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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An electric motor consumes 8.40 kJ of electrical energy in 1.00 min. Part A If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor output, how much torque will this engine develop if you run it at 2900 rpmrpm
Answer:
The torque is 0.31 Nm.
Explanation:
Electrical energy, E = 8400 J
time, t = 1 min
Angular speed, w = 2900 rpm = 303.53 rad/s
efficiency = 2/3 of input power
The toque is given by
[tex]P =\tau w\\\\\frac{2}{3}\times \frac{E}{t}=\tau w\\\\\frac{2}{3}\times \frac{8400}{60}=\tau \times 303.53\\\\\tau =0.31 Nm[/tex]
DL: Activity 2.3 Resistance Complete the questions based on the Resistance presentation. a. All resistors ___________ or _____________ the flow of electrons. b. As resistance __________________ current _______________________
DL: Activity 2.3 Resistance
Complete the questions based on the Resistance presentation.
a. All resistors ___________ or _____________ the flow of electrons.
b. As resistance __________________ current _______________________
Answer:a) limits or opposes
b) increases, decreases
Explanation:Resistors are electrical devices used to resist, limit, oppose or hinder the flow of electrons in a circuit. This resistance causes a reduction in current and an increase in voltage in the circuit. In order words, as resistance increases, the current decreases and voltage increases.
This was further stated by Ohm's law that states that as long as the resistance in a wire/conductor remains constant, the voltage across it is directly proportional to the current flowing through a conductor. i.e
V = IR
Where;
R = constant called resistance
I = current flowing through the wire
V = voltage across the wire
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.
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.
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.
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
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.
A volleyball is served at a speed of 8 / at an angle 35° above the horizontal. What is the speed of the ball when received by the opponent at the same height?
Which element is the biggest contributor to Climate Change?
Answer:
carbon dioxide (CO2)
Explanation:
the burning or combustion of these fossil fuels creates gases that are released into the atmosphere. Of these gases, carbon dioxide (CO2) is the most common and is the gas most responsible for exacerbating the green- house effect that is changing global climate patterns.
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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A toy cannon uses a spring to project a 5.30-g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 8.00 N/m. When the cannon is fired, the spring decompresses and returns to its equilibrium length, causing the ball to speed up, before the ball enters the horizontal barrel of the cannon. The horizontal barrel is 15.0 cm long and it exerts a constant friction force of 0.032 N on the ball. A. With what speed does the projectile leave the barrel of the cannon
Answer:
1.40 m/s
Explanation:
The potential energy of a compressed spring can be expressed as:
[tex]E_{ps}=\dfrac{1}{2}kx^2[/tex]
From above;
k = spring constant
x = distance of the spring (compressed)
From the barrel, the kinetic energy (i.e. the final K.E) of the ball is calculated using the relation:
[tex]E_{kf}= \dfrac{1}{2}mv^2[/tex]
where;
m = the ball mass
v = ball's speed
Equating both equations above, we have:
[tex]E_{ps}- F_fd=E_{kf[/tex]
This can be re-written as:
[tex]\dfrac{1}{2}kx^2 - F_fd=\dfrac{1}{2}mv^2}[/tex]
[tex]v^2 = (\dfrac{k}{m})x^2-\dfrac{2F_fd}{m}[/tex]
[tex]v =\sqrt{ (\dfrac{k}{m})x^2-\dfrac{2F_fd}{m}}[/tex]
replacing the values from the given information:
[tex]v =\sqrt{ (\dfrac{8.00\ N/m}{5.30\times10^{-3} \ kg})(5.00 \ cm \times \dfrac{10^{-2} \ m}{1 \ cm})^2-(\dfrac{2(0.032 \ N)(0.150 \ m)}{5.30\times \dfrac{10^{-3} \ kg}{1 \ g}})}[/tex]
[tex]v = \sqrt{1.962264151}[/tex]
v ≅ 1.40 m/s
The speed at which the projectile leaves the barrel of the cannon will be given as [tex]v=1.40\ \frac{m}{s}[/tex]
What is speed?Speed is defined as the movement of any object with respect to time. It is the ratio of distance and time.
Now it is given in the question:
Mass of ball m = 5.30 g
The deflection of spring = 5 cm
The force constant of spring [tex]k= 8 \ \frac{N}{m^2}[/tex]
The length of the barrel = is 15 cm
The frictional force of the barrel = 0.032 N
Now from the conservation of energy, we can write as
[tex]E_{spring}-E_{friction}=E_{ball}[/tex]
[tex]\dfrac{1}{2} kx^2-F_fd=\dfrac{1}{2} mv^2[/tex]
[tex]v=\sqrt{\dfrac{k}{m}(x^2) -\dfrac{2F_fd}{m} }[/tex]
Now putting the values in the above formula:
[tex]v=\sqrt{\dfrac{8}{5.30\times 10^{-3}}(15\times10^{-2}) -\dfrac{2\times(0.0032)\times (0.015)}{5.30\times 10^{-3}} }[/tex]
[tex]v=1.40\ \frac{m}{s}[/tex]
Thus the speed at which the projectile leaves the barrel of the cannon will be given as [tex]v=1.40\ \frac{m}{s}[/tex]
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A magnetic field of magnitude 0.550 T is directed parallel to the plane of a circular loop of radius 43.0 cm. A current of 5.80 mA is maintained in the loop. What is the magnetic moment of the loop? (Enter the magnitude.)
Answer:
[tex]\mu = 3.36\times 10^{-3}\ A-m^2[/tex]
Explanation:
Given that,
The magnitude of magnetic field, B = 0.55 T
The radus of the loop, r = 43 cm = 0.43 m
The current in the loop, I = 5.8 mA = 0.0058 A
We need to find the magnetic moment of the loop. It is given by the relation as follows :
[tex]\mu = AI\\\\\mu=\pi r^2\times I[/tex]
Put all the values,
[tex]\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2[/tex]
So, the magnetic moment of the loop is equal to[tex]3.36\times 10^{-3}\ A-m^2[/tex].
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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A satellite has a mass of 6463 kg and is in a circular orbit 4.82 × 105 m above the surface of a planet. The period of the orbit is 2.0 hours. The radius of the planet is 4.29 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?
Answer:
The weight of the planet is 29083.5 N .
Explanation:
mass of satellite, m = 6463 kg
height of orbit, h = 4.82 x 10^5 m
period, T = 2 h
radius of planet, R = 4.29 x 10^6 m
Let the acceleration due to gravity at the planet is g.
[tex]T = 2\pi\sqrt\frac{(R+h)^3}{gR^2}\\\\2\times 3600 = 2\times3.14\sqrt\frac{(4.29+0.482)^3\times10^{18}}{g\times 4.29\times 4.29\times 10^{12} }\\\\24.2 g =108.67\\\\g = 4.5 m/s^2[/tex]
The weight of the satellite at the surface of the planet is
W = m g = 6463 x 4.5 = 29083.5 N
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
If v = 5.00 meters/second and makes an angle of 60° with the negative direction of the y–axis, calculate all possible values of vx.
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
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
Q2 A source of frequency 500 Hz emits waves of
wavelength 0.2m. How long does it take the waves to
travel 400m?
Answer:
4 secs
Explanation:
The first step is to calculate the velocity
V= frequency × wavelength
= 500× 0.2
= 100
Therefore the time can be calculated as follows
= distance/velocity
= 400/100
= 4 secs
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]
A truck has a mass of 1.5 x 104 kg. If the truck can reach a maximum acceleration of 1.5 m/s2, what is the net force the truck exerts?
2.25 x 105 N
2.25 x 104 N
2.3 x 104 N
2.3 x 105 N
Answer:
2.25 × 10⁴ NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 1.5 × 10⁴ × 1.5
We have the final answer as
2.25 × 10⁴ NHope this helps you
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
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.
When converting chemical potential energy to kinetic energy some of the energy is lost as heat.
Answer:
When converting chemical potential energy to kinetic energy some of the energy is lost as heat.
What does this circle graph tell you about water on Earth? (2 points)
a pie graph with a big blue section covering seventy one percent and small gray section covering twenty nine percent with a key indicating that blue is water and gray is land
Fresh water covers 71 percent of Earth's surface.
Oceans covers 71 percent of Earth's surface.
Salt water covers 71 percent of Earth's surface.
Water covers 71 percent of Earth's surface.
Answer:
ocean covers 71 percent of the earth
Answer:
the ocean covers 71 percent of Earth's surface.
Explanation:
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]
Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75 cm. Use a value of k = 500 N/m for the spring constant, 10 g for the mass of the ball, and 75 g for the effective mass of the ball holder. Show your work.
Answer:
V = 8.34m/s
Explanation:
Given that
1/2ke^2 = 1/2mv^2 ......1
Where e = 3.75cm = (3.75/100)m
e = 0.0375m
K = 500 N/m
m = 10g = 10/1000
= 0.01kg
Substitute the values into equation 1
0.5×500×(0.0375)^2 = 0.5×0.01×v^2
250×0.001395 = 0.005v^2
0.348 = 0.005v^2
v^2 = 0.348/0.005
v^2 = 69.6
V = √69.6
V = 8.34m/s
The ball launches at the speed of V = 8.34m/s
A steel playground slide is 5.25 m long and is raised 2.75 m on one end. A 45.0 kg child slides down from the top starting at rest. The final speed of the child at the bottom is 6.81 m/s. Find the average force of friction between the child and the slide.
Answer:
[tex]F=32.24N[/tex]
Explanation:
From the question we are told that:
Height [tex]h= 2.75 m[/tex]
Length[tex]l = 5.25 m[/tex]
Mass [tex]m=45kg[/tex]
Final speed [tex]v_f=6.81[/tex]
Generally the equation for Potential Energy P.E is mathematically given by
[tex]P.E=mgh[/tex]
Therefore
Initial potential energy
[tex]P.E_1=45*9.8*2.75 \\\\P.E_1= 1212.75 J[/tex]
Generally the equation for Kinetic Energy K.E is mathematically given by
[tex]K.E=0.5mv^2[/tex]
Therefore
Final kinetic energy
[tex]K.E_2= 1/2*45*6.81*6.81 \\\\K.E_2= 1043.46J[/tex]
Generally the equation for Work_done is mathematically given by
[tex]W=P.E_1-K.E_2\\\\W=169.3[/tex]
Therefore
[tex]F=\frac{W}{d}\\\\F=\frac{169.3}{5.25}[/tex]
[tex]F=32.24N[/tex]