Answer:
C. The solid sphere will reach the bottom first.
E. The solid sphere will reach the bottom with the greater kinetic energy.
Explanation:
To answer this,the sphere that has the smaller mass moment of inertia will be the one that will have the largest acceleration down the plane. And we know that higher acceleration means higher speed.
Now, moment of inertia of both spheres are;
I_solid sphere = (2/5)mr² = 0.4mr²
I_hollow sphere = (2/3)mr² = 0.67 mr²
Now, it is clear that the solid sphere has the smaller mass moment of inertia. This means that the solid sphere will have more acceleration and as well higher speed and will thus reach the bottom first.
Also, the higher the speed, the higher the kinetic energy since K.E = ½mv².
Thus, options C & E are correct.
Which of the following is an expression of Newton's second law? A. The acceleration of an object is determined by its mass and the net force acting on it. O O B. Objects at rest tend to stay at rest and objects in motion tend to stay in motion unless acted on by an unbalanced force. C. Objects at rest tend to stay at rest, and objects in motion tend to slow down until they come to rest. D. Every force is paired with an equal and opposite force.
Answer: A. The acceleration of an object is determined by its mass and the net force acting on it.
Explanation:
Newton's second law of motion explains that the acceleration of an object will depend on two vital variables which are the mass of the object and the net force that's acting on it.
It should be noted that the acceleration of the object directly depend on the net force while it depends inversely on the mass. Therefore, when the force that's acting on such object is increased, then the acceleration will increase as well. On the other hand, when there is an increase in mass, there'll be a reduction in the acceleration.
A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck travels for 20.0s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00s.
(a) How long is the truck in motion?
(b) What is the average velocity of the truck for the motion described?
Answer:
Explanation:
a )
Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .
v = u + a t
20 = 0 + 2 t
t = 20 /2 = 10 s .
Total time = 10 s + 20 s + 5 s = 35 s .
b) Average velocity = Total distance travelled / total time
Distance travelled in first 10 s
S₁ = ut + 1/2 a t²
= 0 + .5 x 2 x 10²
= 100 m
Distance travelled in next 20 s
S₂= 20s x 20 m/s = 400 m
Distance travelled in last 5 s .
deceleration in last 5 s
v = u + at
0 = 20 m/s + a x 5
a = - 4 m/s²
v² = u² - 2 a s
0 = (20 m/s)² - 2 x 4 m/s² x s
s = 50 m
S₃ = 50 m
Total distance = S₁ + S₂ + S₃
= 100 m + 400 m + 50 m
= 550 m .
Average velocity = 550 m / 35 s
= 15.71 m /s .
kinetic energy is associated with energy
True or false
Answer:
true ,the kinetic energy is associated with energy
A 0.200 m wire is moved parallel to a 0.500 T
magnetic field at a speed of 1.50 m/s. What emf is
induced across the ends of the wire?
Answer:
The required emf moved across the wire is zero
Explanation:
For a moving charge particle, the magnetic force can be determined by using the formula;
[tex]\varepsilon = Bvlsin \theta[/tex]
since the wire moves in parallel, the angle [tex]\theta[/tex] between magnetic field and velocity = 0°
B = 0.500 T
v = 1.50 m/s
l = 0.200 m
∴
[tex]\varepsilon = (0.500 \ T )(1.50 \ m/s) \times (0.200 \ m)\times sin (0)[/tex]
[tex]\varepsilon = 0.15\times sin (0)[/tex]
[tex]\varepsilon = 0[/tex]
What is the gain in gravitational potential energy of a body of weight 2000N as it rises from a height of 20m to a height of 25 m above the Earth's surface
Answer:
Explanation:
GPE =weight * height
= 2000*(25-20)
= 10000 J
Gain of Gravitation Potential energy = 10,000 J
What is Gravitation Potential Energy ?
It is the energy acquired by an object due to a change in its position when it is present in a gravitational field .
To calculate gravitational potential energy at height h , formula used is = m g h where m = mass of the body
g = acceleration due to gravity
h = height from the ground
Gain of Gravitation Potential energy = mg(h2 - h1 ) h2 = final height
h1 = initial height
= 2000 (25 -20) (since , mg =weight)
Gain of Gravitation Potential energy = 10,000 J
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A wave with a frequency of 60 hertz would generate 60 wave crests every
Answer:
A wave with a frequency of 60 hertz will have 60 successive wave crests occurring every second
Explanation:
A crest is the highest point the medium rises to in a wave. That means that the crest is a point on the wave where the displacement is at a maximum.
Frequency is the number of oscillations of a wave in one second. Hence, frequency of a wave is the number of successive crests occurring in a second.
Therefore a wave with a frequency of 60 hertz will have 60 successive wave crests occurring every second.
potential Energy is associated with position
True or false
Answer:
yes.
Explanation:
potential Energy is associated with position
Answer:
true
Explanation:
the energy possessed by a body by virtue of its position or configuration is called potential energy.Everything at a height from the earth surface posseses potential energy.
What is the energy stored in food?
Hey there!
The energy that is stored in food is called CHEMICAL ENERGY
So, when humans/people eat foods their bodies transfigure the energy that is stored which is known as the calories.
Whereas the chains (bonds) in the middle of the atoms in the food humans eat that break up the check mal reactions that has been taken into place and newer compounds will be created. Each food that you eat stores unlikeable amounts of energy
Answer: Chemical energy
Good luck on your assignment and enjoy your day!
~Amphitrite1040:)
Preocupada com o aumento da tarifa na conta de luz, uma pessoa resolve economizar diminuindo o tempo de banho de 20 para 15 minutos. Seu chuveiro possui as seguintes especificações: 4200 W e 220V. Sabendo que o kWh custa R$0,30, a economia feita em 10 dias foi de aproximadamente
Does every magnet have a north and south pole?
yes
no
a given mass of an ideal gas law occupies a volume V of a temperature T under a pleasure p, if the pressure is increased to 2 p and temperature 1/2 T what is the percentage change in the volume of the gas
Answer:
75%
Explanation:
We'll begin by calculating the final volume of the gas. This can be obtained as follow:
Initial volume (V₁) = V
Initial temperature (T₁) = T
Initial pressure (P₁) = P
Final pressure (P₂) = 2P
Final temperature (T₂) = ½T
Final volume (V₂) =?
P₁V₁/T₁ = P₂V₂/T₂
PV/T = 2P × V₂ / ½T
Cross multiply
T × 2P × V₂ = PV × ½T
T × 2P × V₂ = PV × ½T
Divide both side by T × 2P
V₂ = (PV × ½T) / T × 2P
V₂ = ¼V
Next, we shall determine the absolute change in the volume of the gas. This can be obtained as follow:
Initial volume (V₁) = V
Final volume (V₂) = ¼V
Absolute change in volume |ΔV| =?
|ΔV| = |V₂ – V₁|
|ΔV| = |¼V – V|
|ΔV| = |0.25V – V|
|ΔV| = 0.75V
Finally, we shall determine percentage change in the volume of the gas. This can be obtained as follow:
Initial volume (V₁) = V
Absolute change in volume |ΔV| = 0.75V
Percentage change =?
Percentage change = |ΔV| / V × 100
Percentage change = 0.75V / V × 100
Percentage change = 0.75 × 100
Percentage change = 75%
Thus, the percentage change in the volume of the gas is 75%
The strength of a magnetic field around a wire carrying a current of 20 A is 0.004 T. What is the strength of the
magnetic field if the current is changed to 40 A?
A 0.002 T
B 0.004 T
C 0.006 T
D 0.008 T
Answer:
D) 0.008
Explanation:
I just did it on Edg22 and I got it correct :D (Picture above)
Option D) 0.008T is the correct answer.
Hence, if the current is increased to the given value, the strength of the magnetic field around the wire carrying the current increases to 0.008T
Given the data in the question;
First scenario
Strength of magnetic field; [tex]B = 0.004T[/tex]Current; [tex]I = 20A[/tex]Second scenario
Strength of magnetic field; [tex]B' =\ ?[/tex]Current; [tex]I' = 40A[/tex]Magnetic FieldMagnetic field is a vector field or region around a magnet or electric charge upon which magnetic force is exerted.
To determine the strength of the magnetic field if the current is changed, we equate the two scenario.
[tex]\frac{B}{I} = \frac{B'}{I'}[/tex]
We substitute our given values into the expression
[tex]\frac{0.004T}{20A} = \frac{B'}{40A}\\ \\ B' * 20A = 0.004T * 40A\\\\B' = \frac{0.004T * 40A}{20A} \\\\B' = \frac{0.004T * 40}{20}\\ \\B' = \frac{0.16T}{20}\\ \\B' = 0.008T[/tex]
Option D) 0.008T is the correct answer.
Hence, if the current is increased to the given value, the strength of the magnetic field around the wire carrying the current increases to 0.008T
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The total power input to a pumped storage power station is 600 MW
The useful power output is 540 MW calculate the efficiency of this pumped storage power station.
Calculate how much power is wasted by the pumped storage power station.
Answer:
60MW wasted
Explanation:
600-540
=60MW
An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N, the spring is stretched by 15.5 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 10.4 cm from that position. (in J)
A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant k is
43.8 N = k (0.155 m) ==> k = (43.8 N) / (0.155 m) ≈ 283 N/m
The total work done on the spring to stretch it to 15.5 cm from equilibrium is
1/2 (283 N/m) (0.155 m)² ≈ 3.39 J
The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be
1/2 (283 N/m) (0.259 m)² ≈ 9.48 J
Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.
A ball of mass=2kg is dropped from h=100m. What is the final velocity when it reaches the ground? *
A)45m/s
B)200m/s
C)32m/s
D)2000m/s
show your work please
Answer:
45m/s
Explanation:
(here I'm taking gravity due to acceleration as 10 m/s^2 and not 9.8 m/s^2 to match the options)
We know that when an object is dropped from a certain height it has initial velocity 0 m/s
so;
u= 0m/s
s= 100m
a= 10m/s^2
Using 4th law of motion;
v^2 - u^2 = 2× a× s
v^2 - 0^2 = 2 × 10 × 100
v^2 = 2000
v = √2000
v = 44.72 m/ s
Final velocity = 45 m/ s (rounding off to 45)
Please feel free to tell me if you have any confusion.
g to generate electricity, solar panels typically absorb visible light. How many photons of light with a frequency of 5045x10 14 hz does a solar panel absorb to create 360 kj
Answer:
the number of photons absorbed by the solar panel is 1.08 x 10²¹
Explanation:
Given;
frequency of each photon absorbed, f = 5045 x 10¹⁴ Hz
energy to be created by the solar panel, E = 360 kJ = 360,000 J
The energy of each photon absorbed is calculated as;
[tex]E_{photon} = hf\\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E_{photon} = (6.626 \times 10^{-34} )(5045 \times 10^{14})\\\\E_{photon} = 3.343 \times 10^{-16} \ J[/tex]
let the number of photons absorbed = n
[tex]n(E_{photon}) = 360,000 \ J\\\\n = \frac{360,000 \ J}{3.343 \times 10^{-16} \ J} \\\\n = 1.08 \times 10^{21} \ photons[/tex]
Therefore, the number of photons absorbed by the solar panel is 1.08 x 10²¹
At what frequency would an inductor of inductance 0.8H have a reactance of 12000^?
Explanation:
The inductive reactance [tex]X_L[/tex] is given by
[tex]X_L = \omega L = 2 \pi fL[/tex]
Solving for f, we get
[tex]f = \dfrac{X_L}{2 \pi L} = \dfrac{12000\:\text{ ohms}}{2\pi (0.8\:H)}[/tex]
[tex]\:\:\:\:\:\:\:= 2387.3\:\text{Hz}[/tex]
A tennis player swings at a ball at a constant speed, taking 0.50 s to rotate her arms and racquet from horizontal to vertical. What acceleration is felt by a small bug at the tip of her racquet if it is 1.3 m from her shoulder
Answer:
the acceleration of the small bug is 12.83 m/s²
Explanation:
Given;
time of motion, t = 0.5 s
radius of the circular path created by his arm, r = 1.3 m
if he rotates his arm from horizontal to vertical, the angular displacement = 90⁰
The centripetal acceleration of the ball is calculated as;
[tex]a_c = \omega^2 r\\\\a_c = (\frac{\theta}{t} )^2 r\\\\[/tex]
[tex]a_c = (\frac{90}{360} \times\frac{ 2\pi }{t} )^2r\\\\a_c = (\frac{\pi}{2t} )^2 r\\\\a_c = \frac{\pi^2r}{4t^2} = \frac{\pi^2 \times1.3 }{4\times 0.5^2} = 12.83 \ m/s^2[/tex]
Therefore, the acceleration of the small bug is 12.83 m/s²
During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What minimum speed did he need at launch if he was traveling at 6.8 m/s at the top of the arc
Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Explanation:
Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.
Now, according to the law of conservation of energy the formula used is as follows.
[tex]mgh = \frac{1}{2} mv^{2}_{y}\\v_{y} = \sqrt{2gh}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.2}\\= 4.85 m/s[/tex]
As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.
[tex]v = \sqrt{v^{2}_{x} + v^{2}_{y}}\\= \sqrt{(6.8)^{2} + (4.85 m/s)^{2}}\\= 11.65 m/s[/tex]
Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.
Genes are arranged on a twisted strands called.
Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curve O B. A car slowing down as it goes around a curve 23 C. A car maintaining constant speed as it goes around a curve D. A car traveling along a straight road
Answer:
Pluto revolves around sun with constant speed in a circular orbit, so it is an example of uniform circular motion.
What is the vibrating source that creates sound waves?
Answer:
Sound is produced when an object vibrates, creating a pressure wave. This pressure wave causes particles in the surrounding medium (air, water, or solid) to have vibrational motion. As the particles vibrate, they move nearby particles, transmitting the sound further through the medium.
A gazelle is running at a constant speed of 19.3 m/s toward a motionless hidden cheetah. At the instant the gazelle passes the cheetah, the cheetah accelerates at a rate of 7.1 m/s/s in pursuit of the gazelle. The gazelle maintains its constant speed. By the time the cheetah reaches a speed of 19.3 m/s to match the gazelle, how far apart are the two animals in units of m
Answer:
the animals are 26.2 meters apart.
Explanation:
Let's define t = 0s as the moment when the cheetah starts accelerating.
The gazelle moves with constant velocity, thus, it is not accelerating, then the acceleration of the gazelle is:
a₁(t) = 0m/s^2
where I will use the subscript "1" to refer to the gazelle and "2" to refer to the cheetah.
for the velocity of the gazelle we just integrate over time to get:
v₁(t) = V0
where V0 is the initial speed of the gazelle, which we know is 19.3 m/s
v₁(t) = 19.3 m/s
To get the position of the gazelle we integrate again:
p₁(t) = ( 19.3 m/s)*t + P0
where P0 is the position of the gazelle at t = 0s, let's define P0 = 0m
p₁(t) = ( 19.3 m/s)*t
The equations that describe the motion of the gazelle are:
a₁(t) = 0m/s^2
v₁(t) = 19.3 m/s
p₁(t) = ( 19.3 m/s)*t
Now let's do the same for the cheetah.
We know that its acceleration is 7.1 m/s^2
then:
a₂(t) = 7.1 m/s^2
for the velocity of the cheetah we integrate:
v₂(t) = (7.1 m/s^2)*t + V0
where v0 is the initial velocity of the cheetah, which we know its zero.
v₂(t) = (7.1 m/s^2)*t
Finally, for the position equation we integrate again, and remember that we have defined the initial position for the gazelle as zero, then the same happens for the cheetah.
p₂(t) = (1/2)*(7.1 m/s^2)*t^2
The equations for the cheetah are:
a₂(t) = 7.1 m/s^2
v₂(t) = (7.1 m/s^2)*t
p₂(t) = (1/2)*(7.1 m/s^2)*t^2
Now, we want to find the distance between both animals when the speed of the cheetah is 19.3 m/s, then first we need to solve:
v₂(t) = (7.1 m/s^2)*t = 19.3 m/s
t = (19.3 m/s)/(7.1 m/s^2) = 2.72s
Now, to find the distance between the two animals, we just compute the difference between the position equations for t = 2.72s
Distance = p₁(2.72s) - p₂(2.72s)
= ( 19.3 m/s)*2.72s - (1/2)*(7.1 m/s^2)*(2.72s)^2
= 26.2 m
So the animals are 26.2 meters apart.
An object of mass m 1 moving with speed v collides with another object of mass m 2 at rest and stick to it. Find the impulse to the second object.
Answer:
(m1+m2)vo=m1v+m2×0
⟹vo=m1+m2m1v
The impulse imparted to second object is equal to change is momentum is -
J=m2vo=m1+m2m2m1v
write the full form of MBBS pliz now
Answer:
The full form of MBBS is Bachelor of Medicine, Bachelor of Surgery
pls mark me brainliest :))
The full form of MBBS in India is 'Bachelor of Medicine, Bachelor of Surgery'. However, MBBS is an abbreviation of Medicinae Baccalaureus Baccalaureus Chirurgiae, which is the term used for this course in Latin.
Calculate the pressure exerted on the floor when an
elephant who weighs 2400N stands on one foot which has
an area of 4m²?
P = F / A
P = 2400 / 4
P = 24 × 100 / 4
P = 6 × 100
P = 600 N/m^2
An electric generator has an 18-cmcm-diameter, 120-turn coil that rotates at 60 HzHz in a uniform magnetic field that is perpendicular to the rotation axis. Part A What magnetic field strength is needed to generate a peak voltage of 330 VV
Explanation:
OK ok nosepo [tex]2825.55[/tex]The magnetic field strength is needed to generate a peak voltage is 0.287 T.
The given parameters;
diameter of the generator, d = 18 cm = 0.18number of turn, N = 120 turnfrequency of the coil, f = 60 Hzmaximum voltage in the coil emf = 330 V;The maximum voltage in the coil is calculated as follows;
[tex]E_{max} = NAB\omega \\\\[/tex]
where;
N is the number of turnsA is the area of the coilB is the magnetic field strengthω is angular speedThe magnetic field strength is needed to generate a peak voltage is calculated as;
[tex]B = \frac{E_{max} }{NA \omega } \\\\B = \frac{E_{max} }{N (\frac{\pi d^2}{4} ) \times 2\pi f}\\\\ B = \frac{330}{120 \times (\frac{\pi \times 0.18^2}{4} ) \times 2\pi \times 60} \\\\B = 0.287 \ T[/tex]
Thus, the magnetic field strength is needed to generate a peak voltage is 0.287 T.
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An astronaut weighing 248 lbs on Earth is on a mission to the Moon and Mars. (a) What would he weigh in newtons when he is on the Moon
Answer:
[tex]W_m=183.495N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=248Ibs[/tex]
Mass of Weight [tex]m= 248*0.453[/tex]
[tex]m=112.344kg[/tex]
Generally the acceleration due to gravity on the Moon is one-sixth that on Earth.
Therefore
The equation for Weight on Moon is given as
[tex]W_m=m*g/6[/tex]
[tex]W_m=122.344*\frac{9.8}{6}[/tex]
[tex]W_m=183.495N[/tex]
Answer:
F ’= 40.9 lb
Explanation:
The weight of a body is the attraction of the Earth on the body
F = [tex]G \frac{m M_e}{R_e^2}[/tex]
F = mg
m = F / g
m = 248/32
m = 7.75 slug
The weight of the body on the moon is the attraction of the body for the satellite
F ’= [tex]G \frac{mM}{R^2}[/tex]
from the tables the mass and radius of the moon are M = 7.34 10²² kg and R = 1.74 10⁶ m
let's reduce the mass to the SI system
m = 7.75 slug (14.59 kg / 1 slug) = 113 kg
F ’= 6.67 10⁻¹¹ 113 7.34 10²² / (1.74 10⁶) ²
F ’= 1.82 10² N
F ’= 1.82 10² N (0.2248 lb / 1 N)
F ’= 40.9 lb
Ftension = 120 N
10 kg
Fg
What is the weight (not mass) of the box?
O 10 kg
0 19 N
98 N
O 98 kg
PLZ HELP
Answer:
98N
Explanation:
El peso se mide en kg y la fuerza no afecta
Consult Interactive Solution 7.10 for a review of problem-solving skills that are involved in this problem. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of 16.0 m/s, while the exiting water stream has a velocity of -16.0 m/s. The mass of water per second that strikes the blade is 36.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.
Answer:
[tex]F=1152N[/tex]
Explanation:
From the question we are told that:
Mass/one sec [tex]m=36.0kg/s[/tex]
Incident water stream [tex]v=16.0m/s[/tex]
Exiting water stream [tex]v'=-16.0m/s[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{mv-mu}{t}[/tex]
[tex]F=\frac{36(-16-16)}{1}[/tex]
[tex]F=1152N[/tex]