Answer:
E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules
Answer:
The answer is D. 2.25 × 1017 J
Explanation:
got it right on edge 2021
what does it mean to do science
Answer:
Doing science could be defined as carrying out scientific processes, like the scientific method, to add to science's body of knowledge.
What are stepdown transformers used for
Answer:
Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.
Explanation:
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Help me with my physics, please
A person jumps out of an airplane above the surface of the Earth, and falls a distance h before opening their parachute. Once the prachute is open the person coasts to the ground a distance d at constant velocity.
a. The work done on the person by the Earth is:
b. The change in gravitational potential energy of the person + Earch system is:
Answer:
a) W_total = mg (2h + d) , b) E_total = - mg (h + d)
Explanation:
a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed
Let's look for work for the part that is in free fall
y = y₀ + v₀ t - ½ g t²
when he jumps out of a plane his vertical speed is zero
y =y₀ - ½ g t²
dy = 0 - ½ g 2t dt
the work in this first part is
W₁ = ∫ F dy
W₁ = mg ∫ g t dt
W₁ = m g² t² / 2
the time it takes to travel the distance y₀-y = h is
y₀-y = ½ g t²
t =[tex]\sqrt{2h/g}[/tex]
we substitute
W₁ = m g² 2h / g
W₁ = m g 2h
now we look for the work for the part with constant speed
since the velocity is constant let's use the uniform motion ratio
W₂ = F d
W₂ = mg d
the total work is
W_total = W₁ + W₂
W_total = 2mgh + m gd
W_total = mg (2h + d)
b) The change in gravitational potential energy
U = mg Δy
in the part with accelerated movement
U₁ = mg h
in the part with uniform movement
U₂ = mg d
the total potential energy is
E_total = U₁ + U₂
E_total = - mg (h + d)
A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 3.30 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged
Answer:
The velocities of the skaters are [tex]v_{1} = 3.280\,\frac{m}{s}[/tex] and [tex]v_{2} = 0.024\,\frac{m}{s}[/tex], respectively.
Explanation:
Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:
First skater
[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex] (1)
Second skater
[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex] (2)
Where:
[tex]m_{1}[/tex] - Mass of the first skater, in kilograms.
[tex]m_{2}[/tex] - Mass of the second skater, in kilograms.
[tex]v_{1,o}[/tex] - Initial velocity of the first skater, in meters per second.
[tex]v_{1}[/tex] - Final velocity of the first skater, in meters per second.
[tex]v_{b}[/tex] - Launch velocity of the meter, in meters per second.
[tex]v_{2}[/tex] - Final velocity of the second skater, in meters per second.
If we know that [tex]m_{1} = 70\,kg[/tex], [tex]m_{b} = 0.043\,kg[/tex], [tex]v_{b} = 32\,\frac{m}{s}[/tex], [tex]m_{2} = 58.5\,kg[/tex] and [tex]v_{1,o} = 3.30\,\frac{m}{s}[/tex], then the velocities of the two people after the snowball is exchanged is:
By (1):
[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex]
[tex]m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}[/tex]
[tex]v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}[/tex]
[tex]v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)[/tex]
[tex]v_{1} = 3.280\,\frac{m}{s}[/tex]
By (2):
[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex]
[tex]v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}[/tex]
[tex]v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}[/tex]
[tex]v_{2} = 0.024\,\frac{m}{s}[/tex]
A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south. If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?
Answer:
E_total = - 50 N / A
Explanation:
The electric field is a vector magnitude whereby
E_total = Eₐ + E_b
where the bold letters indicate vectors, in this case the charges of the two objects A and B are the same and they are on the same line
E_total = - E_a - E_b
the electric field for a point charge is
E_a = [tex]k \ \frac{q_a}{r_a^2 }[/tex]
qₐ= Eₐ rₐ² / k
indicates that Eₐ = 40.0 N / C
qₐ = 40.0 0.250²/9 10⁹
qₐ = 2.777 10⁻¹⁰ C
indicates that the charge of the two points is the same
qₐ = q_b
E_total = - k qₐ / rₐ² - k qₐ / (2 rₐ)²
E_total = [tex]-k \ \frac{q_a}{r_a^2} \ ( 1 + \frac{1}{4} )[/tex]
we calculate
E_total = - 40.0 (5/4)
E_total = - 50 N / A
Name the electrolyte in the chemical method of generating electricity
A source emits sound at a fixed constant frequency f. If you run towards the source, the frequency you hear is
Answer:
increased because as you run into each sound wave the time between each sound decreases meaning the period of each wave decreases to your years and since f=1/T and T is decreasing by greater than 0, f must increase.
Explanation:
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver immediately puts the car in neutral so that it will roll.
What will be the car’s speed as it coasts into the gas station on the other side of the valley?
Answer:
v_f = 17.4 m / s
Explanation:
For this exercise we can use conservation of energy
starting point. On the hill when running out of gas
Em₀ = K + U = ½ m v₀² + m g y₁
final point. Arriving at the gas station
Em_f = K + U = ½ m v_f ² + m g y₂
energy is conserved
Em₀ = Em_f
½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂
v_f ² = v₀² + 2g (y₁ -y₂)
we calculate
v_f ² = 20² + 2 9.8 (10 -15)
v_f = √302
v_f = 17.4 m / s
7. The gravitational potential energy of a body depends on its A speed and position B. mass and volume. C. weight and position D.speed and mass
Answer:
Option "D" is the correct answer to the following question.
Explanation:
The gravitational potential energy of an item is determined by its mass, elevation, and gravitational acceleration. As a result, angular momentum and energy are preserved. The gravitational potential energy, on the other hand, varies with distance. When a consequence, kinetic energy varies during each orbit, resulting in a faster speed as a planet approaches the Sun.
Answer:
SPEED AND MASS
Explanation:
TOOK THE TEST
A magnetohydrodynamic (MHD) drive works by applying a magnetic field to a fluid which is carrying an electric current.
a. True
b. False
Answer:
True
Explanation:
A magnetohydrodynamic drive or MHD accelerator is a method which is used for propelling the vehicles using only by applying the electric and magnetic fields. It has no moving parts. It accelerates an electrically conductive propellant (liquid or gas) with magnetohydrodynamics.
Its working principle is same as an electric motor except that in an MHD drive, the moving rotor is replaced by the fluid acting directly as the propellant.
An MHD accelerator is reversible.
So, the statement is true.
the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is
Answer:
The work done on the block by the spring as it accelerates the block is 4kx².
Explanation:
Let initial distance is x.
It was compressed three times farther and then the block is released, new distance is 3x.
The work done in compressing the spring is given by :
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]
So, the work done on the block by the spring as it accelerates the block is 4kx².
a particle undergoes three consecutive displacement d1=(15i+30j+12k)cm,d2=(23i-14j-5.0k)cm and d3=(-13i+15j)cm find the component of the resultant displacement and magnitude?
Answer:
Explanation:
The density of pure water is 1 gram per 1 milliliter or one cubic cm. By knowing the density of water we can use it in dilution equations or to calculate the specific gravity of other solutions.
It can also help us determine what other substances are made of using the water displacement experiment. This is done by observing how much water is displaced when an object is submerged in the water. As long as you know the density of the water, the mass of the object being submerged and the volume of increase you can calculate the density of the object.
This was done by the great Archimedes in discovering what composed the kings crown.
Two objects attract each other with a gravitational force of magnitude 1.00 3 1028 N when separated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each
Answer:
The mass of each object is 2kg and 3 kg.
Explanation:
Given that,
Gravitational force,[tex]F=1\times 10^{-8}\ N[/tex]
The distance between masses, d = 20 cm = 0.2 m
The total mass of the two objects, M + m = 5 kg
M = 5-m
The formula for the gravitational force is :
[tex]F=G\dfrac{Mm}{d^2}\\\\1\times 10^{-8}=6.67\times 10^{-11}\times \dfrac{(5-m)m}{(0.2)^2}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}=\frac{(5-x)x}{(0.2)^{2}}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}\cdot(0.2)^{2}\\\\5.99=(5-x)x\\\\x=2\ kg\ and\ 3 \ kg[/tex]
So, the mass of each object is 2kg and 3 kg.
You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet
What is true when an object floats in water? A. When an object floats, it exceeds the volume of water available. B. When an object floats, it displaces a volume of water equal to its own volume. C. When an object floats, it does not displace its entire volume.
Answer:
C. When an object floats, it does not displace its entire volume.
Explanation:
Buoyancy can be defined as an upward force which is created by the water displaced by an object.
According to Archimede's principle, it is directly proportional to the amount (weight) of water that is being displaced by an object.
Basically, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up. The buoyancy of an object is given by the formula;
[tex] Fb = pgV [/tex]
[tex] But, \; V = Ah [/tex]
[tex] Hence, \; Fb = pgAh [/tex]
Where;
Fb = buoyant force of a liquid acting on an object.
g = acceleration due to gravity.
p = density of the liquid.
v = volume of the liquid displaced.
h = height of liquid (water) displaced by an object.
A = surface area of the floating object.
The unit of measurement for buoyancy is Newton (N).
Additionally, the density of a fluid is directly proportional to the buoyant force acting on it i.e as the density of a liquid decreases, buoyancy decreases and vice-versa.
Furthermore, an object such as a boat, ship, ferry, canoe, etc, are able to float because the volume of water they displace weigh more than their own weight. Thus, if a boat or any physical object weighs more than the volume of water it displaces, it would sink; otherwise, it floats.
In conclusion, the true statement is that when an object floats, it does not displace its entire volume.
A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).
a. Plot the data as engineering stress versus engineering strain.
b. Compute the modulus of elasticity.
c. Determine the yield strength at a strain offset of 0.002.
d. Determine the tensile strength of this alloy.
e. What is the approximate ductility, in percent elongation?
f. Compute the strain energy density up to yielding (modulus of resilience).
( Load in N Load in lb Length in mm Length in in. 2.000 2.002 2.004 2.006 2.008 2.010 2.020 2.040 2.080 2.120 2.160 2.200 2.240 2.270 2.300 2.330 Fracture 50.800 7330 15,100 3400 23,100 5200 30,400 6850 34,400 7750 38,400 8650 41,3009300 44,800 10,100 46,200 10,400 53, 47,300 10,650 54.864 47,500 10,700 55.880 46,100 10,400 44,800 10,100 42,600 9600 3,400 8200 Fracture Fracture Fracture 50.851 50.902 50.952 51.003 51.054 1650 51.308 51.816 52.832 848 56.896 57.658 58.420 59.182
Answer:
A cylindrical specimen of aluminum having a diameter of 0.505 in. (12.8 mm) and a gauge length of 2.0 in. (50.8 mm) is pulled in tension. Use the load-elongation characteristics tabulated below to complete parts (a) through (f).
A train moving with a uniform speed covers a distance of 120 m in 2 s. Calculate
(i) The speed of the train
(ii) The time it will taketo cover 240 m.
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 120 = (u \times 2) + \frac{1}{2} \times 0 \times {2}^{2} \\ 120 = 2u \\ { \tt{speed = 60 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 240 = (60t) \\ { \tt{time = 4 \: seconds}}[/tex]
Starting with the Ideal Gas Law, show that the relationship between volume and temperature in an adiabatic process is the one given by :
TfVf^γ^-1 = TiVi^γ-1 = Constant
Answer:
hope it helps
explanation:
The velocity of an object increases at a constant rate from 20 m/s to 50 m/s in 10 s.Find the acceleation
Answer:
[tex]{ \bf{v = u + at}} \\ 50 = 20 + (a \times 10) \\ 30 = 10a \\ { \tt{acceleration = 3 \: {ms}^{ - 2} }}[/tex]
Your cell phone typically consumes about 300 mW of power when you text a friend. If the phone is operated using a lithium-ion battery with a voltage of 3.5 V, what is the current (in A) flowing through the cell-phone circuitry under these circumstances
Answer:
I = 0.0857 A
Explanation:
Given that,
Power consumed by the cellphone, P = 300 mW
The voltage of the battery, V = 3.5 V
Let I is the current flowing through the cell-phone. We know that,
P = VI
Where
I is the current
So,
[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]
So, the current flowing the cell-phone is 0.0857 A.
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
As it pulls itself up to a branch, a chimpanzee accelerates upward at 2.4 m/s2 at the instant it exerts a 260-N force downward on the branch.Find the magnitude of the force the chimpanzee exerts on the Earth.
Answer:
[tex]F=208.83N[/tex]
Explanation:
From the question we are told that:
Acceleration [tex]a=2.4m/s^2[/tex]
Force of Branch [tex]F=260N[/tex]
Generally the Newton's equation second law for Force is mathematically given by
[tex]ma=F-mg[/tex]
[tex]m=\frac{260}{2.4+9.8}[/tex]
[tex]m=21.31kg[/tex]
Therefore
[tex]F=mg[/tex]
[tex]F=(21.31)(9.8)[/tex]
[tex]F=208.83N[/tex]
Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by strings with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.
Answer:
(a) 1 : 2
(b) same
Explanation:
Let the mass of puck A is m and the mass of puck B is 2 m.
initial speed for both the pucks is same as u and the distance is same for both is s.
let the tension is T for same.
The kinetic energy is given by
[tex]K = 0.5 mv^2[/tex]
(a) As the speed is same, so the kinetic energy depends on the mass.
So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2
(b) A the distance s same so the final velocities are also same.
(a) The kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b) The final speed of both the puck A and B are same.
Let the mass of puck A is m and the mass of puck B is 2 m.
Initial speed for both the pucks is same as u and the distance is same for both is s.
Let the tension is T for same.
Then, the kinetic energy is given as,
[tex]KE = \dfrac{1}{2}mv^{2}[/tex]
(a)
As the speed is same, so the kinetic energy depends on the mass.
Then,
[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]
So, kinetic energy of A : Kinetic energy of B = 1 : 2.
Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b)
The final speed for the puck is given as,
v = s/t
here, s is the distance covered.
Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.
Thus, we can conclude that the final speed of both the puck A and B are same.
learn more about the kinetic energy here:
https://brainly.com/question/12669551
A uniform ladder of length 24 m and weight w is supported by horizontal floor at A and by a vertical wall at B. It makes an angle 45 degree with the horizontal. The coefficient of friction between ground and ladder is 1/2 and coefficient of friction between ladder and wall is 1/3. If a man whose weight is one-half than the ladder, ascends the ladder, how much length x of the ladder he shall climb before the ladder slips
Answer:
I could not find the answer or do it myself if I did find it I would defenetly share
Help me plssssssss cause I’m struggling
Answer:
I am pretty sure it is C
Explanation:
It can be found all over the universe
suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along
Answer:
Pressure is more in the open container than the closed one.
Explanation:
The pressure due to the fluid at a depth is given by
Pressure = depth x density of fluid x gravity
So, when the container is open, the atmospheric pressure is also add up but when the container is closed only the pressure due to the fluid is there.
So, when the container is open, the pressure is atmospheric pressure + pressure due to the fluid.
hen the container is closed only the pressure due to the fluid is there.
Warm air rises because faster moving molecules tend to move to regions of less
A) density.
B) pressure.
C) both of these
D) none of the above
Answer:
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Explanation:
rshyyjfshfsgfshfsyhrsyhuydtufhr6ra6yris7toe7r9w7rr6w996ryrowosotusuogsuoufsutot
convert 56km/h to m/s.
Explanation:
15.556 metres per second
A 150.0-kg crate rests in the bed of a truck that slows from 50.0 km/h to a stop in 12.0 s. The coefficient of static friction between the crate and the truck bed is 0.645. What is the minimum stopping time for the truck in order to prevent the crate from sliding?
By Newton's second law,
• the net force acting vertically on the crate is 0, and
∑ F = n - mg = 0 ==> n = mg = 1470 N
where n is the magnitude of the normal force; and
• the net force acting in the horizontal direction on the crate is also 0, with
∑ F = f - b = 0 ==> b = f = µn = 0.645 (1470 N) = 948.15 N
where b is the magnitude of the braking force, f is (the maximum) static friction, and µ is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.
Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.
Let a be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for a :
f = ma ==> a = (948.15 N) / (150.0 kg) = 6.321 m/s²
With this acceleration, the truck will come to a stop after time t such that
0 = 50.0 km/h - (6.321 m/s²) t ==> t ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s
and this is the smallest stopping time possible.
