Answer:
F' = F
Explanation:
The gravitational force of attraction between two objects can be given by Newton's Gravitational Law as follows:
[tex]F = \frac{Gm_1m_2}{r^2}[/tex]
where,
F = Force of attraction
G = Universal gravitational costant
m₁ = mass of first object
m₂ = mass of second object
r = distance between objects
Now, if the masses and the distance between them is doubled:
[tex]F' = \frac{G(2m_1)(2m_2)}{(2r)^2}\\\\F' = \frac{Gm_1m_2}{r^2}[/tex]
F' = F
SOMEONE HELP ME PLSS
Answer:
Explanation:
1 minute =60 second
power=work done/time taken
=2500/60
=41.66 watt
work=force *displacement
=500 N * 100 m
=50000 joule
18 of 25
The practice of science can answer only scientific questions. And scientific
questions guide the design of investigations. What must be true of the
possible answers to a scientific question?
A. They agree with all prior experiments,
B. They can be supported by evidence,
C. They are popular with a majority of scientists,
D. They lead to increased funding of scientific research,
Answer:
B
Explanation:
Cause its true
A proton accelerates from rest in a uniform electric field of 595 N/C. At one later moment, its speed is 1.15 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton. (b) Over what time interval does the proton reach this speed? (c) How far does it move in this time interval? (d) What is its kinetic energy at the end of this interval?
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
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].
YO giving 49 dollars if you answer this question .
. ( )
Trace the flow of energy through the food pyramid, label the missing units of energy found at each level by calculating the change in energy
Answer:
Energy Flow
The chemical energy of food is the main source of energy required by all living organisms. This energy is transmitted to different trophic levels along the food chain. This energy flow is based on two different laws of thermodynamics:
First law of thermodynamics, that states that energy can neither be created nor destroyed, it can only change from one form to another.
Second law of thermodynamics, that states that as energy is transferred more and more of it is wasted.
Energy Flow in Ecosystem
The energy flow in the ecosystem is one of the major factors that support the survival of such a great number of organisms. For almost all organisms on earth, the primary source of energy is solar energy. It is amusing to find that we receive less than 50 per cent of the sun’s effective radiation on earth. When we say effective radiation, we mean the radiation, which can be used by plants to carry out photosynthesis.
Explanation:
Energy Flow in Ecosystem
The energy flow in the ecosystem is one of the major factors that support the survival of such a great number of organisms. For almost all organisms on earth, the primary source of energy is solar energy. It is amusing to find that we receive less than 50 per cent of the sun’s effective radiation on earth. When we say effective radiation, we mean the radiation, which can be used by plants to carry out photosynthesis.
Most of the sun’s radiation that falls on the earth is usually reflected back into space by the earth’s atmosphere. This effective radiation is termed as the Photosynthetically Active Radiation (PAR).
Overall, we receive about 40 to 50 percent of the energy having Photosynthetically Active Radiation and only around 2-10 percent of it is used by plants for the process of photosynthesis. Thus, this percent of PAR supports the entire world as plants are the producers in the ecosystem and all the other organisms are either directly or indirectly dependent on them for their survival.
The energy flow takes place via the food chain and food web. During the process of energy flow in the ecosystem, plants being the producers absorb sunlight with the help of the chloroplasts and a part of it is transformed into chemical energy in the process of photosynthesis.
This energy is stored in various organic products in the plants and passed on to the primary consumers in the food chain when the herbivores consume (primary consumers) the plants as food. Then conversion of chemical energy stored in plant products into kinetic energy occurs, degradation of energy will occur through its conversion into heat.
Then followed by the secondary consumers. When these herbivores are ingested by carnivores of the first order (secondary consumers) further degradation will occur. Finally, when tertiary consumers consume the carnivores, energy will again be degraded. Thus, the energy flow is unidirectional in nature.
Moreover, in a food chain, the energy flow follows the 10 percent law. According to this law, only 10 percent of energy is transferred from one trophic level to the other; rest is lost into the atmosphere. This is clearly explained in the following figure and is represented as an energy pyramid.
Two blocks A and B with mA = 2.9 kg and mB = 0.87 kg are connected by a string of negligible mass. They rest on a frictionless horizontal surface. You pull on block A with a horizontal force of 6 N. Determine the tension in the string connecting the two blocks.
Answer:
Tension = 1.38 N
Explanation:
Given that:
Mass of block A [tex]m_A[/tex] = 2.9 kg
mass of block B [tex]m_B[/tex] = 0.87 kg
Force F = 6 N
Assume a = acceleration of the blocks.
Then:
[tex]m_A[/tex] (a) + [tex]m_B[/tex] (a) = 6
2.9a + 0.87a = 6
3.77a = 6
a = 6/3.77
a = 1.59 m/s²
Suppose T to be the tension in the string.
If we take a look at the forces acting on the first block, then:
F - T = [tex]m_A[/tex] (a)
T = F - [tex]m_A[/tex] (a)
T = 6 - 2.9(1.59)
T = 6 - 4.62
T = 1.38 N
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.
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.
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
200N force was used to move 150N block on an inclined plane of length 12m with a height of 4m. The efficiency of the inclined plane will be
Answer:
42.1%
Explanation:
Applying,
E(%) = (M.A/V.R)×100............. Equation 1
Where E(%) = Percentage efficiency, M.A = mechanical advantage, V.R = Velocity ratio.
But,
M.A = L/E............... Equation 2
Where L = Load, E = Effort.
From the question,
Given: L = 150 N, E = 200 N
Substitute these values into equation 2
M.A = 200/150
M.A = 1.33
Also,
V.R for Inclined plane = 1/sin∅ =
V.R = 1/sin∅............... Equation 2
Where ∅ = angle of the inclined plane.
Where,
tan∅ = 4/12
∅ = tan⁻¹(4/12)
∅ = 18.43°
Therefore,
V.R = 1/sin18.43
V.R = 3.16.
Substituting the value of M.A and V.R into equation 1
E(%) = (1.33/3.16)×100
E(%) = 42.1%
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]
21. The value of 'g on the surface of earth is 9.8 m/s. If mass of Jupiter is 319 times the mass of the
arth and its radius is 11 times the radius of the earth, find the value of 'g' on the surface of Jupiter.
What will be the weight of a man having mass 50 kg on the surface of Jupiter?
Vhy? Also write solve step by step
.
The answer is in the photo. Hope you are satisfied.Have a nice day ☺️
To solve this we must be knowing each and every concept related to acceleration due to gravity and its calculation. Therefore, the value of 'g' on the surface of Jupiter is 25.8m/s².
What is acceleration due to gravity?
Gravitational acceleration is indeed a vector, which implies it has a direction and a magnitude. The acceleration caused by gravity at the Earth's surface is denoted by the symbol g. It possesses a standard value of 9.80665 m/s².
Mathematically, the formula for acceleration due to gravity can be given as
g = G M /R²
G=6.674 x 10⁻¹¹ m³ kg⁻¹s⁻²
Mass of Jupiter=319×mass of earth
Radius of Jupiter=11×radius of earth
substituting all the given values in the above equation, we get
g = 6.674 x 10⁻¹¹× 319×6×10²⁴ /(11×6378 10³)²
On calculations, we get
g=25.8m/s²
Therefore, the value of 'g' on the surface of Jupiter is 25.8m/s².
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hệ thống điện bao gồm những phần tử nào
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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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 car pulled by a tow truck has an acceleration of 2.0 m/s^2. What is the mass of the car if the net force on the car is 3,500 N?
Answer:
ans: 1750 kg
Explanation:
F = ma
so ,
m = F/a
= 3500 / 2
= 1750 kg
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]
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?
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:
A toy plane weighing 1.5 kg is flying in the air at 8 m/s. The player
turns down its speed to 6m/s. What is the work done on it?
Use the work-energy theorem. The total work done on the plane is equal to the change in its kinetic energy:
W = ∆K
W = 1/2 (1.5 kg) (6 m/s)² - 1/2 (1.6 kg) (8 m/s)²
W = -21 J
(Note that the work must be negative since the "braking" force opposes the direction in which the plane is moving.)
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.
If a diffuse gas of atoms is trapped, cooled, and compressed, the de Broglie wavelengths of some atoms can become larger than the distance between atoms. The overlap between atoms leads to new states of matter. In one experiment, 87Rb atoms were cooled to 20 nK.
Required:
a. What is their de Broglie wavelength at this temperature?
b. How many times larger is this than the 0.47 nm diameter of the atoms?
Answer:
a) λ = 0.5346 10⁻⁹ m, b) 1.13
Explanation:
a) The wave-particle duality is
λ = [tex]\frac{\hbar }{p}[/tex]
the expression for energy kinetic is
E = p² / 2m
energy is also related to temperature
E = [tex]\frac{3}{2} \ kT[/tex]
we substitute
p =
λ =
Rubidium's mass is m = 85 uma
we calculate
λ =
λ=
λ = 0.5346 10⁻⁹ m
b) the relationship of the two lengths is
λ /0.47 nm = 0.53 / 0.47 = 1.13
You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity
of 0.50 m/s. What is the work done by gravity on the book?
1) +10 J
2) -10
3) +5.0
4) -5.0
The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).
The force exerted on the book by gravity has magnitude
F = mg = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N
You raise the book 1.0 m in the opposite direction, so the work done is
W = (10 N) (-1.0 m) = -10 J
The work done by gravity on the book is -10 Joules. The correct option is 2.
What is gravity?The force of attraction felt by a person at the center of a planet or Earth is called as the gravity.
Given are the three positions. The Earth has the acceleration due to gravity, g = 9.81 m/s²,
The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).
The force exerted on the book by gravity has magnitude
F = mg = 10 N x( -9.80 m/s² )= -9.8 N
F ≈ -10 N
The book is raised by 1.0 m in the opposite direction, so the work done is
W =- 10 N x 1.0 m = -10 J
Thus, the correct option is 2.
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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
find the equivalent resistance.
Answer:
Explanation:
Note the connection between D-L-H shorts out the resistors on the right. Those 4 resistors can be ignored.
The connection between D-L-K means the 2 resistors between C-D and C-K are in parallel. The equivalent resistance of 2 6-ohm resistors in parallel is 3 ohms.
Similarly the connection between K-L-H means the 2 resistors between J-K and J-H are in parallel. The equivalent resistance of 2 6-ohm resistors in parallel is 3 ohms.
Adding the resistors in series together:
A-C-L-J-D = 6 + 3 + 3 + 6
= 18 ohms
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
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 car with a mass of 850kg is moving at a speed of 72km/h when colliding with a concrete wall until it stops. After the collision, the car travels 2.0m before stopping. Determine the resultant of the forces acting on the car.
Answer:
Explanation:
The vehicle is experiencing a large force created by the concrete wall.
Equation
vf^2 = vi^2 + 2*a * d
Givens
vf = 0 The car eventually does stop.
vi = 72 km/hr * [ 1000 m/ km] * [1 hour / 3600 seconds]
vi = 20 meters / second
a = ?
m = 850 kg
Solution
vf^2 = vi^2 + 2a*d
0 = 20 m/s + 2* 2 *a
-20 m/s = 4a
-20/4 = a
a = - 5 m/s^2 The minus sign tells you the vehicle is slowing down. It sure should be.
Force = m * a
F = - 850 * (-5)
F = - 4250 N
The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.
3. How can a generator that otherwise produces AC
current be modified to produce DC current?