Answer:
[tex]473.56\ \text{J}[/tex]
Explanation:
T = Time taken to complete one revolution = 0.3 s
m = Mass of sphere = 15 kg
r = Radius of sphere = 0.6 m
I = Moment of inertia of sphere = [tex]\dfrac{2}{5}mr^2[/tex]
Angular speed of the sphere is
[tex]\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.3}\\\Rightarrow \omega=20.94\ \text{rad/s}[/tex]
Rotational kinetic energy is given by
[tex]K_r=\dfrac{1}{2}I\omega^2=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2\\\Rightarrow K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 15\times 0.6^2\times 20.94^2\\\Rightarrow K_r=473.56\ \text{J}[/tex]
The rotational kinetic energy of the sphere is [tex]473.56\ \text{J}[/tex].
Question 9 of 15
Locate the polyatomic ion in the compound MgSO4-
A. Mgs
B. o
C. Mg
D. SO4
Answer:
d
Explanation:
What happens to the energy of a rubber band when it is stretched?
Which formula is used to find an object's acceleration?
a= Δt – Δν
a= Δv + Δt
a= Δv/ Δt
a= Δt/Δv
Answer:
its the third one
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a large toy rocket to the back of a sled and take the modified sled to a large, flat, snowy field. You ignite the rocket and observe that the sled accelerates from rest in the forward direction at a rate of 11.5 m/s^2 for a time period of 3.30 s. After this time period, the rocket engine abruptly shuts off, and the sled subsequently undergoes a constant backward acceleration due to friction of 4.15 m/s^2.
Required:
a. After the rocket turns off, how much time does it take for the sled to come to a stop?
b. By the time the sled finally comes to a rest, how far has it traveled from its starting point?
Answer:
a) t = 9.2s
b) Δx = 242.2 m
Explanation:
a)
In order to find the time that the sled traveled since the rocket was turned off, we need to find the first the speed that it had at that moment.Applying the definition of accceleration, since we know that the sled started from rest, we can find the value of the final speed (for this part) as follows:[tex]v_{f1} = a_{1} * t_{1} = 11.5m/s2* 3.30 s = 38.0 m/s (1)[/tex]
This speed, is just the initial speed for the second part, so we can find the time traveled from the moment the rocket was turned off until it came to an stop, as follows:[tex]t_{2} = \frac{v_{f1}}{a_{2} } = \frac{38m/s}{4.15m/s} = 9.2 s (2)[/tex]
b)
We need to find find first the displacement when the sled was accelerating.Assuming the acceleration is constant, since it started from rest, we can use the following kinematic equation:[tex]v_{f1} ^{2} = 2* a_{1} * x_{1} (3)[/tex]
Solving for x₁:[tex]x_{1} =\frac{v_{f1}^{2} }{2*a_{1}} =\frac{(38m/s)^{2} }{2*11.5m/s2} =62.8 m (4)[/tex]
In the same way, we can use the same equation, replacing the values of the final speed (which becomes zero), initial speed (which is the same as vf1), and a, which becomes -4.15 m/s2 as it is backwards.[tex]-v_{f1} ^{2} = 2* a_{2} * x_{2} (5)[/tex]
Solving for x₂:[tex]x_{2} =\frac{-v_{f1}^{2} }{2*a_{2}} =\frac{-(38m/s)^{2} }{2*(-4.15m/s)^2} =174.0 m (6)[/tex]
Δx = x₁ + x₂ = 68.2 m + 174.0 m = 242.2 m (7)The only way that heat can travel through outer space is ______
convection
radiation
conduction
none of the above
plssssssssssss answer correctly
a toy of mass 600 is whirled by a child in a horizontal circle using a string of length 2m with a linear speed of 5 m/s determine the angular velocity of the toy?
Explanation:
angular velocity = velocity/radius
= 5/2
= 2.5 rad/s
What is a overly-simplified definition of Einstein's theory of general relativity?
Answer:
the laws of physics are the same for all non-accelerating observers
Explanation:
1. A silicon BJT is connected as shown in Fig 1, where RC = 3.6 k 2. VBE = 0.8 V. (10%)
(a) Predict Ic and specify Rp to establish Vce at 5 V.(5%)
(b) The BJT is said to be in forward-reverse bias. Explain what is meant by this. (5%)
Answer:
The circuit is missing attached below is the required circuit
answer :
a) Ic = 1.944 mA
Rp = 288.66 kΩ
b) The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased
Explanation:
Rc = 3.6 kΩ
VBE = 0.8 v
1) predict Ic and specify Rp to establish Vce at 5 V
we will apply Kirchhoff's voltage law to resolve this
solution attached below
b ) The BJT is said to be in Forward reverse bias because The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased
Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q1 and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.250 m. What is the net force on 92?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
-5.00 x 10-6
-5.00 x 10-6 C
91
92
93
0.500 m
0.250 m
q1 = -5.00 x 10-6 C
q2 = -5.00 x 10-6 C
q3 = -5.00 x 10-6 C
E1 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.5^2) = 180000 N/C to the left
E2 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.25^2) = 720000 N/C to the right
E net = 720000 - 180000 = 540000 N/C to the right
F = qE
F = (-5 x 10^6 C)(540000 N/C) = - 2.7 N
The force on q2 is 2.7 N to the left.
The net electrostatic force on the q2 is 2.7N owards left
The equation for electrostatic force is
[tex]F= k\frac{q_{1}q_{2} }{r^{2} }[/tex]
where k = [tex]9*10^{9} Nm^{2}/C^{2}[/tex] and r is the distance separating charges q1 and q2.
the force has to be calculated on a charge q2 = -5.0 ×[tex]10^{-6}[/tex] C by the charges q1= -5.0 ×[tex]10^{-6}[/tex] C and q3= -5.0 ×[tex]10^{-6}[/tex] C
distance between q1 and q2 is 0.5 m = 5×[tex]10^{-1}[/tex]m
distance between q2 and q3 is 0.25 m = 25×[tex]10^{-2}[/tex]m
force due to charge q1
[tex]F_{1}[/tex] = 9×[tex]10^{9}[/tex]×(-5)×(-5)×[tex]10^{-12}[/tex]/25×[tex]10^{-2}[/tex] N = +0.9N = 0.9N towards right
[tex]F_{2}[/tex] = 9×[tex]10^{9}[/tex]×(-50)×(-4)×[tex]10^{-12}[/tex]/625×[tex]10^{-4}[/tex] N = -3.6N = 3.6N towards left
hence net force F = [tex]F_{1}+F_{2}[/tex]
= 0.9N - 3.6N = -2.7N
F = 2.7 N towards left
Learn more about electrostatic force:
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A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1) Calculate the final velocity of the 2.00-kg ball. (Express your answer to three significant figures.)
Answer:
The final velocity of the 2kg ball is 1.270 m/s
Explanation:
According to Newton's second and third laws of motion
Newton's second law state that "the rate of change of momentum is proportional to the applied force and takes place in the direction of that force".
Newton's third law state that "for every action, there must be an equal and opposite reaction".
The combinations of these two laws resulted in an elastic collision
Given that:
m1 = 2kg
u1 = 2.20m/s
m2 = 4.00kg
u2 = 0m/s
An Elastic collision is when kinetic energy before = kinetic energy after
E.K before = [tex]1/2mv^{2}[/tex]
E.K before = 1/2 * 2 * (2.20)^2
E.K = 1/2 * 2 * 4.84
E.K before = 4.84j
E.K after = 1/2 x (4 + 2)v^2
E.K after = 1/2(6v^2)
E.K after = 3v^2
Since E.K before = E.K after
4.84 = 3v^2
Divide through by 3
4.84/3 = 3v^2/3
1.6133 = v^2
[tex]V = \sqrt{1.6133} \\V = 1.270 m/s[/tex]
how does temperature affect brownian motion
Answer: This (random) thermal motion of the particles due to the temperature is also called Brownian motion. ... The higher the temperature, the faster the diffusion will be, because the stronger the molecule movement and thus the “mixing”.
Explanation:
After supper, your mother runs the warm pan under cold water. The pan cools off quickly. This is an example of -
conduction
convection
radiation
Answer:
conduction (the heat is transferring to the air)
5
Select the correct answer.
What is the current in a parallel circuit which has two resistors (17.2 ohms and
22.4 ohms) and a power source of 6.0 volts?
ОА.
0.30 amps
OB.
9.8 amps
OC.
0.61 amps
D.
1.2 amps
Reset
Next
Answer:
Current in a parallel circuit = 0.61 amps (Approx)
Explanation:
Given:
Voltage V = 6 volt
Two resistors = 17.2 , 22.4 in parallel circuit
Find:
Current in a parallel circuit
Computation:
1/R = 1/r1 + 1 / r2
1/R = 1/17.2 + 1 / 22.4
R = 9.73 ohms (Approx)
Current in a parallel circuit = V / R
Current in a parallel circuit = 6 / 9.73
Current in a parallel circuit = 0.61 amps (Approx)
An elevator suspended by a vertical cable is moving downward at a constant speed. The tension in the cable must be A) greater than the weight of the elevator. B) less than the weight of the elevator. C) equal to the weight of the elevator.
Answer:
(C) because the elevator is not accelerating
Note F = M a = M g (the resultant force on the elevator is due to gravity)
or Fup = Fc the force exerted on the elevator by the cable
and Fdown = Fe the force exerted on the elevator by gravity
F = M a = Fup - Fdown = zero resultant force on elevator
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 86.0 kg skydiver falling in a pike (headfirst) position with a surface area of 0.145 m2. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7. For each answer, enter a number.)
Answer:
terminal velocity is;
v = 117.54 m/s
v = 423.144 km/hr
Explanation:
Given the data in the question;
we know that, the force on a body due to gravity is;
[tex]F_g[/tex] = mg
where m is mass and g is acceleration due to gravity
Force of drag is;
[tex]F_d[/tex] = [tex]\frac{1}{2}[/tex]pCAv²
where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.
Terminal velocity is reach when the force of gravity is equal to the force of drag.
[tex]F_g = F_d[/tex]
mg = [tex]\frac{1}{2}[/tex]pCAv²
we solve for v
v = √( 2mg / pCA )
so we substitute in our values
v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )
v = √( 1685.6 / 0.122015 )
v = √( 13814.6949 )
v = 117.54 m/s
v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr
Therefore terminal velocity is;
v = 117.54 m/s
v = 423.144 km/hr
In 2014, physicists from FOM Foundation at the University of Amsterdam introduced a new hypothesis of how the Pyramids at Giza were built. The group of physicists suggestedthat ancient Egyptians wetted sand in an effort toreduce friction and then pulled the 3000 kg stoneblocks to their final resting place. 15 men couldmove a block at a rate of 0.5m/sby pulling a largerope angled at 30owith respect to the plane anda tension of 7,200 N.
Required:
a. What is net work done on block?
b. What is speed of blck after it moved .25m?
c. What is work done by block if kinetic friction coefficient is 0.3?
d. What is net work including friction?
Answer:
The correct answer is:
(a) 0
(b) 0.5 m/s
(c) 7740 N
(d) 0
Explanation:
The given values are:
mass,
m = 3000 kg
Tension,
T = 7,200 N
Angle,
= 30°
(a)
Even as the block speed becomes unchanged, the kinetic energy (KE) will adjust as well:
⇒ [tex]\Delta K =0[/tex]
By using the theorem of energy, the net work done will be:
⇒ [tex]\Delta K =0[/tex]
(b)
According to the question, After 0.25 m the block is moving with the constant speed
= 0.5 m/s.
(c)
The given kinetic friction coefficient is:
u = 0.3
The friction force will be:
= [tex]u(mg-Tsin30^{\circ})[/tex]
On substituting the values, we get,
= [tex]0.3[(3000\times 9.8)-(7200\times 0.5)][/tex]
= [tex]0.3[29400-3600][/tex]
= [tex]0.3\times 25800[/tex]
= [tex]7,740 \ N[/tex]
(d)
On including the friction,
The net work will be:
⇒ [tex]\Delta K=0[/tex]
Which of the following explains a projectile's parabolic motion? Choose all that apply
The law of inertia
acting on the x axis
The acceleration on
the x axis
The applied force
keeping the
projectile moving
The downward
force of gravity
A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with a radius of 15.0 m. She has a mass of 80.0 kg and her motorcycle has mass 30.0 kg. What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere
Answer:
v = 12.1 m/s
Explanation:
When at the top of the circle, there are two forces acting on the combined mass of the rider and the motorcycle.These are the force of gravity (downward) and the normal force, which is directed from the surface away from it, perpendicular to the surface.In this case, as the motorcycle runs in the interior of the circle, at the top point this force is completely vertical, and is also downward.Since the motorcycle is moving in a vertical circle, there must be a force, keeping the object moving around a circle.This force is the centripetal force, aims towards the center of the circle, and is just the net force aiming in this direction at any point.At the top point, this force is just the sum of the normal force and the weight of the mass of the rider and the motorcycle combined, as follows (we take the direction towards the center as positive):[tex]F_{c} = N + m*g (1)[/tex]
Now, we know that the centripetal force is related with the tangential speed at this point and the radius of the circle as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (2)[/tex]
Since the normal force takes any value as needed to make (1) equal to (2), if the speed diminishes, it will be needed less force to keep the equality valid.In the limit, when the motorcyvle tires barely touch the surface, this normal force becomes zero.In this condition, from (1) and (2), we can find the minimum possible value of the speed that still keeps the motorcycle touching the surface, as follows:[tex]v_{min} =\sqrt{r*g} =\sqrt{15.0m*9.8m/s2} = 12.1 m/s (3)[/tex]You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light when it was hit from behind by car A, of mass 1500 kg. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.30 m long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.65.
(a) What was the speed of car A just before the collision?
(b) If the speed limit was 35 mph, was car A speeding, and if so, by how many miles per hour was it exceeding the speed limit?
Answer:
Explanation:
Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .
work done by friction = 13377 x 7.30 = 97652.1 J
If v be the common velocity of both the cars after collision
kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²
= 1800 v²
so , applying work - energy theory ,
1800 v² = 97652.1
v² = 54.25
v = 7.365 m /s
This is the common velocity of both the cars .
To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 7.365
v₁ = 17.676 m /s
= 63.63 mph .
( b )
yes Car A was crossing speed limit by a difference of
63.63 - 35 = 28.63 mph.
(a) The speed of car A just before the collision is 51.58 mph.
(b) With the given speed limit of 35 miles per hour, car A was crossing the speed limit by 16.58 mph.
What is collision?
The event when two objects strike each other from either direction, then such event is known as a collision. During the collision, the speed of colliding objects may vary according to the direction of the approach.
Given data -
The mass of car A is, mA = 1500 kg.
The mass of car B is, mB = 2100 kg.
The length of the skid mark is, d = 7.30 m.
The coefficient of kinetic friction between tires and road is, [tex]\mu = 0.65[/tex].
(a)
The combined kinetic energy of both cars is,
[tex]KE_{T}=\dfrac{1}{2} (mA+mB)v^{2}\\\\KE_{T}=\dfrac{1}{2} (1500+2100)v^{2}\\\\KE_{T}=1800v^{2}[/tex]
Applying the work-energy principle as,
Work done due to kinetic friction = Combined kinetic energy of cars
[tex]F \times d = KE_{T}\\\\(\mu \times (mA+mB)\times g) \times d = KE_{T}\\\\(0.65 \times (1500+2100)\times 9.8) \times 7.30 = 1800v^{2}\\\\v = 9.64 \;\rm m/s[/tex]
Converting into mph as,
[tex]v = 9.64 \times 2.23\\\\v = 21.49 \;\rm mph[/tex]
To know the speed of car A , we shall apply the law of conservation of momentum. Let the speed of car A before collision be v₁.
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 21.49
v₁ = 51.58 mph
Thus, we can conclude that the speed of car A just before the collision is 51.58 mph.
(b)
With the given speed limit of 35 mph, the obtained speed of car A before the collision is 51.58 mph. Clearly, car A is crossing the speed limit. And the difference is,
= 51.58 - 35 = 28.63 mph.
= 16.58 mph
Thus, we can conclude that car A was crossing the speed limit by 16.58 mph.
Learn more about the average speed here:
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A student picks up two spheres that are the same size. One is made of steel and the other is made of aluminum. The student notices that the steel sphere feels much heavier than the aluminum sphere. He then holds one sphere in each hand at eye level and lets go of them at the same time. They fall to the floor. Which ball, if any, will hit the ground first and for what reason
Answer:
They will fall at the same time. This is because gravity accelerates all objects at the same speed, Earth's gravity being approximately 9.8m/s²
They'll both fall at the same time. This is because gravity accelerates everything at the same rate, with Earth's gravity being approximately 9.8 m/s2.
What is gravity with some instances?The energy that holds the gases inside the sun together. the force that causes a ball to fall after being thrown into the air the force that causes a car to coast downhill even when the gas pedal is not depressed the force that causes a glass to shatterGravity, also known as gravitation, affects all material objects in the universe. Gravity attracts any two objects or particles with nonzero mass toward one another. Gravity affects everything from subatomic particles to galaxy clusters. Gravity is the attraction force between two objects. It's what causes things to fall and keeps us from floating away into space. Gravity is a fundamental natural force.To learn more about gravity, refer to:
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#SPJ2
Question 11 of 15
Which of the following compounds is carbon tetrahydride?
A. CaH4
B. CH4
C. CaH
D. CAH
the correct answer would be "B"
Which of the following electromagnets is the strongest? Why?
Answer:
Bitter Magnet inside a superconducting magnet
Explanation:
Since there are no options available, generally, the electromagnet that is considered the strongest is the Bitter Magnet inside a superconducting magnet.
This electromagnet produces 45 Tesla units which is a result of bitter magnet producing 33.5 Tesla and the superconducting coil produces the additional 11.5 Tesla.
Hence, justifying that the greater the current in the coil the stronger the electromagnet.
What is the displacement for a driver who travels 10 km to get to a point that is 4 km from his starting point?
4 km
10 km
6 km
14 km
Answer:
6km
Explanation:
Four bicyclists travel different distances and times along a straight path. Which cyclist traveled with the greatest average
speed?
A
B
Cyclist 2 travels
87 min 22 s
Cyclist 4 travels
108 min 24 s
D
Cyclist 1 travels
95 m in 27 s
Cyclist 3 travels
106 m in 26 s
Answer:
The cyclist with the greatest average speed is Cyclist 4 with average speed of 4.5 m/s
Explanation:
Given;
Cyclist 1 travels 9 m in 27 s
Cyclist 2 travels 87 m in 22 s
Cyclist 3 travels 106 m in 26 s
Cyclist 4 travels 108 m in 24 s
Determine the average speed of the cyclists as follows;
Average speed of Cyclist 1: v = 9/27 = 0.33 m/s
Average speed of Cyclist 2: v = 87/22 = 3.96 m/s
Average speed of Cyclist 3: v = 106/26 = 4.08 m/s
Average speed of Cyclist 4: v = 108/24 = 4.5 m/s
Therefore, the cyclist with the greatest average speed is Cyclist 4 with average speed of 4.5 m/s
2) How much work is required to pull a sled 15
meters if you use 30N of force?
2 people
Explanation:
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being investigated is immersed in a magnetic field and oriented so that its surface is perpendicular to the field. In a particular measurement, the magnetic field strength was 0.685 T, the strip was 0.107 mm thick, the current along the strip was 2.25 A, and the Hall voltage between the strip's edges was 2.59 mV.Find the density nof mobile electrons in the material. The elementary charge is 1.602×10−19 C.
Answer:
the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
Explanation:
Given the data in the question;
we make use of the following expression;
hall Voltage VH = IB / ned
where I = 2.25 A
B = 0.685 T
d = 0.107 mm = 0.107 × 10⁻³ m
e = 1.602×10⁻¹⁹ C
VH = 2.59 mV = 2.59 × 10⁻³ volt
n is the electron density
so from the form; VH = IB / ned
VHned = IB
n = IB / VHed
so we substitute
n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )
n = 1.54125 / 4.4396226 × 10⁻²⁶
n = 3.4716 × 10²⁵ m⁻³
Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³
A load of mass 120kg is raised
vertically through a height of 2m in
30s by a machine whose efficiency,
is 100% Calculate the power generated
by the machine
Answer: P = 120 kg·9.81 m/s² · 2 m / 30 s = 78 W
Explanation: power P = Work done / time
Work is lifting work = mgh in which g = 9.81 m/s²
Time 30 s
What impulse occurs when a cart that is originally at rest experiences an average force of
N for 2.5 s? *
(10 Points)
25 N
25 Nm
25 Ns
25 kg m/s
Explanation:
What impulse occurs when a cart that is originally at rest experiences an average force of
N for 2.5 s? *
(10 Points)
25 N
25 Nm
25 Ns
25 kg m/s
The slope of a displacement time graph for a uniform motion represent what
Answer:
Velocity.
Explanation:
The slope of a displacement time graph for a uniform motion represent the gradient of the line i.e. the velocity of the object.
The velocity of an object is given by :
v = d/t
Where
d is displacement
t is time
Hence, the slope of the displacement-time graph gives the velocity of the object.
A trapeze artist swings in simple harmonic motion with a period of 3.8 s.
Calculate the length of the cables supporting the trapeze. (g=9.81 m/s2)