Thus equating initial and final angular momentum we solve for final angular velocity of rod as
[tex]\omega _{f}=\frac{4.496\times 10^{-3}}{0.01447}\\\\\omega _{f}=0.3107rad/sec[/tex]
part b)
When the rings leave the system we again conserve the angular momentum just before the rings leave the system and the instant when they just leave
[tex]\therefore 0.01447=\frac{1}{12}ml^{2}w\\\\\therefore w=\frac{4.582\times 10^{-3}}{3.068\times10^{-4} }\\\\w_{final}=14.93rad/sec[/tex]
Answer:
A)[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].
B)[tex]\omega_{f}=140\frac{rev}{min}[/tex].
Explanation:
A)
For this problem, we will use the conservation of angular momentum.
[tex]L_{0}=L_{f}\\[/tex].
In the beginning, we have that
[tex]L_{0}=I_{0}\omega_{0}\\[/tex]
where [tex]I_{0}[/tex] is the inertia moment of all the system at the starting position, this is the inertia moment of the rod plus the inertia moment of each ring ([tex]mr^{2}[/tex], with [tex]r[/tex] the distance from the ring to the fixed axis and, [tex]m[/tex] its mass) at the starting position and, [tex]\omega_{0}[/tex] is the initial angular velocity. So
[tex]L_{0}=(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}[/tex].
When the rings are at the ends of the rod the angular momentum becomes
[tex]L_{f}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex],
where [tex]r_{f}[/tex] is the distance from the fixed axis to the end of the rod (the final position of the rings).
Using conservation of angular momentum we get
[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2}+2mr_{f}^{2})\omega_{f}[/tex].
thus
[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2}+2mr_{f}^{2})}[/tex]
computing this last expresion we get
[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2+2(0.200)(0.19)^{2})}[/tex]
[tex]\omega_{f}=2.92\frac{rev}{min}[/tex].
B)
Again we use the conservation of angular momentum. The initial angular momentum if the same as before. The final angular momentum will be
[tex]L_{f}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex],
this time we will not take into account the inertia moment of the rings because they are no longer part of the system (they leave the rod).
[tex](\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}=(\frac{1}{12}Ml^{2})\omega_{f}[/tex].
thus
[tex]\omega_{f}=\frac{(\frac{1}{12}Ml^{2}+2mr_{0}^{2})\omega_{0}}{(\frac{1}{12}Ml^{2})}[/tex]
computing this last expresion we get
[tex]\omega_{f}=\frac{(\frac{1}{12}(2.55*10^{-2})(0.380)^{2}+2(0.200)(4.80*10^{-2})^{2})(35.0)}{(\frac{1}{12}(2.55*10^{-2})(0.380)^2})[/tex]
[tex]\omega_{f}=140\frac{rev}{min}[/tex].
Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface. When she lands on Noveria, a distant planet in our galaxy, she weighs 356 N. What is the acceleration of gravity on Noveria in m/s2? Round to one decimal place. (hint: first find her mass!)
Answer:
Acceleration of gravity on Noveria = 4.4 m/s²
Explanation:
Commander Shepard, an N7 spectre for Earth, weighs 799 N on the Earth's surface.
We have weight, W = mg
Acceleration due to gravity, g = 9.81m/s²
799 = m x 9.81
Mass of Shepard, m = 81.45 kg
She lands on Noveria, a distant planet in our galaxy, she weighs 356 N.
We have weight, W = mg'
356 = 81.45 xg'
Acceleration of gravity on Noveria, g' = 4.4 m/s²
What frequency must a sound wave have in air to have the same wavelength as a 750 Hz sound wave in a platinum bar? Vair = 340 m/s, Vpt = 2800 m/s,
Answer:
f = 91.1 Hz
Explanation:
As we know that it is given here that wavelength of sound in platinum is same as the wavelength of sound in air
so we can use the formula of wavelength in two mediums
[tex]\frac{speed}{frequency} = \frac{speed}{frequency}[/tex]
now it is given that
[tex]v_{air} = 340 m/s[/tex]
[tex]v_{pt} = 2800 m/s[/tex]
frequency of sound in platinum is 750 Hz
now frequency of sound in air = f
now from above formula
[tex]\frac{340}{f} = \frac{2800}{750}[/tex]
[tex]f = 91.1 Hz[/tex]
A table-top fan has radius of 0.5 m. It starts to rotate from rest to 800 rpm within 30 seconds. Determine: 1) the angular momentum 2) how many revolutions have it gone through after 2 minutes? 3) If the rotational inertia of each blade around the center is 0.3 kg.m^ 2, what is the magnitude of the torque provided by the motor of the fan, assuming no friction?
Answer:
Part a)
L = 25.13 kg m^2/s
Part b)
N = 3200 rev
Part c)
torque = 0.837 Nm
Explanation:
Part a)
As we know that angular frequency is given as
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(800/60)[/tex]
[tex]\omega = 83.77 rad/s[/tex]
now the angular momentum is given as
[tex]L = I\omega[/tex]
[tex]L = (0.3)(83.77)[/tex]
[tex]L = 25.13 kg m^2/s[/tex]
Part b)
angular acceleration of fan is given as
[tex]\alpha = 2\pi\frac{800/60}{30}[/tex]
[tex]\alpha = 2.79 rad/s^2[/tex]
total number of revolutions are
[tex]N = \frac{1}{4\pi}\alpha t^2[/tex]
[tex]N = \frac{1}{4\pi}(2.79)(120^2)[/tex]
[tex]N = 3200 rev[/tex]
Part c)
Torque is given as
[tex]\tau = I\alpha[/tex]
[tex]\tau = (0.30)(2.79) = 0.837 Nm[/tex]
The acceleration of a particle is given by a = −ks2 , where a is in meters per second squared, k is a constant, and s is in meters. Determine the velocity of the particle as a function of its position s. Evaluate your expression for s = 5 m if k = 0.1 m−1 s−2 and the initial conditions at time t = 0 are s0 = 3 m and v0 = 10 m /s
Answer:
[tex]v = \sqrt{v_0^2 - \frac{2k}{3}(s^3 - s_0^3)}[/tex]
v = 9.67 m/s
Explanation:
As we know that acceleration is rate of change in velocity
so it is defined as
[tex]a = \frac{dv}{dt}[/tex]
[tex]a = v\frac{dv}{ds}[/tex]
here we know that
[tex]a = - ks^2 = v\frac{dv}{ds}[/tex]
now we have
[tex]vdv = - ks^2ds[/tex]
integrate both sides we have
[tex]\int vdv = -k \int s^2ds[/tex]
[tex]\frac{v^2}{2} - \frac{v_0^2}{2} = -k(\frac{s^3}{3} - \frac{s_0^3}{3})[/tex]
[tex]v^2 = v_0^2 - \frac{2k}{3}(s^3 - s_0^3)[/tex]
here we know that
[tex]v_0 = 10 m/s[/tex]
[tex]s_0 = 3 m[/tex]
[tex]v^2 = 10^2 - \frac{2(0.10)}{3}(5^3 - 3^3)[/tex]
[tex]v = 9.67 m/s[/tex]
A car is traveling at a speed of 70 mph, a long a straight road, when the driver slams on the brakes and the car m and the car a,=- 2.00 accelerates at comes to a complete stop. How much time passes between the time when the brakes were applied and the instant the car comes to a complete stop and how far did the car travel in that time? (1609 meters 1 mile, 3600 seconds 1 hour) a.) 7.28 s and 422.10 m. b.) 15.64 s and 244.70 m. c.) 12.32 s and 428.73 m. d.) 14.53 s and 210.98 m. e.) None of the above.
Answer:
Option B is the correct answer.
Explanation:
Speed of car = 70mph [tex]=70\times \frac{1609}{3600}=31.28m/s[/tex]
Deceleration of car = 2 m/s²
We have equation of motion
v = u + at
0 = 31.28 - 2 t
t = 15.64 seconds.
We also have
[tex]s=ut+\frac{1}{2}at^2=31.28\times 15.64-\frac{1}{2}\times 2\times 15.64^2=244.70m[/tex]
Option B is the correct answer.
A jet airplane with a 75.0 m wingspan is flying at 255 m/s. What emf is induced between the wing tips in V if the vertical component of the Earth's magnetic field is 3.00 x 10^-5 T?
Answer:
The emf is 0.574 volt.
Explanation:
Given that,
Magnetic field [tex]B =3.00\times10^{-5}\ T[/tex]
Length = 75.0 m
Velocity = 255 m/s
We need to calculate the emf
Using formula of emf
[tex]\epsilon=Blv[/tex]
Where, v = velocity
l = length
B = magnetic field
Put the value into the formula
[tex]\epsilon= 3.00\times10^{-5}\times75.0\times255[/tex]
[tex]\epsilon=0.574\ volt[/tex]
Hence, The emf is 0.574 volt.
The emf induced between the wingtips of a jet airplane flying at 255 m/s with a 75.0 m wingspan in Earth's magnetic field of strength 3.00 x 10^-5 T is 0.57 V. This is calculated using the formula for motional emf: E = Blv.
Explanation:To calculate the emf induced between the airplane wings due to Earth's magnetic field, we use the formula for motional emf: E = Blv, where B is the magnetic field strength, l is the length of the conductor moving through the field (in this case, the wingspan of the airplane), and v is the velocity of the conductor.
Substituting the given values into the formula we get:
E = (3.00 x 10^-5 T) * (75.0 m) * (255 m/s) = 0.57 V
This means that a 0.57 Volt emf is induced between the wingtips of the airplane due to the Earth's magnetic field. This is a direct application of Faraday's law of electromagnetic induction, which states that a changing magnetic field will induce an emf in a conductor.
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The velocity field of a steady flow is given as V=(U0+bx)i?b y j , where U0 and b are constant. Find the acceleration in the x- and y-directions if U0=1m/ s and b=0.3m .
Explanation:
The velocity field o a steady flow is given by :
[tex]V=(U_o+bx)i[/tex]
Where
U₀ and b are constant
We know that, the acceleration is given by :
[tex]a=\dfrac{dV}{dt}[/tex]
[tex]a=\dfrac{d((U_o+bx)i)}{dt}[/tex]
a = b i
a = 0.3 m/s²
So, acceleration corresponding to this velocity is 0.3 m/s². Hence, this is the required solution.
By how much does a 64.3 kg mountain climber stretch her 0.910 cm diameter nylon rope when she hangs 32.2 m below a rock outcropping? (For nylon, Y = 1.35 â 10^9 Pa. Enter your answer in centimeters.)
Answer:
23 cm
Explanation:
m = 64.3 kg, diameter = 0.910 cm
radius, r = 0.455 cm = 4.55 x 10^-3 m
A = 3.14 x (4.55 x 10^-3)^2 = 6.5 x 10^-5 m^2
L = 32.2 m
Y = 1.35 x 10^9 Pa
Let the stretched length is l.
Use the formula for Young's modulus
Y = mg L / A l
l = m g L / A Y
l = (64.3 x 9.8 x 32.2) / (6.5 x 10^-5 x 1.35 x 10^9)
l = 0.23 m
l = 23 cm
A long wire carries a current toward the south in a magnetic field that is directed vertically upward. What is the direction of the magnetic force on the wire?
The magnetic force on a wire carrying current towards the south under a magnetic field directed vertically upwards will point towards the East. In order to determine this, use the right-hand rule.
Explanation:The direction of the magnetic force on a current-carrying wire under a magnetic field can be deduced using the right-hand rule. In this case, with the current flowing towards the south and the magnetic field directed vertically upward, you would point your right thumb in the direction of the current (southwards) and curl your fingers in the direction of the magnetic field (upwards). The palm of your hand will then face toward the direction of the force. In this case, the force would be pointing toward the East.
The right-hand rule is a vital principle in the study of electromagnetism as it aids in identifying the direction of various quantities in magnetic fields. The magnetic force on a current-carrying wire represents the phenomenon underlying the working of many electric motors.
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An airplane is moving horizontally with a constant velocity of (+ 115 m/s) at altitude of 1050 m. The directions to the right and upward have been chosen as the positive directions. The plane releases a "care package' that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground
Answer:
The time required for the package to hit the ground = 14.63 seconds.
Explanation:
Considering vertical motion of care package:-
Initial velocity, u = 0 m/s
Acceleration , a = 9.81 m/s²
Displacement, s = 1050 m
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
1050 = 0 x t + 0.5 x 9.81 x t²
t = 14.63 seconds
The time required for the package to hit the ground = 14.63 seconds.
To determine the time required for the care package to hit the ground, we can separate the vertical and horizontal components of the motion. By using the equation h = (1/2)gt^2 and plugging in the values h = 1050 m and g = 9.8 m/s^2, we find that it will take approximately 14.63 seconds for the care package to hit the ground.
Explanation:To determine the time required for the care package to hit the ground, we can separate the vertical and horizontal components of the motion. Since the horizontal velocity of the airplane does not change, it will not affect the time of flight. The vertical motion can be analyzed using the equation of motion for free fall, which is h = (1/2)gt^2, where h is the initial vertical displacement, g is the acceleration due to gravity, and t is the time of flight. In this case, the initial vertical displacement is 1050 meters and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging these values into the equation, we can solve for t and find the time required for the care package to hit the ground.
By using the equation h = (1/2)gt^2 and plugging in the values h = 1050 m and g = 9.8 m/s^2, we have 1050 = (1/2)(9.8)t^2. Solving for t, we get t^2 = (1050) / (0.5)(9.8) = 214.29. Taking the square root of both sides, we find t ≈ 14.63 seconds. Therefore, it will take approximately 14.63 seconds for the care package to hit the ground.
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An electron enters a magnetic field of 0.66 T with a velocity perpendicular to the direction of the field. At what frequency does the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)
Answer:
1.85 x 10^10 cycles per second
Explanation:
B = 0.66 T, theta = 90 degree, q = 1.6 x 10^-19 C,
The time period of electron is given by
T = 2 π m / B q
Frequency is teh reciprocal of time period.
f = 1 /T
f = B q / (2 π m)
f = (0.66 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31)
f = 1.85 x 10^10 cycles per second
A rifle is aimed horizontally at a target 30m away. The bullet hits the target 2.9 cm below the aiming point. What is the bullet's speed ( in the unit of m/s) as it emerges from the rifle.
Answer:
Speed of bullet = 389.61 m/s.
Explanation:
Considering the vertical motion of bullet
Initial vertical speed = 0 m/s
Vertical displacement = 2.9 cm = 0.029 m
Vertical acceleration = 9.81 m/s²
Substituting in s = ut + 0.5at²
0.029 = 0 x t + 0.5 x 9.81 x t²
t = 0.077 s
So ball hits the target after 0.077 s.
Now considering the vertical motion of bullet
Time = 0.077 s
Horizontal displacement = 30 m
Horizontal acceleration = 0 m/s²
Substituting in s = ut + 0.5at²
30 = u x 0.077 + 0.5 x 0 x 0.077²
u = 389.61 m/s
Speed of bullet = 389.61 m/s.
Answer:
390 m/s
Explanation:
Let the horizontal speed be u.
Horizontal distance , x = 30 m
Vertical distance, y = 2.9 cm
Let time taken be t.
Use second equation of motion in vertical direction
H = uy × t + 1/2 gt^2
0.029 = 0.5 × 9.8 × t^2
t = 0.077 s
Horizontal distance = horizontal velocity × time
30 = u × 0.077
u = 390 m/s
If a person is standing erect and flexes the trunk on the hip, the center of mass will move __________ and the line of gravity moves __________ within the base of support.
Answer:
anterior
anterior
Explanation:
In the given question is asked that
If a person is standing erect and flexes the trunk on the hip, the center of mass will move ___________ and the line of gravity moves___________ within the base of support.
The current answer to the blanks will be
anterior
anterior
hope this helps any further query can be asked in comment section.
As a car increases velocity, does its kinetic energy or momentum increase faster? Explain.
Explanation:
The kinetic energy of an object is associated with its motion. Mathematically, it is given by :
[tex]E_k=\dfrac{1}{2}mv^2[/tex]........(1)
Where
m is the mass of the car
v is the velocity of the car
And the momentum of an object is equal to the product of mass and velocity i.e.
[tex]momentum(p)=mass(m)\times velocity(v)[/tex]..........(2)
From equation (1) and (2) it is clear that the kinetic energy and the momentum is directly proportional to the velocity of the object. As a car increases velocity, its kinetic energy or momentum increase faster.
A heavy fuel oil has a specific gravity of 0.918. How much will 100 gallons(300 liters) of this oil weigh? INCLUDE UNITS!
Answer:
275.4 kg
Explanation:
specific gravity = 0.918
Density = 0.918 g/cm^3 = 918 kg/m^3
Volume = 100 gallons = 300 litres = 300 x 10^-3 m^3 = 0.3 m^3
Mass = volume x density = 0.3 x 918 = 275.4 kg
A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. What is the magnitude of the acceleration of the ball when it is at its highest point?
Answer:
at the highest point of the path the acceleration of ball is same as acceleration due to gravity
Explanation:
At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest
So here we will say that at the highest point of the path the speed of the ball comes to zero
now by the force diagram we can say that net force on the ball due to gravity is given by
[tex]F_g = mg[/tex]
now the acceleration of ball is given as
[tex]a = \frac{F_g}{m}[/tex]
[tex]a = \frac{mg}{m} = g[/tex]
so at the highest point of the path the acceleration of ball is same as acceleration due to gravity
Final answer:
The magnitude of the acceleration of a tennis ball at its highest point, in an evacuated chamber and under the influence of gravity alone, is 9.8 m/s². This remains constant regardless of the ball's position in its trajectory.
Explanation:
The question asks about the magnitude of the acceleration of a tennis ball at its highest point when it is shot vertically upward in an evacuated chamber with an initial speed. This scenario is governed by the principles of classical mechanics, specifically under the domain of gravity's influence on objects in motion. Regardless of whether the object is moving up or down or is at its peak height, the acceleration due to gravity on Earth remains constant.
Acceleration at the highest point, or at any point during its motion under gravity alone (assuming no air resistance), is -9.8 m/s2. It's essential to understand that the negative sign indicates the direction of acceleration, which is towards the center of the Earth but does not affect the magnitude of acceleration. Hence, the magnitude of acceleration is 9.8 m/s2, indicating that gravity is the only force acting on the ball throughout its trajectory.
If a 54 kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?
Answer:
Power, P = 600 watts
Explanation:
It is given that,
Mass of sprinter, m = 54 kg
Speed, v = 10 m/s
Time taken, t = 3 s
We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.
[tex]P=\dfrac{W}{t}[/tex]
Work done is equal to the change in kinetic energy of the sprinter.
[tex]P=\dfrac{\dfrac{1}{2}mv^2}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}[/tex]
P = 900 watts
So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.
Final answer:
The average power generated is 900 Watts, which is approximately 1.21 horsepower.
Explanation:
The question involves calculating the average power generated by a sprinter during acceleration. Power is defined as the work done per unit time. To find the power, we need to first calculate the work done, which, in the case of the sprinter, is the kinetic energy gained as they accelerate to the final speed.
In this case, we can calculate the kinetic energy (KE) of the sprinter using the formula KE = 0.5 × mass ×[tex]velocity^2[/tex]. The mass of the sprinter is given as 54 kg and the final velocity is 10 m/s. Thus, KE = 0.5 × 54 kg × [tex](10 m/s)^2[/tex] = 2700 Joules.
Now, the power can be found by dividing the work done by the time it takes to do that work. The sprinter takes 3 seconds to achieve this final velocity. So, the average power P is P = work/time = 2700 Joules / 3 s = 900 Watts.
Converting this power to horsepower (1 horsepower = 746 Watts), we get approximately P = 900 W × 1 hp/746 W = 1.21 horsepower.
A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 2.10 cm distant from the first, in a time Interval of 1.10 x 10-6 s. (a) Find the magnitude of the electric field N/C (b) Find the speed of the proton when it strikes the negatively charged plate m/s
Answer:
a)
365.3 N/C
b)
3.85 x 10⁴ m/s
Explanation:
q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷ kg
d = distance between the plates = 2.10 cm = 0.021 m
t = time interval = 1.10 x 10⁻⁶ sec
E = magnitude of electric field
v₀ = initial velocity = 0 m/s
a = acceleration of the proton
Using the equation
d = v₀ t + (0.5) a t²
0.021 = (0) (1.10 x 10⁻⁶) + (0.5) a (1.10 x 10⁻⁶)²
a = 3.5 x 10¹⁰ m/s²
acceleration of the proton inside electric field is given as
[tex]a =\frac{qE}{m}[/tex]
[tex]3.5\times 10^{10} =\frac{(1.6\times 10^{-19})E}{1.67\times 10^{-27}}[/tex]
E = 365.3 N/C
b)
v = final speed of the proton
using the equation
v = v₀ + a t
v = 0 + (3.5 x 10¹⁰) (1.10 x 10⁻⁶)
v = 3.85 x 10⁴ m/s
A water trough is 8 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 80 cm wide at the top, and has height 60 cm. If the trough is being filled with water at the rate of 0.3 m3/min how fast is the water level rising when the water is 30 cm deep?
Answer:It is rising at a rate of [tex]7.5cm/min[/tex]
Explanation:
We have volume of trapezoid equals
[tex]V=Area\times Length\\\\V=\frac{1}{2}(a+b)h\times L[/tex]
Thus at any time 't' we have
[tex]V(t)=\frac{1}{2}(a(t)+b(t))h(t)\times L\\\\\therefore V(t)=\frac{1}{2}(20+b(t))\times h(t)\times L[/tex]
Differentiating both sides with respect to time we get
[tex]\frac{dV(t)}{dt}=\frac{1}{2}b'(t)h(t)L+\frac{1}{2}(20+b(t))\times h'(t)L[/tex]
Applying values we have
[tex]b(t)=20+h(t)\\b'(t)=h'(t)[/tex]
Thus we have
[tex]\frac{dV(t)}{dt}=\frac{1}{2}h'(t)h(t)L+\frac{1}{2}(20+20+h(t))\times h'(t)L\\\\2V'(t)=h'(t)L[h(t)+(40+h(t))]\\\\\therefore h'(t)=\frac{2V'(t)}{L(h(t)+(40+h(t)))}[/tex]
Applying values we get
[tex]h'(t)=0.075m/min=7.5cm/min[/tex]
If the intensity of light that is incident on a piece of metal is increased, what else will be increased? Choose all that apply. number of electrons ejected stopping voltage cutoff frequency frequency KEmax work function wavelength
Answer:
explained
Explanation:
When the intensity of light is increased on a piece of metal only the number of electron ejected will increase because all other things independent of intensity of light.
Light below certain frequency will not cause any electron emission no matter how intense.
The intensity produces more electron but does not change the maximum kinetic energy of electrons.
Work function is independent of the intensity of light, because it is an intrinsic property of a material.
Motor oil, with a viscosity of 0.25 N ∙ s/m 2, is flowing through a tube that has a radius of 5.0 mm and is 25 cm long. The drop in pressure is 300 kPa. What is the volume of oil flowing through the tube per second?
Answer:
1.18 x 10^-3 m^3/s
Explanation:
η = 0.25 N s/m^2, radius, r = 5 mm = 0.005 m, l = 25 cm = 0.25 m
P = 300 kPa = 300,000 Pa, Volume of flow = ?
By use of Poiseuillie's equation
Volume of flow = π P r^4 / (8 η l)
Volume of flow = (3.14 x 300000 x 0.005^4) / (8 x 0.25 x 0.25)
Volume of flow = 1.18 x 10^-3 m^3/s
Using Poiseuille's Law for viscous flow, the volume of oil flowing through the tube per second is calculated to be 1.8 liters/sec.
Explanation:The volume of oil flowing through the tube per second can be calculated using Poiseuille's Law, which is an equation for viscous flow. This Law states that the flow rate of fluid through a pipe (Q) is directly proportional to the fourth power of the radius (r) of the tube, the pressure difference (ΔP) and inversely proportional to the viscosity (η) of the fluid, and the length (L) of the tube.
Using Poiseuille's Law, we have:
Q = (π/8) * (ΔP/r^4L) * η
By substituting the provided values:
Q = (π/8) * (300,000 Pa/(0.005 m)^4 * 0.25 m) * 0.25 N ∙ s/m^2
This yields: Q = 0.0018 m^3/s, or 1.8 liters/sec.
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What hanging mass will stretch a 3.0-m-long, 0.32 mm - diameter steel wire by 1.3 mm ? The Young's modulus of steel is 20×10^10 N/m^2.
Answer:
0.71 kg
Explanation:
L = length of the steel wire = 3.0 m
d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m
Area of cross-section of the steel wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.32 x 10⁻³)²
A = 8.04 x 10⁻⁸ m²
ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m
Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²
m = mass hanging
F = weight of the mass hanging
Young's modulus of steel is given as
[tex]Y = \frac{FL}{A\Delta L}[/tex]
[tex]20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}[/tex]
F = 6.968 N
Weight of the hanging mass is given as
F = mg
6.968 = m (9.8)
m = 0.71 kg
A 0.12-kg ball is moving at 6 m/s when it is hit by a bat, causing it to reverse direction and have a speed of 14 m/s. What is the change in the magnitude of the momentum of the ball? A. 0.39 kg*m/s B. 0.42 kg*m/s C. 1.3 kg*m/s D. 2.4 kg*m/s
Explanation:
0.05
momentum =Mv
then,
Mv=mv
v2=0.12*6/14
v2=0.05kgm/s
The change in the magnitude of the momentum of the ball is 2.4 kg·m/s.
Explanation:To calculate the change in the magnitude of the momentum of the ball, we need to find the initial momentum and the final momentum of the ball. Momentum is calculated by multiplying the mass of an object by its velocity. Let's start by calculating the initial momentum:
Initial momentum = mass × initial velocity
Final momentum = mass × final velocity
Change in momentum = final momentum - initial momentum
Plugging in the given values:
Initial momentum = 0.12 kg × 6 m/s
Final momentum = 0.12 kg × (-14 m/s) (since the ball changes direction)
Change in momentum = (-1.68 kg·m/s) - (0.72 kg·m/s) = -2.4 kg·m/s.
The change in the magnitude of the momentum of the ball is 2.4 kg·m/s. However, since the question is asking for the magnitude, we take the absolute value of the change in momentum, which is 2.4 kg·m/s.
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Two slits are illuminated by a 602 nm light. The angle between the zeroth-order bright band at the center of the screen and the fourth-order bright band is 17.4 ◦ . If the screen is 138 cm from the double-slit, how far apart is this bright?
Answer:
y = 43.2 cm
Explanation:
As we know by the formula of diffraction we have
[tex]a sin\theta = N\lambda[/tex]
here we have
[tex]\theta = 17.4[/tex]
N = 4
[tex]\lambda = 602 nm[/tex]
now we have
[tex]a sin(17.4) = 4(602\times 10^{-9})[/tex]
[tex]a = 8.05 \times 10^{-6} m[/tex]
now the distance of screen is 138 cm
now we can say
[tex]\frac{y}{L} = tan\theta[/tex]
[tex]y = L tan(\theta)[/tex]
[tex]y = 138 tan(17.4) = 43.2 cm[/tex]
How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from 30.0ºC to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W
Answer:
Part a)
[tex]Q = 2.47 \times 10^6[/tex]
Part b)
t = 4950.3 s
Explanation:
As we know that heat required to raise the temperature of container and water in it is given as
[tex]Q = m_1s_2\Delta T_1 + m_2s_2\Delta T_2[/tex]
here we know that
[tex]m_1 = 0.750[/tex]
[tex]s_1 = 900[/tex]
[tex]m_2 = 2.50 kg[/tex]
[tex]s_2 = 4186[/tex]
[tex]\Delta T_1 = \Delta T_2 = 100 - 30 = 70^oC[/tex]
now we have
[tex]Q_1 = 0.750(900)(70) + (2.5)(4186)(70) = 779800 J[/tex]
now heat require to boil the water
[tex]Q = mL[/tex]
here
m = 0.750 kg
[tex]L = 2.25 \times 10^6 J/kg[/tex]
now we have
[tex]Q_2 = 0.750(2.25 \times 10^6) = 1.7 \times 10^6 J[/tex]
Now total heat required is given as
[tex]Q = Q_1 + Q_2[/tex]
[tex]Q = 779800 + 1.7 \times 10^6 = 2.47 \times 10^6 J[/tex]
Part b)
Time taken to heat the water is given as
[tex]t = \frac{Q}{P}[/tex]
here we know that
power = 500 W
now we have
[tex]t = \frac{2.47 \times 10^6}{500} = 4950.3 s[/tex]
A 1500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500-kg van traveling north at a speed of 20.0 m/s. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (i.e. they stick together).
From conservation of linear momentum, the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.
COLLISIONThere are four types of collision
Elastic collisionPerfectly elastic collisionInelastic collisionPerfectly inelastic collisionIn elastic collision, both momentum and energy are conserved. While in inelastic collision, only momentum is conserved.
From the given question, the following parameters are given.
[tex]m_{1}[/tex] = 1500kg[tex]v_{1}[/tex] = 25 m/s[tex]m_{2}[/tex] = 2500 kg[tex]v_{2}[/tex] = 20 m/sSince the collision is inelastic, they will both move with a common velocity after collision.
Horizontal component
[tex]m_{1}[/tex][tex]v_{1}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex] ) V
1500 x 25 = (1500 + 2500) V
37500 = 4000V
V = 37500 / 4000
V = 9.375 m/s
Vertical component
[tex]m_{2}[/tex][tex]v_{2}[/tex] = ([tex]m_{1}[/tex] + [tex]m_{2}[/tex])V
2500 x 20 = (1500 + 2500)V
50000 = 4000V
V = 50000 / 4000
V = 12.5 m/s
The net velocity will be the magnitude of the velocity of the wreckage after collision
V = [tex]\sqrt{9.4^{2} + 12.5^{2} }[/tex]
V = [tex]\sqrt{244.61}[/tex]
V = 15.6 m/s
The direction will be
Tan Ф = 12.5 / 9.4
Ф = [tex]Tan^{-1}[/tex](1.329)
Ф = 53 degrees.
Therefore, the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.
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The velocity of the wreckage after the collision is 32.015 m/s in the direction 38.66° north of east.
Explanation:This problem is related to the conservation of linear momentum which states the total momentum of an isolated system remains constant if no external forces act on it. Momentum is a vector quantity having both magnitude and direction. The total initial and final momentum in both horizontal (x) and vertical (y) directions must be equal.
Initial momentum of the car is (1500 kg)*(25.0 m/s) = 37500 kg*m/s in the east direction whereas initial momentum of the van is (2500 kg)*(20.0 m/s) = 50000 kg*m/s in the north direction.
Since the collision is perfectly inelastic, the two vehicles stick together after the collision and move as one. Therefore, the final momentum is the vector sum of the individual momenta. We therefore calculate the magnitude of the resultant velocity using Pythagoras theorem, √[(25.0 m/s)² + (20.0 m/s)²] = 32.015 m/s.
To find the direction of the final velocity, we use the tangent of the angle which is equal to the vertical component divided by the horizontal component, which gives us an angle of 38.66° north of east.
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Light with a wavelength of 488 nm is incident on a single slit with a width of 6.23 x 10^-4 m. If the screen is 2.7 m away, what is the distance to the first antinodal line?
Answer:
The distance to the first anti-nodal line is [tex]2.11\times10^{-3}\ m[/tex].
Explanation:
Given that,
Wavelength = 488 nm
Width [tex]d=6.23\times10^{-4}\ m[/tex]
Distance D =2.7 m
We need to calculate the distance to the first anti-nodal line
Using formula of the distance for first anti-nodal line
[tex]y_{n}=\dfrac{n\lambda D}{d}[/tex].....(I)
Where, n = number of fringe
d = width
D = distance from the screen
[tex]\lambda[/tex]=wavelength of light
Put the all value in the equation (I)
[tex]y_{n}=\dfrac{488\times10^{-9}\times2.7}{6.23\times10^{-4}}[/tex]
[tex]y_{n}=2.11\times10^{-3}\ m[/tex]
Hence, The distance to the first anti-nodal line is [tex]2.11\times10^{-3}\ m[/tex].
A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature of the gas is raised to 235 degrees kelvin, what is the new pressure of the gas? (Assume that there's no change in volume.) A. 854.46 kPa B. 510 kPa C. 3.636 kPa D. 1.702 kPa
Answer:
A) 854.46 kPa
Explanation:
P₁ = initial pressure of the gas = 400 kPa
P₂ = final pressure of the gas = ?
T₁ = initial temperature of the gas = 110 K
T₂ = final temperature of the gas = 235 K
Using the equation
[tex]\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}[/tex]
Inserting the values
[tex]\frac{400}{110}=\frac{P_{2}}{235}[/tex]
P₂ = 854.46 kPa