need help pleaseee,question is in the pic
Explanation:
For engine 1,
Energy removed = 239 J
Energy added = 567 J
[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]
For engine 2,
Energy removed = 457 J
Energy added = 789 J
[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]
For engine 3,
Energy removed = 422 J
Energy added = 1038 J
[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]
So, the engine 2 has the highest thermal efficiency.
A 1,760 W toaster, a 1,420 W electric frying pan, and an 85 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.) (a) What current (in A) is drawn by each device
Answer:
Toaster = I = 14.67 A
Frying Pan = 11.83 A
Lamp = 0.71 A
Explanation:
The electric power is given as:
[tex]P = VI\\\\I = \frac{P}{V}[/tex]
where,
I = current
P = Power
V = Voltage = 120 V
FOR TOASTER:
P = 1760 W
Therefore,
[tex]I = \frac{1760\ W}{120\ V}[/tex]
I = 14.67 A
FOR FRYING PAN:
P = 1420 W
Therefore,
[tex]I = \frac{1420\ W}{120\ V}[/tex]
I = 11.83 A
FOR LAMP:
P = 85 W
Therefore,
[tex]I = \frac{85\ W}{120\ V}[/tex]
I = 0.71 A
A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a wall at the bottom of the incline. A light string attached to the block runs over a frictionless pulley to a 60.0-g suspended mass. The suspended mass is given an initial downward speed of 1.60 m/s .
How far does it drop before coming to rest? (Assume the spring is unlimited in how far it can stretch.)
Express your answer using two significant figures.
Answer:
0.5
Explanation:
because the block is attached to the pulley of the string
The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C?
Answer:
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
Explanation:
The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:
[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)
Where:
[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.
[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.
[tex]\Delta P[/tex] - Pressure change, in pascals.
If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:
[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]
[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]
[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]
A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.
A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the instant you make measurements on the glider, it is moving at 0.835 m/sm/s and is 4.00 cmcm from its equilibrium point.
Required:
a. Use energy conservation to find the amplitude of the motion.
b. Use energy conservation to find the maximum speed of the glider.
c. What is the angular frequency of the oscillations?
(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is
W = - (1/2 kx ² - 1/2 k (0.0400 m)²)
(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)
By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so
W = ∆K = 0 - 1/2 m (0.835 m/s)²
Solve for x :
- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²
==> x ≈ 0.0493 m ≈ 4.93 cm
(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is
W = - 1/2 k (0.0400 m)²
If v is the glider's maximum speed, then by the work-energy theorem,
W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²
Solve for v :
- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²
==> v ≈ 1.43 m/s
(c) The angular frequency of the glider's oscillation is
√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz
The amplitude of the motion is 0.049 cm. The maximum speed of the glider is 1.429 m/s. The angular frequency of the oscillation is 29.02 rad/s
From the given information;
the mass of the glider = 190 gForce constant k = 160 N/mthe horizontal speed of the glider [tex]v_x[/tex] = 0.835 m/sthe distance away from the equilibrium = 4.0 cm = 0.04 mUsing energy conservation E, the amplitude of the motion can be calculated by using the formula:
[tex]\mathbf{E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{E = \dfrac{1}{2}(0.19 \ kg )\times (0.835)^2 + \dfrac{1}{2}(160) (0.04)^2}[/tex]
[tex]\mathbf{E =0.194 \ J}[/tex]
Similarly, we know that:
[tex]\mathbf{E = \dfrac{1}{2}kA^2}[/tex]
Making amplitude A the subject, we have:
[tex]\mathbf{A = \sqrt{\dfrac{2E}{k}}}[/tex]
[tex]\mathbf{A = \sqrt{\dfrac{2(0.194)}{160}}}[/tex]
[tex]\mathbf{A =0.049 \ cm}[/tex]
Again, using the energy conservation, the maximum speed of the glider can be calculated by using the formula:
[tex]\mathbf{E =\dfrac{1}{2} mv^2 _{max}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2E}{m}}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2\times 0.194}{0.19}}}[/tex]
[tex]\mathbf{v _{max} = 1.429 \ m/s}[/tex]
The angular frequency of the oscillation can be computed by using the expression:
[tex]\mathbf{\omega = \sqrt{\dfrac{k}{m}}}[/tex]
[tex]\mathbf{\omega = \sqrt{\dfrac{160}{0.19}}}[/tex]
ω = 29.02 rad/s
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water contracts on freezing is it incorrect or conrrect
Answer:
hope it helps
much as you can
Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle θ if the resultant force is directed vertically upward.
Answer:
how to solve this problem ???????
The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.
Resultant of the two forces
The resultant of the two forces is determined by resolving the force into x and y component as shown below;
[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]
where;
F1 = 500 NF2 = 600 NValue of Angle θThe value of Angle θ is determined from equation (1)
-500sinθ + 600sin(30) = 0
500sinθ = 600sin(30)
500sinθ = 300
sinθ = 3/5
θ = 36.87⁰
Resultant of the two forcesThe resultant of the forces is determined using the second equation;
500cosθ + 600cos(30) = R
500 x cos(36.87) + 600 x cos(30) = R
919.6 N = R
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A
Fluids in which the shear stress must reach
certain minimum value(yield stress)
before flow commences are called
Answer:
Plastic
Explanation:
Shear Modulus can be defined as the ratio of shear stress to shear strain with respect to a physical object.
This ultimately implies that, Shear Modulus arises as a result of the application of a shear force on an object or body which eventually leads to its deformation. Thus, this phenomenon is simply used by scientists to measure or determine the rigidity of an object or body.
Fluids in which the shear stress must reach certain minimum value (yield stress) before flow commences are called plastic. Thus, a plastic would only begin to flow when its shear stress attain a certain minimum value (yield stress). The unit of measurement of yield stress is usually mega pascal (MPa).
Four equal-value resistors are in series with a 5 V battery, and 2.23 mA are measured. What isthe value of each resistor
Answer:
560.54 Ω
Explanation:
Applying,
V = IR'............... Equation 1
Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors
make R' the subject of the equation
R' = V/I............ Equation 2
From the question,
Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A
Substitute these values into equation 2
R' = 5/(2.23×10⁻³ )
R' = 2242.15 Ω
Since the fours resistor are connected in series and they are equal,
Therefore the values of each resistor is
R = R'/4
R = 2242.15/4
R = 560.54 Ω
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
The thrust F of a screw propeller is known to depend upon the diameter d, Speed of advance v, fluid density e, revolution per second N, and the coefficient of viscosity M, of the fluid. Find the expression for F, in terms of the quantities
Answer:
[tex]{ \bf{F = { \tt{ \frac{4}{3} \pi {r}^{3}v gM}}}}[/tex]
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
find the weight of a body of mass 200kg on the earth at a latitude 30°.(R=6400 km ,g=9.8m/s²,ω=7.27×10⁻⁵ rad/sec)
Answer:
................ftf6x
Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitude and directions of their resultant
Answer:
19N to the south
Explanation:
F =10N + 5N + 4N
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info
Answer:
The correct option is (E).
Explanation:
Given that,
Mass of object 1, m₁ = 1 kg
Mass of object 2, m₂ = 2 kg
They collides after the collision. We need to find the speed of the two boxes after the collision.
The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.
So, the information is not enough.
A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take the mass of the Earth to be 5.97 x 10^24 kg and its radius 6.38 x10^6 m.
Required:
What is the orbital period of this GPS satellite?
Answer:
[tex]T=66262.4s[/tex]
Explanation:
From the question we are told that:
Altitude [tex]A=2.90 *10^7[/tex]
Mass [tex]m=5.97 * 10^{24} kg[/tex]
Radius [tex]r=6.38 *10^6 m.[/tex]
Generally the equation for Satellite Speed is mathematically given by
[tex]V=(\frac{GM}{d} )^{0.5}[/tex]
[tex]V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}[/tex]
[tex]V=3354.83m/s[/tex]
Therefore
Period T is Given as
[tex]T=\frac{2 \pi *a}{V}[/tex]
[tex]T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}[/tex]
[tex]T=66262.4s[/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.
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Of the following, which have the highest frequency in the electromagnetic
spectrum?
A. Visible light
B. Infrared waves
C. Ultraviolet rays
D. X-rays
A cylindrical container with a cross sectional area of 65.2 cm^2 holds a fluid of density 806 kg/m^3. At the bottom of the container the pressure is 116 kPa.
(a) What is the depth of the fluid?
(b) Find the pressure at the bottom of the container after an additional 2.05 X 10^-3 m^3 of this fluid is added to the container. Assume that no fluid spills out of the container.
An ice skater with a mass of 50 kg is gliding acrossthe ice at a speed of 8 m/s when herfriend comes up from behind and gives her a push,causing her speed to increase to 12m/s. How much work did the friend do on the skater
Answer:
[tex]W=2KJ[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=50kg[/tex]
Initial Velocity [tex]v_1=8m/s[/tex]
Final Velocity [tex]v_2=12m/s[/tex]
Generally the equation for Work-done is mathematically given by
W=\triangle K.E
Therefore
[tex]W=0.5M(v_2^2-v_1^2)[/tex]
[tex]W=0.5*50(12^2-8^2)[/tex]
[tex]W=2KJ[/tex]
(a) What is the maximum frictional force (in N) in the knee joint of a person who supports 45.0 kg of her mass on that knee if the coefficient of static friction is 0.016
Answer:
f = 7.06 N
Explanation:
The maximum frictional force on the knee joint of the person can be given by the following formula:
[tex]f = \mu R = \mu W \\[/tex]
where,
f = maximum frictional force = ?
μ = static friction coefficient = 0.016
W = Weight load on knee = mg
m = mass supported by knee = 45 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = \mu mg\\f = (0.016)(45\ kg)(9.81\ m/s^2)\\[/tex]
f = 7.06 N
What is the length of the x-component of the vector shown below?
у
6
28°
Answer:
Explanation:
6cos28
=5.3 N
b) Two skaters collide and grab on to each other on a frictionless ice. One of them, of mass 80 kg, is moving to the right at 5.0 m/s, while the other of mass 70 kg is moving to the left at 2.0 m/s. What are the magnitude and direction of the two skaters just after they collide
Answer:
The two skaters move with a speed of 1.73 m/s after the collision in the right direction.
Explanation:
Given that,
The mas of skater 1, m₁ = 80 kg
The speed of skater 1, u₁ = 5 m/s (right)
The mass of skater 2, m₂ = 70 kg
The speed of skater 2, u₂ = -2 m/s (left)
Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
Put all the values,
[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]
So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.
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:
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Click Stop Using the slider set the following: coeff of restitution to 1.00 A velocity (m/s) to 6.0 A mass (kg) to 6.0 B velocity (m/s) to 0.0 Calculate what range can the mass of B be to cause mass A to bounce off after the collision. Calculate what range can the mass of B be to cause mass A to continue forward after the collision. Check your calculations with the simulation. What are the ranges of B mass (kg)
Answer:
[tex]M_b=6kg[/tex]
Explanation:
From the question we are told that:
Coefficient of restitution [tex]\mu=1.00[/tex]
Mass A [tex]M_a=6kg[/tex]
Initial Velocity of A [tex]U_a=6m/s[/tex]
Initial Velocity of B [tex]U_b=0m/s[/tex]
Generally the equation for Coefficient of restitution is mathematically given by
[tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]
[tex]1=\frac{v_B}{6}[/tex]
[tex]V_b=6*1[/tex]
[tex]V_b=6m/s[/tex]
Generally the equation for conservation of linear momentum is mathematically given by
[tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]
[tex]6*6+=M_b*6[/tex]
[tex]M_b=6kg[/tex]
Drag the titles to the correct boxes to complete the pairs.
The north pole of magnet A will __?____ the south pole of magnet B
Answer:
A will attract
B will repare
1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
How do you find the product of gamma decay?
Answer:
The mass and atomic numbers don't change
Explanation:
An excited atom relaxes to the ground state emitting a photon...called a gamma ray.
The answer is that the mass and atomic numbers don't change.
In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.
To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.
During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.
The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).
For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.
It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).
Thus, the product nucleus remains unchanged in terms of atomic number and mass number.
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A horse gallops a distance of 10 kilometers in a time of 30 minutes its average speed is?
Answer:
20 km/hr
Explanation:
Distance = 10km
Time = 30 minutes = 1/2 hour
Average Speed = Total distance / Total Time Taken
= 10 ÷ 1/2
= 10 x 2
= 20 km/hr
Average speed = (distance covered) / (time to cover the distance)
Average speed = (10 km) / (30 minutes)
Average speed = 1/3 km/min
Most people would probably want to see it in a more convenient, more familiar unit, such as km/hour or m/second.
(10 km / 30 min) x (60 min / hour) = (10 x 60 / 30) (km-min / min-hour)
Average speed = 20 km/hour
AvgSpd = (10 km / 30 min) x (1,000 m / km) x (min / 60 sec)
AvgSpd = (10x1,000 / 30x60) (km-m-min / min-km-sec)
Averge Speed = 5.56 m/s
