Answer:
A
Explanation:
Ke = 1/2 MV^2
The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbed to the top of the building
Answer:
579600J
Explanation:
Given parameters:
Height of the building = 828m
Weight of the man = 700N
Unknown:
Work done by the man = ?
Solution:
The work done by the man is the same as the potential energy expended.
Work done:
Work done = Weight x height = 700 x 828
Work done = 579600J
The law of ____ states that energy in a system can change forms but can never be created or destroyed.
Answer:
The Law of Conversation
A piece of aluminum with a mass of 3.05 g initially at a temperature of 10.8 °C is heated to a temperature of 20.
Assume that the specific heat of aluminum is 0.901 J/(g°C).
How much heat was needed for this temperature change to take place?
Answer:
25.3J
Explanation:
Given parameters:
Mass of aluminum = 3.05g
Initial temperature = 10.8 °C
Final temperature = 20 °C
Specific heat = 0.9J/g °C
Unknown:
Amount of heat needed for the temperature to change = ?
Solution:
To solve this problem, we use the expression:
H = m C Ф
H is the amount of heat
m is the mass
C is the specific heat capacity
Ф is the change in temperature
H = 3.05 x 0.901 x (20 - 10.8) = 25.3J
7 x 7 x 7 x.........x 7 = n (30 times)
7 is multiplied 30 times to get a number n. Which of the following will be a factor of n?
which phenomenon supports the particular model of light?
Answer:
photoelectric effect
Explanation:
The photoelectric effect supports a particle theory of light in that it behaves like an elastic collision (one that conserves mechanical energy) between two particles, the photon of light and the electron of the metal.
An 8.0-kg object moving with an initial velocity of 8.0 m/s on a surface comes to rest due to friction after it travels a horizontal distance of 11 m. What is the coefficient of kinetic friction between the object and the surface
Answer:
μk = 0.3
Explanation:
According the work-energy theorem, the net work done on an object, is equal to the change in the kinetic energy of the object.In this case, assuming that the object was not accelerating when moving at 8.0 m/s, the only force doing work on it, is the kinetic friction force, acting through the 11 m distance before causing the object to come to rest.The kinetic friction force, is given by the following expression:[tex]f_{frk} = \mu_{k} * F_{n} (1)[/tex]
where μk = coefficient of kinetic friction, Fn = normal force.Assuming that the surface is horizontal, since the normal force is always perpendicular to the surface, and the object is not accelerating in the vertical direction, this means that the normal force must be equal and opposite to the force that gravity exerts on the object, as follows:[tex]F_{n} = m*g (2)[/tex]
The friction force and the horizontal displacement produced by it have opposite directions, so the angle between the force and the displacement is 180º.So, we can express the work done by the kinetic friction force, as follows:[tex]W_{ffr} = -F_{fr} * d = - \mu_{k}* m*g*d (3)[/tex]
We have already said that (3) must be equal to the change in the kinetic energy, ΔK, as follows:[tex]\Delta K = K_{f} - K_{o} (4)[/tex]
Since the object comes to rest, Kf = 0.Replacing K₀ = 1/2*m*v₀² in (4), we have:[tex]\Delta K = 0 - \frac{1}{2} *m * v_{o} ^{2} = - \frac{1}{2} *m * v_{o} ^{2} (5)[/tex]
From (3) and (5), since they are equal each other, we get:[tex]W_{ffr} = \mu_{k}* g*d = \frac{1}{2} * v_{o} ^{2} (6)[/tex]
Replacing by the givens, and solving for μk, we get:[tex]\mu_{k} = \frac{1}{2} * v_{o} ^{2} *\frac{1}{g*d} =\frac{(8.0m/s)^{2}}{2*9.8m/s2*11m} = 0.3 (7)[/tex]
What must the charge (sign and magnitude) of a particle of mass 1.40 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/CN/C
Answer:
the charge of the particle is -2.144 x 10⁻⁵ C.
Explanation:
The force acting on the particle is calculated as;
F = EQ = mg
[tex]Q = \frac{mg}{E}[/tex]
where;
Q is magnitude of the charge of the particle
[tex]Q = \frac{(1.4\times 10^{-3})(9.8)}{640} \\\\Q = 2.144 \ \times \ 10^{-5} \ C[/tex]
since the magnetic field is acting downward, the force must be acting upward in opposite direction.
Thus, the charge of the particle will be -2.144 x 10⁻⁵ C.
Match the choices to the appropriate blank . Use each choice only once.
1. about 100,000 light-years
2. about 5 AU
3. about 8 light-years
4. about 1.5 AU
5. less than 0.01 AU
a. Mars is_____from the Sun.
b. Jupiter is_____from the Sun.
c. The star Sirius is_____from the Sun.
d. The diameter of the Milky Way Galaxy is_____.
e. The distance from Earth to the Moon is_____.
Answer:
1. about 1.5 AU
2. about 5 AU
3. about 8 light-years
4. about 100,000 light-years
5. less than 0.01 AU
Explanation:
a. Mars is about 1.5 AU from the Sun.
b. Jupiter is about 5 AU from the Sun.
c. The star Sirius is about 8 light-years from the Sun.
d. The diameter of the Milky Way Galaxy is about 100,000 light-years.
e. The distance from Earth to the Moon is less than 0.01 AU.
Note: AU is an acronym for Astronomical Unit and it is a standard unit by astronomers to illustrate the distance between the planetary bodies found in the solar system.
Earth has a mass of 5.98 1024 kg and a radius of 6.38 106 m. Assume it is a uniform solid sphere. The distance of Earth from the Sun is 1.50 1011 m. (Assume Earth completes a single rotation in 24.0 hours and orbits the Sun once every 365 Earth days.) (a) What is the rotational kinetic energy of Earth on its axis
Answer:
Explanation
Rotational kinetic energy of the earth = 1/2 Iω²
where I is moment of inertia of the earth and ω is angular velocity .
I = 2/5 m R² , m is mass and R is radius of the earth .
I = 2/5 x 5.98 x 10²⁴ x ( 6.38 x 10⁶ )²
=97.36 x 10³⁶
ω = 1 / T
T = 24 x 60 x 60 = 86400 s
ω = 1 / 86400
= 11.57 x 10⁻⁶ rad / s
Rotational Kinetic energy = 1/2 Iω²
= .5 x 97.36 x 10³⁶ x (11.57 x 10⁻⁶ )²
= 6516.54 x 10²⁴ J
Your destroyer has a RADAR antenna height of 40 m. Using RADAR, what is the maximum detection range of a patrol boat with a mast height of 11 m above the water?
Answer:
The maximum detection range is 39.75 km
Explanation:
Given that;
Antenna height h1 = 40 m
Target height ( patrol boat mast ) h2 = 11 m
Using RADAR, what is the maximum detection range = ?
Using RADAR
we know that; Maximum detection range = (√17h1 + √17h2) km
where h1 and h2 are heights of the antenna and target height in meters
so we substitute in our values
Maximum detection range = (√(17 × 40) + √(17 × 11)) km
Maximum detection range = (√680 + √187) km
Maximum detection range = (26.0768 + 13.6747) km
Maximum detection range = (26.0768 + 13.6747) km
Maximum detection range = 39.75 km
Therefore, The maximum detection range is 39.75 km
Electron cloud configuration for
Answer:
electrons are located around the nucleus of an atom.
Explanation:
Electron configurations describe where electrons are located around the nucleus of an atom. For example, the electron configuration of lithium, 1s²2s¹, tells us that lithium has two electrons in the 1s subshell and one electron in the 2s subshell.
A police car is traveling north on a straight road at a constant 20.0 m/sm/s. An SUV traveling north at 30.0 m/sm/s passes the police car. The driver of the SUV suspects he may be exceeding the speed limit, so just as he passes the police car he lets the SUV slow down at a constant 1.90 m/s2m/s2. How much time elapses from when the SUV passes the police car to when the police car passes the SUV
Answer:
elapsed time is 10.53 sec
Explanation:
Given that;
velocity of the car [tex]V_{P}[/tex] = 20.0m/s
Initial velocity of the SUV [tex]U_{suv}[/tex] = 30.0 m/s
SUV slow down at a constant a = 1.90 m/s²
Now, Let Vs represent the final velocity of the SUV and after time t when the police can cross it and d represent the distance both of the cars covers between the two crossing
so;
d = [tex]V_{P}[/tex] × t
also d = [tex]U_{suv}[/tex] t - 1/2 at²
so
[tex]V_{P}[/tex] × t = [tex]U_{suv}[/tex] t - 1/2 at²
we substitute
20 × t = 30 × t - 1/2 × 1.9 × t²
20t = 30t - 1/2 × 1.9 × t²
1/2 × 1.9 × t² = 30t - 20t
1/2 × 1/9 × t² = 10t
t = 2 × 10 / 1.9 ( as t ≠ 0 )
t = 20 / 1.9
t = 10.53 sec
Therefore, elapsed time is 10.53 sec
The friction force that acts on objects that are at rest is___________
Answer:
static friction
Explanation:
static friction is the friction force that acts on objects at rest
Two objects travel the same distance. The one that is moving faster will:
Take more time to go the distance
Take less time to go the same distance
Take the same time as the slower object
None of the above
Answer: take less time to go the same distance
Explanation:
Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.
A woman is pushing a stroller with a baby with a mass of 8.18 kg. If the stroller is accelerating at 88.2 m/s2. How much force is she exerting?
Answer:
well you want to add 88.2 and s2.
Explanation:
The small spheres that are moving through the circuit are the electric current. Current is the flow or movement of electrons. Describe how the current moves in the simulation.
Answer:
When the circuit is connected properly, the current starts flowing from one end of the battery to the other end.
Explanation:
hope this helps!
Answer:
When the circuit is connected properly, the current starts flowing from one end of the battery to the other end.
Explanation:
Edmentum answer.
Why are soft materials used in theaters and auditoriums?
Answer:
The roof and walls of the auditorium or cinema hall are generally covered with sound absorbent materials like draperies or compressed fibreboard to reduce reverberation. These materials reduce the formation of echoes by absorbing sound waves.
Explanation:
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A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter for 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will doublethe pressure of the gas inside the cylinder.
Answer:
the local atmospheric pressure is 93.63 kPa
the mass of the weights is 156.9 kg
Explanation:
Given that;
Initial pressure of gas = 100 kPa
mass of piston = 10 kg and diameter = 14 cm = 0.14 m
g = 9.81 m/s²
Now,
P_gas = P_atm + P_piston
100 = P_atm + P_piston --------- let this equation 1
P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²
P_piston = 98.1 / (π/4×( 0.14 )²)
P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa
now, from equation 1
100 = P_atm + P_piston
we substitute
100 = P_atm + 6.37
P_atm = 100 - 6.37
P_atm = 93.63 kPa
Therefore, the local atmospheric pressure is 93.63 kPa
Now for pressure of the gas in the cylinder ⇒ 2×initial pressure
Pgas_2 = 2 × 100 = 200 kPa
Pgas_2 = P_atm + P_piston + P_weight
Pgas_2 = P_gas + P_weight
we substitute
200 kPa = 100 kPa + P_weight
P_weight = 200 kPa - 100 kPa
P_weight = 100 kPa = 100,000 Pa
Also;
P_weight = M×g / A
100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)
100,000 × 0.01539 = M × 9.81
1539 = M × 9.81
M = 1539 / 9.81
M = 156.9 kg
Therefore, the mass of the weights is 156.9 kg
What is the sum of internal energy
6. What is the lowest temperature on the Kelvin scale? What happens to matter when it
reaches this temperature?
7. What is different about the degrees on the Fahrenheit and Kelvin scales and the Celsius
and Kelvin scales?
The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive plate, and an electron is held stationary at the negative plate. The plate separation is 4.22 cm. At the same moment, both particles are released.
A. Calculate the distance (in cm) from the positive plate at which the two pass each other.
B. Repeat part (a) for a sodlum lon (Nat) and a chlorlde lon (CI).
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
Data Given:
Electric Field between two parallel plates = 628 N/C
Separation = 4.22 cm
a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.
Solution:
First of all:
Force on proton due to the Electric field between the plates is:
[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E
and, we know that, F = ma
So,
[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E
a = [tex]\frac{q_{p}.E }{m_{p} }[/tex] Equation 1
So,
The distance covered by the electron is:
S = ut + 1/2[tex]at^{2}[/tex]
Here, u = 0.
S = 1/2[tex]at^{2}[/tex]
Put equation 1 into the above equation:
S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex] )[tex]t^{2}[/tex] Equation 2
So,
Similarly, the distance covered by electron will be:
(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex] )[tex]t^{2}[/tex] Equation 3
We know that the charge of electron is equal to the charge of proton so,
[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q
By dividing the equation 2 by equation 3, we get:
[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]
Solve the above equation for S,
S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S
So,
S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]
Plugging in the values,
As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]
S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]
S = 0.002298 cm (Distance from the positive plate at which the two pass each other)
b) In this part, we to calculate distance for Sodium ion and chloride ion as above.
So,
we already have the equation, we need to put the values in it.
So,
S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]
As we know the mass of chlorine is 35.5 and of sodium is 23
S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]
S = 2.56 cm
Fluids
A = 2804 cm3 B = 2862 cm2 C = 2916 cm3
Three separate fluids, A, B, and C have been selected at random and each initially fills a 3000 cm3 volume at atmospheric pressure. A gage pressure of 6 x 107 N/m2 is then applied to each fluid. The final volume is given below. Determine which fluids were selected from the given list.
Acetone E = 0.92 GPa Glycerin E = 4.35 GP
Water E = 2.15 GPa Mercury E = 28.5 GPa
Benzene E = 1.05 GPa Sulfuric Acid E = 3.0 GPa
Ethyl Alcohol E = 1.06 GPa Gasoline E = 1.3 GPa
Petrol E = 1.45 GPa Seawater E = 2.34 GPa
Answer:
Explanation:
Fluid A :
Δ V = Change in volume = (3000 - 2804) x 10⁻⁶ m³ = 196 x 10⁻⁶ m³
volume strain = Δ V / V = 196 x 10⁻⁶ / 3000 x 10⁻⁶
= .06533
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .06533 = 91.84 x 10⁷ Pa = .92 GPa .
It is Acetone .
Fluid B :
Δ V = Change in volume = (3000 - 2862) x 10⁻⁶ m³ = 138 x 10⁻⁶ m³
volume strain = Δ V / V = 138 x 10⁻⁶ / 3000 x 10⁻⁶
= .046
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .046 = 130.43 x 10⁷ Pa = 1.3 GPa .
It is Gasoline .
Fluid C :
Δ V = Change in volume = (3000 - 2916) x 10⁻⁶ m³ = 84 x 10⁻⁶ m³
volume strain = Δ V / V = 84 x 10⁻⁶ / 3000 x 10⁻⁶
= .028
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .028 = 214.28 x 10⁷ Pa = 2.14 GPa .
It is Water .
A roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds. What is the acceleration of the rider?
A. 1.33 m/s2
B. 3 m/s2
C. 5.33 m/s2
D. 4 m/s2
Answer:27
Explanation:
If a roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds, then the acceleration of the rider would be 4 m / s² , therefore the correct answer is option D.
What are the three equations of motion?There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem, A roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds.
By using the first equation of the motion,
v = u + at
16 = 4 + 3a
a = 16 -4 / 3
= 12 / 3
= 4 m / s²
Thus, the acceleration of the rider would be 4 m / s².
Learn more about equations of motion here, refer to the link given below ;
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The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a parallel beam of white light in the 400 nm to 700 nm wavelength band. What is the angular width of the gap between the first order spectrum and the second order spectrum
Answer:
3.28 degree
Explanation:
We are given that
Distance between the ruled lines on a diffraction grating, d=1900nm=[tex]1900\times 10^{-9}m[/tex]
Where [tex]1nm=10^{-9} m[/tex]
[tex]\lambda_2=400nm=400\times10^{-9}m[/tex]
[tex]\lambda_1=700nm=700\times 10^{-9}m[/tex]
We have to find the angular width of the gap between the first order spectrum and the second order spectrum.
We know that
[tex]\theta=sin^{-1}(\frac{m\lambda}{d})[/tex]
Using the formula
m=1
[tex]\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})[/tex]
[tex]\theta=21.62^{\circ}[/tex]
Now, m=2
[tex]\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})[/tex]
[tex]\theta_2=24.90^{\circ}[/tex]
[tex]\Delta \theta=\theta_2-\theta_1[/tex]
[tex]\Delta \theta=24.90-21.62[/tex]
[tex]\Delta \theta=3.28^{\circ}[/tex]
Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree
3. A 70 kg person climbs a 6 m ladder. How much work is required by the person?
Answer:
4116J
Explanation:
Given parameters:
Mass of the person = 70kg
Height of the ladder = 6m
Unknown:
Work done = ?
Solution:
The work done by the person climbing is the same as the potential energy.
Work done is the force applied to move a body through a distance;
So;
Potential energy = mass x acceleration due to gravity x height
Therefore;
Potential energy = 70 x 9.8 x 6 = 4116J
Two boxes on opposite ends of a massless board that is 3.0 m long. The board is supported in the middle by a fulcrum. The box on the left has a mass (m1) of 25 kg, and the box on the right has a mass (m2) of 15kg. How far should the fulcrum be positioned from the left side of the board in order to balance the masses horizontally?
a. 0.38 m
b. 1.1 m
c. 0.60 m
d. 1.9 m
Answer:
b. 1.1 m
Explanation:
It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,
[tex]s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)[/tex]
where,
s₁ = distance of fulcrum from left mass
s₂ = distance of fulcrum from right mass
In order to achieve balance, the torque due to both masses must be equal:
[tex]T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}[/tex]
s₁ = 1.1 m
Hence, the correct option is:
b. 1.1 m
How is energy transferred when
hitting a nail?
Answer:
kinetic energy
kinetic energy
What is displacement?
a. The distance an object travels.
b. The distance between the starting point and the ending point of an object's
journey.
C. The amount of time it takes an object to travel to a destination.
d. The path in which an object travels.
Answer:
displacement is the distance between the starting point and the ending point of an object's journey
The pickup truck has a changing velocity because the pickup truck
A.can accelerate faster than the other two vehicles
B.is traveling in the opposite direction from the other two vehicles
C.is traveling on a curve in the road
D.needs a large amount of force to move
please get right i need awnser today
Answer:
C. Is traveling on a curve in the road
Hope this helps :3
The pick up truck has a changing velocity because, it is travelling on a curve in the road. A change in direction results in its change in velocity because, velocity is a vector quantity.
What is velocity ?Velocity is a physical quantity that measures the distance covered by an object per unit time. It is a vector quantity, thus having magnitude as well direction.
The rate of change in velocity is called acceleration of the object. Like velocity, acceleration also is a vector quantity. Thus, a change in magnitude or direction or change in both for velocity make the object to accelerate.
Here, all the three vehicles are travelling with the same velocity. But, the truck is moving to a curve on the road. The curvature in the path will make a change in its velocity.
Find more on velocity:
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a plane passes over Point A with a velocity of 8,000 m/s north. Forty seconds later it passes over Point B with a velocity of 10,000 m/s north. Which is the plane's acceleration from A to B ?
Acceleration = (change in velocity) / (time for the change)
Change in velocity = (ending velocity) - (starting velocity)
Change in the plane's velocity = (10,000 m/s north) - (8,000 m/s north)
Change in the plane's velocity = 2,000 m/s north
Time for the change = 40 seconds
Acceleration = (2,000 m/s north) / (40 seconds)
Acceleration = 50 m/s² north