Answer:
t = 1.62 h
Explanation:
A flat mirror fulfills the law of reflection where the incident angle is equal to the reflected angle.
θ_i = θ_r
If we use trigonometry to find the angles, the mirror is at a height of L = 1.87 m, and the reflected rays reach a distance x1 = 3.56 m
tan θ₁ = x₁ / L
tan θ₁ = [tex]\frac{3.56}{1.87}[/tex]
θ₁ = tan⁻¹ 1.90
θ₁ = 62.29º
for the second case x₂ = 1.46 m
tan θ₂ = x₂ / L
θ₂ = tan⁻¹ [tex]\frac{1.46}{1.87}[/tex]
θ₂ = 37.98º
the difference in degree traveled is
Δθ = θ₁- θ₂
Δθ = 62.29 - 37.98
Δθ = 24.31º
as in the exercise they indicate that every 15º there is an hour
t = 24.31º (1h / 15º)
t = 1.62 h
True or false? A system must contain more than one object.
Answer:
true
Explanation:
normally -No system has ever performed well with one object.
A system must contain more than one object is a true statement.
What is system?A system is a group of interacting or interrelated objects that act according to a set of rules to form a unified whole.
Normally, no system has ever performed well with one object.
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30. Easy Guided Online Tutorial One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is 25 m/s. The masses of the two objects are 3.0 and 8.0 kg. Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object is the one moving initially.
Answer:
[tex]18.18\ \text{m/s}[/tex]
[tex]6.82\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of large object = 8 kg
[tex]m_2[/tex] = Mass of smaller object = 3 kg
When large mass is moving
[tex]u_1[/tex] = 25 m/s
[tex]u_2[/tex] = 0
For completely inelastic collision we have the relation
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 25+3\times 0}{8+3}\\\Rightarrow v=18.18\ \text{m/s}[/tex]
Speed of the combined mass when the larger object is moving is [tex]18.18\ \text{m/s}[/tex]
When smaller mass is moving
[tex]u_1[/tex] = 0
[tex]u_2[/tex] = 25 m/s
[tex]v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{8\times 0+3\times 25}{8+3}\\\Rightarrow v=6.82\ \text{m/s}[/tex]
Speed of the combined mass when the smaller object is moving is [tex]6.82\ \text{m/s}[/tex]
A 744 N force is applied to an object to reach an acceleration of 24 m/s2. What is the objects mass?
31kg
Explanation:
F = ma
m = F/a
m = 744N/24m/s^2
m = 31kg
(*Newton's Second Law*)
If you wrap 150 coils of heavy wire around a big iron nail and attach the ends of the wire to a 6.0v battery, you have a A) radio B) electromagnet C) galvanometer D) ammeter
Answer:
B
Explanation:
Because of the voltage attached to the iron nail
Badll
Which of the following is an example of
the Law of Inertia?
A. Sitting in a chair and breaking it
B. Throwing a ball in outer space and it goes on forever
unless acted upon by another force
C. Eating a salad to bring chemical energy into the body
D. Driving a car on a track
Help me, 100 points to answer right, answer without context will be reported
1. In the situation below, a tractor pulls a 850 sledge along a ramp of height ℎ = 1 and large = 30 °. If the tractor applies a constant force to the sledge = 6750 , at an angle = 36.9 °, determine the total work performed by all forces on the sledge to move it along the ramp. The coefficient of kinetic friction between the sledge and the plane is = 0.3. Tip: for the calculation, remember that only the components of the windows that are parallel to the direction of travel contribute to the work. Disregard the dimensions of the sled.
2) When firing a 2 projectile at a 1.4 bloco block, initially at rest, it is observed that the projectile is stuck in the block and the system moves together for a distance = 0.1 before stop. If the coefficient of kinetic friction between the block and the surface is = 0.25, determine what was the velocity of the projectile in the instant before impact. Tip: here you must use the conservation of linear momentum and also energy, considering the work done by the frictional force
2.4 What is the radiation error of a temperature measurement?
I
Answer:
diameter of the wire = 0.05 in =0.05 /12 =4.167 *10 ^-3 ft
area of cross section of the wire = A = 22/7 * ( d /2 ) ^2 =0.786 * ( 4.167 *10 ^-3 ) ^2 =1.365 *10 ^-5 ft2
E =...
Explanation:
A concave lens cannot produce a real image.
A. True
B. False
Answer:
B. False
A concave mirror and a converging lens will only produce a real image if the object is located beyond the focal point.
~Hoped this helped~
~Brainiliest?~
50 points help please
Answer:
C?
Explanation:
Yep,It's C all right.
Answer:
Yep,It's C all right.
Explanation:
which two options describes behaviors of particles that are related to the chemical properties of the materials
a- forming hydrogen bonds between them
b- reacting quickly with water
c- having a high mass
d- forming bonds with other atoms
Answer:
The two correct answers are B.) reacting quickly with water, and D.) forming bonds with other atoms.
Explanation:
I took the quiz on a.pex and these were correct.
can someone pls tell me what a force diagram is
An Atwood's machine consists of two masses, m1 and m2, connected by a string that passes over a pulley. If the pulley is a disk of radius R and mass M , find the acceleration of the masses. Express your answer in terms of the variables m1, m2, R, M, and appropriate constants.
Answer:
Explanation:
Suppose m₂ is greater than m₁ and it is going down . m₁ will be going up.
Let tension in string be T₁ and T₂ . Let common acceleration of system be a
For motion of m₁
T₁-m₁g = m₁a ----- (1)
For motion of m₂
m₂g- T₂ = m₂a ------- (2)
For motion of pulley
(T₂-T₁ )R represents net torque
(T₂-T₁ )R = I x α where I is moment of inertia of disc and α is angular acceleration of disc
(T₂-T₁ )R = 1/2 M R² x a / R
(T₂-T₁ ) = M a /2
Adding (1) and (2)
(m₂-m₁)g = (m₂+m₁)a + (T₂-T₁ )
(m₂-m₁)g = (m₂+m₁)a + Ma/2
(m₂-m₁)g = (m₂+m₁+ 0.5M)a
a = (m₂-m₁)g / ( (m₂+m₁+ 0.5M)
The acceleration of the masses is [tex]\dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex].
Given to us
Masses = m₁, m₂
The radius of the pulley = R
Mass of the pulley = M
Assumption
Let the mass m₁ > m₂. therefore, the mass m₁ is going down due to its weight while m₂ is going up.Assuming the tension in the string be T₁ and T₂, respectively.Also, the common acceleration in the system is a.Tensions in stringsWe know the acceleration due to gravity is denoted by g,
Tension in string 1, T₁[tex]T_1 = m_1(a+g)[/tex]......... equation 1
Tension in string 2, T₂[tex]T_2 = m_2(a-g)[/tex]......... equation 2
Inertia and acceleration of the pulleyThe inertia of the pulley, [tex]I = \dfrac{1}{2}MR^2[/tex]
Acceleration of the pulley, [tex]a = {\alpha }\times {R}[/tex]
Torque in the pulley[tex](T_2-T_1)R = I \times \alpha \\\\[/tex]
Substitute the values we get,
[tex][m_2(a-g)-m_1(a+g)]R = \dfrac{1}{2} MR^2 \times \dfrac{a}{R}\\\\[/tex]
[tex][m_2a-m_2g-m_1a-m_1g]R= \dfrac{1}{2} MR \times a[/tex]
[tex]m_2a-m_1a-m_2g-m_1g= \dfrac{1}{2} M \times a[/tex]
[tex]a(m_2-m_1)-g(m_2+m_1)= 0.5M \times a[/tex]
[tex]a(m_2-m_1)-0.5Ma=g(m_2+m_1)\\\\a(m_2-m_1-0.5M) = g(m_2+m_1)\\\\a = \dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex]
Hence, the acceleration of the masses is [tex]\dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex].
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A twin-sized air mattress used for camping has dimensions of 100 cm by 194 cm by 14 cm when blown up. The weight of the mattress is 4 kg. How heavy a person (in N) could the air mattress hold if it is placed in freshwater
Answer:
[tex]2625.156\ \text{N}[/tex]
Explanation:
Dimensions of mattress 100 cm by 194 cm by 14 cm
[tex]m_m[/tex] = Mass of mattress = 4 kg
[tex]\rho[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Volume of mattress
[tex]V=100\times 194\times 14=271600\ \text{cm}^3=0.2716\ \text{m}^3[/tex]
Weight of water displaced is equal to the buoyant force
Mass of water
[tex]m=\rho V\\\Rightarrow m=1000\times 0.2716\\\Rightarrow m=271.6\ \text{kg}[/tex]
Mass of person would be
[tex]m_p=m-m_m=271.6-4\\\Rightarrow m_p=267.6\ \text{kg}[/tex]
Weight of the person would be
[tex]w=m_pg\\\Rightarrow w=267.6\times 9.81\\\Rightarrow w=2625.156\ \text{N}[/tex]
The air mattress could hold a person that weighs up to [tex]2625.156\ \text{N}[/tex].
When the disks collide and stick together, their temperature rises. Calculate the increase in internal energy of the disks, assuming that the process is so fast that there is insufficient time for there to be much transfer of energy to the ice due to a temperature difference. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks.)
Answer:
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex]
Explanation:
This is an interesting problem, no data is given, so the result is a general expression.
Suppose that the disks are initially rotating with angular velocity w₁ and w₂, as well as that they have radii r₁ and r₂ and masses m₁ and m₂
we start the problem finding odl final angular velocity of the discs together, for this we define a system formed by the two discs, in this case the torques during the collision are internal and the angular momentum is conserved
initial instant. Just before the crash
L₀ = L₁ + L₂
with
L₁ = I₁ w₁
the moment of inertia of a disc with an axis passing through its center is
I₁ = ½ m₁ r₁²
we substitute
I₀ = ½ m₁ r₁² w₁ + ½ m₂ r₂² w₂
final instant. Right after the crash
L_f = I w
in angular momentum it is a scalar quantity, so it is additive
I = I₁ + I₂
angular momentum is conserved
L₀ = L_f
I₁ w₁ + I₂ w₂ = I w
w = [tex]\frac{ I_1 w_1 + I_2 w_2 }{I}[/tex] (1)
We already have the angular velocities of the system, let's find the kinetic energy of it
initial
K₀ = K₁ + K₂ = ½ I₁ w₁² + ½ I₂ w₂²
final
K_f = K = ½ I w²
the variation of the kinetic energy is the loss in the increase of the temperature of the system, they indicate us that we neglect the other possible losses
ΔK = K_f -K₀
ΔK = ½ I w² - (½ I₁ w₁² + ½ I₂ w₂²) (2)
In this chaos we know all the values for which the numerical value of ΔK can be calculated, the symbolic substitution gives expressions with complicated
Now if all this variation of energy turns into heat
Q = ΔK
m_{total} c_e ΔT = ΔK
where the specific heat of the bear discs must be known, suppose they are of the same material
ΔT = [tex]\frac{\Delta K}{(m_1+m_2) c_e }[/tex] (3)
to make a special case, we suppose some data
the discs have the same mass and radius, disc 2 is initially at rest and the discs are made of bronze that has c_e = 380 J / kg ºC
we look for the angular velocity
I₁ = I₂ = I₀
I = 2 I₀
we substitute in 1
w = [tex]\frac{I_o w_1 + I_o 0 }{2I_o}[/tex] I₀ w₁ + I₀ 0 / 2Io
w = w₁ /2
we look for the variation of the kinetic energy with 2
ΔK = ½ (2I₀) (w₁ /2)² - (½ I₀ w₁² + ½ I₀ 0)
ΔK = ¼ I₀ w₁² -½ I₀ w₁²
ΔK = - ¼ I₀ w₁²
the negative sign indicates that the kinetic energy decreases
We look for the change in Temperature with the expression 3
ΔT = [tex]\frac{ \Delta K}{(m_1 +m_2) c_e}[/tex]ΔK / (m1 + m2) ce
ΔT = [tex]\frac{ \frac{1}{4} I_o w_1^2 }{ 2m c_e}[/tex]
ΔT = [tex]\frac{1}{8} \frac{ (\frac{1}{2} m r_1^2 ) w_1^2 }{ m c_e}[/tex]
ΔT = [tex]\frac{1}{16} r_1^2 w_1^2 / c_e[/tex]
in this expression all the terms are contained
The increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
What is internal energy?The energy contained within a thermodynamic system is known as its internal energy. It's the amount of energy required to build or prepare a system in any given internal state.
The given data in the problem is;
[tex]\rm \omega_1[/tex] is the angular velocity of disk 1
[tex]\rm \omega_2[/tex] is the angular velocity of disk 2
r₁ is the radius of disk 1
r₂ is the radius of disk 2
m₁ is the mass of disk 1
m₂ is the mass of disk 2
Momentum before the collision;
[tex]\rm L_1 = I_1 \omega_1[/tex]
The moment of inertia of disc 1
[tex]\rm i_1 = \frac{1}{2} m_1r_1^2[/tex]
The momentum gets conserved;
[tex]\rm L_0 = L_f \\\\ I_1 \omega_1 + I_2\omega_2 = I \omega \\\\ \rm \omega= \frac{I_1 \omega_1 + I_2\omega_2}{I}[/tex]
The change in the kinetic energy is;
[tex]\traingle KE= K_f - K_0 \\\\ \traingle KE= \frac{1}{2} I \omega^2-(\frac{1}{2} I_1\omega_1^2 + (\frac{1}{2} I_2\omega_2^2 )[/tex]
The change in the energy gets converted into heat;
[tex]\rm Q= \triangle k \\\\\ m_{total } c_e dt = \triangle k[/tex]
The change in the temperature is
[tex]\triangle T= \frac{\triangle k }{(m_1+m_2)c_e}[/tex]
The internal energy change is found by;
[tex]\rm \triangle E = mc_v dt[/tex]
[tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex]
Hence the increase in internal energy of the disks will be [tex]\rm \triangle E= mc\frac{\triangle k }{(m_1+m_2)c_e}[/tex].
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A person drops a pebble of mass m1 from a height h, and it hits the floor with kinetic energy KE. The person drops another pebble of mass m2 from a height of 4h, and it hits the floor with the same kinetic energy KE. How do the masses of the pebbles compare
Hello,
QUESTION)✔ We have: KE = PE (potential energy)
PE = m x g x h
The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2
PE1 = PE2 ⇔ PE1/PE2 = 1
[tex]\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\ \frac{m_1}{m_2\times 4} = 1 \\ \\ \frac{m_1}{m_2} = 4[/tex]
The mass m1 is therefore 4 times greater than that of the stone of mass m2.
BRAINLEST FOR CORRECT ANSWER PLEASE
Which has more momentum: a 3 kg sledgehammer swung at 1.5 m/s OR a 4 kg sledgehammer swung at 0.9 m/s? SHOW YOUR WORK
Answer:
Sledgehammer A has more momentum
Explanation:
Given:
Mass of Sledgehammer A = 3 Kg
Swing speed = 1.5 m/s
Mass of Sledgehammer B = 4 Kg
Swing speed = 0.9 m/s
Find:
More momentum
Computation:
Momentum = mv
Momentum sledgehammer A = 3 x 1.5
Momentum sledgehammer A = 4.5 kg⋅m/s
Momentum sledgehammer B = 4 x 0.9
Momentum sledgehammer B = 3.6 kg⋅m/s
Sledgehammer A has more momentum
Can you please help me with this worksheet I will give BRAINLISET to whoever answers first?
An ideal gas undergoes an adiabatic expansion, a process in which no heat flows into or out of the gas. As a result, (a) the temperature of the gas remains constant and the pressure decreases. (b) both the temperature and pressure of the gas decrease. (c) the temperature of the gas decreases and the pressure increases. (d) both the temperature and volume of the gas increase. (e) both the temperature and pressure of the gas increase. Group of answer choices a b c d e
Answer:
(b) both the temperature and pressure of the gas decrease.
Explanation:
An ideal gas undergoes an adiabatic expansion, a process in which no heat flows into or out of the gas. As a result, both the temperature and pressure of the gas decrease.
Gay Lussac states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Gay Lussac's law is given by;
[tex] PT = K[/tex]
Also, according to the first law of thermodynamics which states that energy cannot be created or destroyed but can only be transformed from one form to another. Thus, the ideal gas does work on the environment with respect to the volume and temperature.
An airplane of mass 13300 kg is flying in a straight line at a constant altitude and with a speed of 560.0 km/hr. The force that keeps the airplane in the air is provided entirely by the aerodynamic lift generated by the wings. The direction of this force is perpendicular to the wing surface. Calculate the magnitude of the lift generated by the wings of this airplane.
Answer:
The magnitude of the lift generated by the wings of the airplane is 130,340 N.
Explanation:
Given;
mass of the airplane, m = 13,300 kg
speed of the airplane, v = 560 km/h = 155.56 m/s
The magnitude of the lift generated by the wings of the airplane is calculated as;
[tex]F_l = mg\\\\where;\\\\F_l \ is \ the \ magnitude \ of \ the \ lift \ generated\\g \ is \ acceleration \ due \ to \ gravity = 9.8 \ m/s^2\\\\F_l = 13,300 \times \ 9.8\\\\F_l = 130,340 \ N[/tex]
Therefore, the magnitude of the lift generated by the wings of the airplane is 130,340 N.
How much work is done in pushing an object 7.0 m across a floor with a force of 50 N and then
pushing it back to its original position? How much power is used if this work is done in 20 sec?
Answer:
35/2 J/s
Explanation:
Just use the 2 formulas
Work done = Force * distance moved
Power = Work done/time
WD = 7 * 50 = 350
Power = 350 / 20
= 35/2 J/s
Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle.
Answer:
The amount of torque is 0.03 N.m.
Explanation:
To find the amount of torque we need to use the following equation:
[tex] \tau = \vec {r} \times \vec{F} = rFsin(\theta) [/tex] (1)
Where:
r: is the radius = 1x10⁻² m
F: is the force = 6 N
θ: is the angle = 30°
By entering the above values into equation (1) we have:
[tex]\tau = 1 \cdot 10^{-2} m*6 N*sin(30) = 0.03 N.m[/tex]
Therefore, the amount of torque is 0.03 N.m.
I hope it helps you!
Which of the following would produce the most power?
b
ОООО
A mass of 10 kilograms lifted 10 meters in 10 seconds
A mass of 5 kilograms lifted 10 meters in 5 seconds
A mass of 10 kilograms lifted 10 meters in 5 seconds
A mass of 5 kilograms lifted 5 meters in 10 seconds
d
Answer:
A mass of 10 kilograms lifted 10 meters in 5 seconds.
Explanation:
Power can be defined as the energy required to do work per unit time.
Mathematically, it is given by the formula;
[tex] Power = \frac {Energy}{time} [/tex]
But Energy = mgh
Substituting into the equation, we have
[tex] Power = \frac {mgh}{time} [/tex]
Given the following data;
Mass = 10kg
Height = 10m
Time = 5 seconds
We know that acceleration due to gravity is equal to 9.8 m/s²
[tex] Power = \frac {10*9.8*10}{5} = 490 Watts [/tex]
Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.
ASAP 20 POINTS!!
The air also contained a small amount of argon
As the temperature of the air decreased from 20C to -190 C the argon changed
Explain the changes in arrangement and movement of the particles of the argon as the temperature of the air decreased
Answer:
See explanation
Explanation:
Let us recall that temperature is a measure of the average kinetic energy of the molecules of a body.The higher the temperature, the higher the kinetic energy of the molecules of the body.
As temperature decreases, the kinetic energy of the molecules of a substance also decreases rapidly and the magnitude of intermolecular interaction between molecules of the substance increases.
Hence, as argon gas is cooled from 20°C to -190°C the kinetic energy of the gas molecules decreases an the magnitude of intermolecular interaction increases hence the gas changes into liquid and subsequently changes into a solid at -190°C.
Which of the following parallel plate diagrams would have the greatest electric fields between them?
The greatest electric field between them is Option B.
An electric field is the physical subject that surrounds electrically charged particles and exerts a force on all other charged debris inside the area, either attracting or repelling them. It also refers back to the bodily field of a machine of charged particles.
Reasoning:-
Electric field = Kq/r²
Since the distance is inversely proportional to the square root of the distance between them. therefore decreasing the distance electric field will be the greatest. Hence Option B.
The electrical field is defined mathematically as a vector discipline that can be related to each point in space, the force in step with unit charge exerted on a fine take a look at the price at rest at that factor. the electric field is generated by the electrical rate or through time-varying magnetic fields. electric fields provide us with the pushing force we need to set off the contemporary float.
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Match the descriptions with the graphs !
Answer:
Graph 1 matches with B, 2 with A, and 3 with C.
Explanation:
Graph 2 shows a car whose distance part of the graph is not going up or down, while the time going up. That means that the car is stopped. Graph 1 shows a straight line, meaning that the car is traveling at a constant speed. Graph 3 is a curved line, meaning the speed of the car is changing somehow, and since the line is becoming more horizontal, the car is getting slower.
A beam of protons is directed in a straight line along the z direction through a region of space in which there are crossed electric and magnetic fields The electric field is 550 V m in the y direction and the protons move at a constant speed of 105 m s 1 What must be the magnitude of the magnetic field such that the beam of protons continues along its straight line trajectory Express your answer using two significant figures
Answer:
B = 5.23 T
Explanation:
Given that,
Electric field, E = 550 V/m
The speed of the proton, v = 105 m/s
We need to find the magnitude of the magnetic field such that the beam of protons continues along its straight-line trajectory.
To move in a straight line, the magnitude of the electric force from the field and the magnetic field must be equal i.e.
[tex]qE=qvB\\\\B=\dfrac{E}{v}\\\\B=\dfrac{550}{105}\\\\B=5.23\ T[/tex]
So, the magnitude of the magnetic field is equal to 5.23 T.
At the end of the previous experiment, aclumsy scientist drops the coil, while still in the magnetic field, and still oriented with its plane perpendicular to the magnetic field, denting it and changing its shape to a semi-circle. The new shape has the same perimeter, but a different area, and it takes 0.036s to deform. What isthe average induced EMF during this mishap
Answer:
hello your question has some missing parts below are the missing parts
A Circular, 10-turn coil has a radius of 10.7 cm and is oriented with its plane perpendicular to a 0.2-T magnetic field.
answer : 1 volt
Explanation:
Determine the Average induced EMF during this mishap
A' = A/2 ( for a semi circle )
where A = [tex]\frac{\pi r^2}{2}[/tex]
To determine the Average induced EMF apply the relation below
| E | = η * [tex]\frac{\beta A}{T}[/tex] ----- ( 1 )
Replace A in equation 1 with A = [tex]\frac{\pi r^2}{2}[/tex]
hence equation becomes : | E | = η * βπr^2 / 2T'
where : T' = 0.0365 , β = 0.2 , η = 10 , r = 0.107
∴| E | = 0.999 ≈ 1volts
A car travels at a constant speed of 25 m/s. Find the power supplied by the engine if it can supply a maximum force of 18,000 N
Answer:
720
Explanation:
The way you change the speed of a wave is to:
a
Change it's medium
b
Change it's energy
c
Transfer it to a new position
d
Apply a force
Answer:
transfer it to a new position
Tom has a mass of 50,000 g and runs up a flight of stairs 4 m high in 12.5 seconds.
Calculate Tom’s power. (g = 10 m/s2)
Answer:
160 watts.
Explanation:
Remark
Power = Work / Time
Work = F * d
Note: Since he is running up stairs he is doing work against gravity.
Givens
m = 50000 g kg / 1000 grmsm = 50000 / 1000 = 50 kgh = 4 mtime = 12.5 secondsg = 10 m/s^2Formula
P = W * d/tW = m*g *d / tSolution
P = 50kg * 10 m/s^2 * 4 m / 12.5 P = 160 watts.