Answer:
Friction force always acts tangent to the surface at points of contact. Friction force acts opposite to the direction of motion.
Explanation:
A 24 cm radius aluminum ball is immersed in water. Calculate the thrust you suffer and the force. Knowing that the density of aluminum is 2698.4 kg / m3
Answer:
W =1562.53 N
Explanation:
It is given that,
Radius of the aluminium ball, r = 24 cm = 0.24 m
The density of Aluminium, [tex]d=2698.4\ kg/m^3[/tex]
We need to find the thrust and the force. The mass of the liquid displaced is given by :
[tex]m=dV[/tex]
V is volume
Weight of the displaced liquid
W = mg
[tex]W=dVg[/tex]
So,
[tex]W=dg\times \dfrac{4}{3}\pi r^3\\\\W=2698.4\times 10\times \dfrac{4}{3}\times \pi \times (0.24)^3\\\\W=1562.53\ N[/tex]
So, the thrust and the force is 1562.53 N.
A ray of light is projected into a glass tube that is surrounded by air. The glass has an index of refraction of 1.50 and air has an index of refraction of 1.00. At what minimum angle will light in the glass tube be totally reflected at the glass/air interface?
Answer:
θ = 41.8º
Explanation:
This is an internal total reflection exercise, the equation that describes this process is
sin θ = n₂ / n₁
where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂
θ = sin⁻¹ n₂ / n₁
let's calculate
θ = sin⁻¹ (1.00 / 1.50)
θ = 41.8º
a) A conductor carrying a current I = 12.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A magnetic force per unit length of 0.110 N/m acts on the conductor in the negative y direction. Determine the magnitude of the magnetic field in the region through which the current passes.
Answer:
8.8 mT
Explanation:
Current through the conductor = 12.5 A
Magnetic force per unit length on the wire = 0.110 N/m
Recall that the magnetic force per unit length on a current carrying conductor is in a uniform magnetic field is
[tex]\frac{F}{l}= IBsin\alpha[/tex]
where B is the magnetic field magnitude
[tex]I[/tex] is the current in the conductor
α is the angle the conductor makes with the magnetic field ( since it is perpendicular in this case, α is 90°)
imputing values into the equation, we'll have
0.11 = 12.5 x B x sin 90°
but sin 90° = 1, therefore, we have
0.11 = 12.5B
B = 0.11/12 = 8.8 x 10^-3 T
or rather = 8.8 mT
The following passage has not been edited. There is an error in each line. Write the
incorrect word and the correction in your answer sheet against the correct question
number. The first one has been done as an example. ( 1 x 4 = 4 )
Community service sensitize people to Error: sensitize ; Correction: sensitizes
other‟s needs and supports inclusive (a) Error: _______ ; Correction: ______
development to the underprivileged (b) Error: _______ ; Correction: ______
sections with society. Courses about social (c) Error: ______ ; Correction: _______
work prepares frontline workers to (d) Error: _______ ; Correction: ______
Answer:
(a) Error: Other's ; Correction: Others'
(b) Error: to ; Correction: for
(c) Error: with ; Correction: of
(d) Error: prepares ; Correction: prepare
Explanation:
a) The error is in the word "other's" as the position of apostrophe is wrong, so the correct word will be "others'", it shows plural nouns.
b) The error is in the word "to", so the correct word will be "for" as for is use to talk about a purpose.
c) The error is in the word "with" and the correct word will be "of" as of indicates relationships between other words including things that made of other things.
d) The error is in the word "preapres" and the correct word will be "prepare".
To a stationary observer, a bus moves south with a speed of 12 m/s. A man
inside walks toward the back of the bus with a speed of 0.5 m/s relative to
the bus. What is the velocity of the man according to a stationary observer?
A. 11 m/s south
B. 12.5 m/s south
C. 11.5 m/s south
D. 0.5 m/s south
ANSWER
C 11.5 m/s
EXPLANATION
Answer:
11.5m/s south
Explanation:
Online classes
Your car's 32.5 W headlight and 2.00 kW starter are ordinarily connected in parallel in a 12.0 V system. What power (in W) would one headlight and the starter consume if connected in series to a 12.0 V battery
Answer:
Explanation:
the resistance of a electrical device
R = V² / P where V is volt and P is power .
The devices are in parallel so same volt will apply on them
So R₁ = 12² / 32.5 = 4.431 ohm
R₂ = 12² / 2 x 10³ = .072 ohm
when they are in series
Common Current in them = 12 / 4.431 + .072
= 2.6649 A
power consumed by first device when they are in series
= current² x resistance
= 2.6649² x 4.431 = 31.46 W
power consumed by other
= 2.6649² x .072 = .511 W
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 7 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 17 m2 and 0.20 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Answer:
88 %
Explanation:
The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by
P = KAΔT/d
The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m
The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m
The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂
Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT
So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂
Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is
P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)
= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)
= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)
= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]
= 1/(0.003927/0.0288 + 1)
= 1/(0.1364 + 1)
= 1/1.1364
= 0.88.
The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %
The percentage of heat lost by window of the total heat is 88 %
PLEASE HELP ASAP. IT'S URGENT
Answer:
Q1 acceleration = 16m/s²
Q2 Net force = 9N North
Explanation:
Q1 Using the formula F=ma
Q2 R = F1 + F2
The angle of incidence of a ray of light striking an equilateral triangular prisms ABC of refracting angle 60o is 40o. Calculate:
(i)the angle of refraction at the first face
(ii)the angle of emergence
Answer:
1: the refracted angle in the first face is equal to the incident angle that is 60degrees
2. Emergence Angle is 42degrees
Explanation:
Pls see attached file
Two kilograms of nitrogen (N2) at 25°C is contained in a 0.62 m3 rigid tank. This tank is connected by a valve to a 0.16 m3 rigid tank containing 0.8 kg of oxygen (O2) at 127°C. The valve is opened, and the gases are allowed to mix, achieving an equilibrium state at 87°C.
initial pressures of N2 is 5.7293 bar and O2 is 5.2 bar.
the final pressure is 6.44 bar.
the magnitude of the heat transfer for the process is 162.8 kJ, and the direction of energy flow is going in.
What is the entropy change for the mixing process, in kJ/K?
Answer:
Explanation:
For entropy change the formula is
ΔS = ΔQ / T
ΔQ = Δ H
ΔS = Δ H / T
Given
Δ H = + 162.8 kJ
We can take equilibrium temperature as average temperature of the whole process
So, T = 273 + 87 = 360 K
ΔS = Δ H / T
= 162.8 kJ / 360
= + 0.508 kJ / K .
When the magnitude of the heat transfer for the process is 162.8 kJ, Then the entropy change for the mixing process, in kJ/K is = + 0.508 kJ / K
What is Entropy change?
For The entropy change, the formula is
Then ΔS = ΔQ / T
After that ΔQ = Δ H
Then ΔS = Δ H / T
Given as per question are:
Then Δ H = + 162.8 kJ
Now We can take equilibrium temperature as average temperature of the whole process are:
So, T is = 273 + 87 = 360 K
Then ΔS = Δ H / T
After that = 162.8 kJ / 360
Therefore, = + 0.508 kJ / K.
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Let us treat a helicopter rotor blade as a long thin
rod, as shown in Fig. 8–49. (a) If each of the three rotor
helicopter blades is 3.75 m long and has a mass of 135 kg,
calculate the moment of inertia of the three rotor blades
about the axis of rotation. (b) How much torque must the
motor apply to bring the blades from rest up to a speed
of 6.0 rev/s in 8.0 s?
Rotor
Answer:
(a) 1900 kg m²
(b) 8950 Nm
Explanation:
(a) The moment of inertia of a rod about its end is I = ⅓mL².
For 3 rods of mass m = 135 kg and length L = 3.75 m, the total moment of inertia is:
I = 3 (⅓ (135 kg) (3.75 m)²)
I = 1900 kg m²
(b) Net torque = moment of inertia × angular acceleration
∑τ = Iα
First, find the angular acceleration.
ω₀ = 0 rad/s
ω = 6.0 rev/s (2π rad/rev) = 37.7 rad/s
t = 8.0 s
α = (37.7 rad/s − 0 rad/s) / 8.0 rad/s = 4.71 rad/s²
∑τ = Iα
∑τ = (1900 kg m²) (4.71 rad/s²)
∑τ = 8950 kg m² / s²
∑τ = 8950 Nm
When we double the distance between a source of light and the
surface on which it falls, the amount of illumination on the surface
decreases to
(what fraction) of the
original illumination.
The amount of lighting on a surface drops to (1/4) of the initial illumination when the distance between a light source and the surface it falls on is doubled.
What is illumination?The amount of light or luminous flux that falls on a surface is known as illumination. It is expressed as lux or lumens per square meter.
The square of the distance has an inverse relationship with the light intensity;
[tex]\rm I = \frac{1}{r^2}[/tex]
Where,
I is the light intensity
r is the distance
Let r is the distance and I is the sound intensity for case 2;
r' = 2r
[tex]\rm I' = \frac{1}{(2r)^2} \\\\ I' = \frac{1}{4r^2} \\\\ I' = \frac{I}{4}[/tex]
When we double the distance between a source of light and the surface on which it falls, the amount of illumination on the surface decreases to(1/2) of the original illumination.
Hence the value of the fraction is 1/4.
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What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10 24 kg and radius of the earth is 6.4 × 10 6 m.)
Answer:
Explanation:
just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)
b. i. In a hydraulic press, a force of 200N is applied to master piston of area 25cm. If
the press is designed to produce a force of 5000N, determine the area of the slave
piston.
(4marks)
Answer:
625 cm²
Explanation:
The pressure is the same on both pistons.
F/A = F/A
200 N / 25 cm² = 5000 N / A
A = 625 cm²
a 6 letter word a way of explaining an object or event using a set of facts
Explanation:
A theory is a way of explaining an object or event using a set of facts.
Two trains run in the opposite direction with speeds of v1 = 15 m / s and v2 = 20 m / s. A passenger on the first train (the one on v1) notes that train 2 takes 6 s to pass on its side. What is the length of the second train? (The passenger is supposed to be immobile looking through the window)
Answer:
210 m
Explanation:
The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.
It takes 6 seconds for the train to pass, so the length of the train is:
(35 m/s) (6 s) = 210 m
A square loop 24.0 cm on a side has a resistance of 6.10Ω. It is initially in a 0.665-T magnetic field, with its plane perpendicular to magnetic field B but is removed from the field in 40.0ms. Calculate the electric energy dissipated in this process.
Answer:
[tex]E=6.01\times 10^{-3}\ J[/tex]
Explanation:
It is given that,
Side of a square loop, l = 24 cm = 0.24 m
Resistance of loop, R = 6.1 ohms
Initial magnetic field is 0.665 T and final magnetic field is 0 as the field is removed in 40 ms
We need to find the electrical energy dissipated in this process.
Due to change in magnetic field, the loop will induce a voltage. The induced voltage is given by :
[tex]V=-\dfrac{dB}{dt}\\\\V=\dfrac{BA}{t}[/tex]
If I is induced current then,
[tex]V=IR[/tex]
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{BA}{tR}[/tex]
Power is given by voltage times current. So,
[tex]P=\dfrac{(BA)^2}{(t^2R)}[/tex]
Now, energy is given by the product of power and time. So,
[tex]E=\dfrac{(BA)^2}{(t^2R)}\times t\\\\E=\dfrac{(BA)^2}{(tR)}[/tex]
Now putting all the values in above formula. So,
[tex]E=\dfrac{(0.665\times (0.24)^2)^2}{(40\times 10^{-3}\times 6.1)}\\\\E=6.01\times 10^{-3}\ J[/tex]
So, the electrical energy of [tex]6.01\times 10^{-3}\ J[/tex] is dissipated in this process.
The electrical energy dissipated throughout this process will be "6.01 × 10⁻³ J".
Magnetic fieldAccording to the question,
Square loop's side, l = 24 cm or,
= 0.24 m
Loop's resistance, R = 6.1 ohms
Initial magnetic field = 0.665 T
Final magnetic field = 0
We know the relation,
→ V = - [tex]\frac{dB}{dt}[/tex]
= [tex]\frac{BA}{t}[/tex]
Also we know,
Current, V = IR
I = [tex]\frac{V}{R}[/tex]
= [tex]\frac{BA}{tR}[/tex]
Now, Energy, E = [tex]\frac{(BA)^2}{t^2R}[/tex] × t or,
= [tex]\frac{(BA)^2}{tR}[/tex]
By substituting the values,
= [tex]\frac{(0.665\times (0.24)^2)^2}{40\times 10^{-3}\times 6.1}[/tex]
= 6.01 × 10⁻³ J
Thus the response above is correct.
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3.) [15 points] A physics teacher is on the west side of a small lake and wants to swim across and up at a point directly across from his starting point. He notices that there is a current in the lake and
that a leaf floating by him travels 4.2m [S] In 5.0s. He is able to swim 1.9 m/s in calm water,
(a) What direction will he have to swim in order to arrive at a point directly across from his position?
Answer:
The teacher should swim in a direction 29.24° North of East
Explanation:
Given that the there is a water current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;
The speed of the water current is given by the speed of the floating leaf traveling with the water current
Distance traveled by the leaf = 4.2 m South
Time of travel of the leaf = 5.0 s
Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current
Swimming peed of the teacher, v = 1.9 m/s
To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.
Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim
Therefore;
1.9 × sin(θ) = 0.84
sin(θ) = 0.84/1.9 = 0.44
θ = 26.24°
That is the teacher should swim in a direction 29.24° North of East.
To cross the lake the teacher has to swim in a direction 29.24° North of the East
Finding the direction of speed required:
The speed of the water current can be derived from the speed of the floating leaf :
The distance traveled by the leaf L = 4.2 m South
Time taken T = 5s
So, the speed of the leaf is:
u = 4.2/5
u = 0.84 m/s South
So, the speed of the current is 0.84 m/s South
Now, it is given that the speed of the teacher is, v = 1.9 m/s East
To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.
Let, the speed of the teacher makes an angle of θ from the EAST.
So, the North component is given by:
v(north) = vsinθ
it must be equal to the speed of the current:
vsinθ = u
1.9 × sinθ = 0.84
sinθ = 0.84/1.9
sinθ = 0.44
θ = 26.24°
The teacher should swim in a direction 29.24° North of East.
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calculate the energy dissipated by an electric iron which draws a current of 5A from a240v power supply for 25minutes
Explanation:
Given,
I = 5 A
V = 240 V
T = 25 mins = 1500 sec
Now,
Energy dissipated = IVT= 5×240×1500 = 1800000 J
A very thin film of soap, of thickness 170 nm, in between air seems dark. On the other hand, when placed on top of glass some visible light is seen to shine from the film. How can this happen and what is the smallest visible light that creates constructive interference when we place the film on top of glass
Answer:
λ₀ = 2 d n
Explanation:
A soap film is a layer where the lus is reflected on the surface and on the inside of the film, these two reflected rays can interfere with each other either constructively or destructively.
Let's analyze the general conditions of this interference,
* When the ray of light reaches the surface of the film it is reflected, as the index of refraction of the air is less than the index of the film, the reflected ray has a phase change of 180º
* When the ray penetrates the film, its wavelength changes due to the refractive index of the film.
λ = λ₀ / n
where lick is the wavelength in the vacuum or air and n index of refraction of the film, in general this interference is observed perpendicular to the film, so the sine veils 1. the expression for constructive interference taking in what previous remains
2d = (m + ½) λ
the expression for destructive interference remains
2d = m λ
2d = m λ₀ / n
When the film is placed on a glass plate whose index of refraction is greater than the index of refraction of the film, in the reflection in the lower part of the film another phase difference of 180º is created, for which we have a difference of total phase of 180 +180 = 360º, which is equivalent to no phase difference, therefore the two previous equations are interchanged.
Therefore where we had destructive interference now a cosntructive interference happens we can see the reflected light.
Find us the wavelength that this constructive interference creates
2d n = m λ₀
λ₀ = 2 d n / m
To find the minimum wavelength, suppose we observe the first interference pattern m = 1
λ₀ = 2 d n
where d is the thickness of the film and n the index of refraction of the same
Iron man wears an awesome ironsuit.He is flying over high current carrying wire. Will he be affected?
Answer:
According to super hero logic , nothing will happen to him.
But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.
He may / may not be affected but his suit will be damaged for sure as it is made of metal.
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The graph shows a wave that oscillates with a frequency of 60 Hz. Based on the information given in the diagram, what is the speed of the wave?
Answer:
900 cm/s or 9 m/s.
Explanation:
Data obtained from the question include the following:
Length (L) = 30 cm
frequency (f) = 60 Hz
Velocity (v) =.?
Next, we shall determine the wavelength (λ).
This is illustrated below:
Since the wave have 4 node, the wavelength of the wave will be:
λ = 2L/4
Length (L) = 30 cm
wavelength (λ) =.?
λ = 2L/4
λ = 2×30/4
λ = 60/4
λ = 15 cm
Therefore, the wavelength (λ) is 15 cm
Now, we can obtain the speed of the wave as follow:
wavelength (λ) = 15 cm
frequency (f) = 60 Hz
Velocity (v) =.?
v = λf
v = 15 × 60
v = 900 cm/s
Thus, converting 900 cm/s to m/s
We have:
100 cm/s = 1 m/s
900 cm/s = 900/100 = 9 m/s
Therefore, the speed of the wave is 900 cm/s or 9 m/s.
On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the average speed for the entire journey is 40km/h?
Answer:
v = 46.67 km/h
Explanation:
We will use the following formula throughout this numerical:
s = vt
where,
s = distance covered
v = speed
t = time taken
FOR FIRST 30 km:
s = 30 km
v = 30 km/h
t = t₃₀ = ?
Therefore,
30 km = (30 km/h)(t₃₀)
t₃₀ = (30 km)/(30 km/h)
t₃₀ = 1 h
FOR TOTAL 100 km:
s = 100 km
v = 40 km/h (Average Speed)
t = total time = ?
Therefore,
100 km = (40 km/h)(t)
t = (100 km)/(40 km/h)
t = 2.5 h
FOR LAST 70 km:
s = 70 km
t₇₀ = t - t₃₀ = 2.5 h - 1 h = 1.5 h
v = v₇₀ = ?
Therefore,
70 km = v(1.5 h)
v = 70 km/1.5 h
v = 46.67 km/h
Two long, straight wires are parallel and 10 cm apart. One carries a current of 2.0 A, the other a current of 5.0 A. If the two currents flow in opposite directions, what is the magnitude and direction of the force per unit length of one wire on the other
Answer:
The magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.
Explanation:
Given;
distance between the two parallel wires, r = 10 cm = 0.1 m
current in the first wire, I₁ = 2A
current in the second wire, I₂ = 5 A
The force per unit length on each wire can be calculated as;
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r} \\\\\frac{F}{L} = \frac{4\pi*10^{-7}*2*5}{2\pi *0.1} \\\\\frac{F}{L} = 2 *10^{-5} \ N[/tex]/ m
The direction of the force between the two wires is attractive since the current in the two wires are in opposite direction.
Therefore, the magnitude and direction of the force per unit length of one wire on the other is 2 x 10⁻⁵ N/m, attractive force.
A load of 500N is carried by 200N effort in a simple machine having load distance 3m Calculate effort distance.
Answer:
2.5 mExplanation:
Load ( L ) = 500 N
Effort ( E ) = 200 N
Load distance ( LD ) = 3 m
Effort distance ( ED ) = ?
Now, Let's find the Effort distance ( ED )
We know that,
Output work = Input work
i.e L × LD = E × ED
plug the values
[tex]500 \times 3 = 200 \times ED[/tex]
multiply the numbers
[tex]1500 = 200 \times ED[/tex]
Swipe the sides of the equation
[tex]200 \: ED \: = 500[/tex]
Divide both sides of the equation by 200
[tex] \frac{200 \: ED}{200} = \frac{500}{200} [/tex]
Calculate
[tex]ED\: = 2.5 \: m[/tex]
Hope this helps..
best regards!!
PLEASE HELP ME ASAP. IT'S VERY IMPORTANT
Answer:
1) a. 52.41 m/s
b. The skier will be going 15.35 m/s slower
2) 103.68 m
3) 35,127 J
4) a. 88.825 kJ
(b) 16.36 %
5) 3,071.12 J
Explanation:
1) a. The given height of the hill, h = 140.0 m
The mass of the skier at the top of the hill, m = 85.0 kg
The acceleration due to gravity, g = 9.81 m/s²
The initial potential energy, P.E of the skier = m×g×h = 85.0×140.0×9.81 = 116739 J
From the principle of conservation of energy, we have;
The potential energy, P.E. lost = The gain in kinetic energy, K.E.
m×g×h = 1/2×m×v²
116739 J = 1/2×85.0×v²
v² = 116739/(1/2*85.0)= 2746.8 m²/s²
v = √(2746.8 m²/s²) = 52.41 m/s
b. From 70 m up, we have;
The initial potential energy, P.E., of the skier is now = 85.0×70×9.81 = 58,369.5 J
The potential energy, P.E. lost = The gain in kinetic energy, K.E.
58,369.5 J = 1/2×85.0×v²
v² = 58,369.5/(1/2*85.0) = 1373.4 m²/s²
v = 37.06 m/s
The skier will be going 52.41 - 37.06 = 15.35 m/s slower
The skier will be going 15.35 m/s slower
2) From the principle of conservation of energy, the amount of work done (energy used) = The (potential) energy gained by the load
The amount of work done by the electric hoist = 356,000 J
The mass of the load = 350.0 kg
The height to which the load is raised = h
The potential energy gained by the load = m×g×h = 350.0×9.81×h
356,000 J = 350.0×9.81×h
h = 356,000/(350.0*9.81) = 103.68 m
The height to which the load is lifted= 103.68 m
3) The initial potential energy of the roller coaster cart = 600*35.0*9.81 = 206010 J
The final potential energy = 600*28.0*9.81= 164808 J
The velocity at point 3 = 4.5 m/s
The kinetic energy at point 3 = 1/2*600*4.5^2 = 6075 J
The total energy at point 3 = 164808 + 6075 = 170,883 J
The energy loss = The initial potential energy at point 1 - Total energy at point 3
The energy loss = 206010 - 170,883 = 35,127 J
The heat energy due to friction that must have been produced between points 1 and 3 = 35,127 J
4) a. The heat energy absorbed = mass × specific heat capacity for water, [tex]C_{water}[/tex] × Temperature change
The mass of the water = 2.5×10² g = 0.25 kg
[tex]C_{water}[/tex] = 4,180 J/(kg·°C)
Initial temperature = 10.0°C
Final temperature = 95°C
The temperature change = 95.0°C - 10.0°C = 85.0°C
The heat energy absorbed = 0.25*4,180* 85 = 88,825 J = 88.825 kJ
(b) The percentage efficiency = (Heat absorbed/(Heat supplied)) × 100
The heat supplied = 543 kJ
The efficiency = (88.825/543) × 100 = 16.36 %
5) The mass of the box = 115 kg
Force acting on the rope = 255 N
The angle of inclination of the force to the horizontal = 24.5°
The distance the box is displaced = 15.0 m to the right
The work done = Force applied × distance moved in the direction of the force
The work done = Force applied × distance moved in the direction of the force
Given that the load moves a distance 15.0 m to the right,we have;
The component of the force acting in the direction of the movement of the load (to the right) is 225 × cos(24.5°) = 204.74 N
The work done = 204.7*15 = 3071.12 J
The amount of work done = 3,071.12 J
A plane drops a package for delivery. The plane is flying horizontally at a speed of 120m/s,and the package travels 255 m horizontally during the drop. We can ignore air resistance.What is the package's vertical displacement during the drop?
Answer:
Package's vertical displacement(s) = 22.12 meter
Explanation:
Given:
Speed of plane = 120 m/s
Total distance = 255 m
Find:
Package's vertical displacement(s)
Computation:
Time taken = Distance / Speed
Time taken = Total distance / Speed of plane
Time taken = 255 / 120
Time taken = 2.125 s
Acceleration due to gravity(g) = 9.8 m/s²
Initial velocity (u) = 0
So,
Package's vertical displacement(s) = ut + (1/2)gt²
Package's vertical displacement(s) = (0)(2.125) + (1/2)(9.8)(2.125)²
Package's vertical displacement(s) = 22.12 meter
Answer: -22.1
Explanation:
I just did the Khan Academy and that was the answer, not the one provided by that one person. :)))
Please answer this question
Explanation:
m = kg. v=m/s. g=m/s^2. h= m
>>1/2mv^2=mgh
>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m
>>1/2mv^2=mgh>> kg*(m/s)^2= kg*m/s^2*m>>kg m^2/s^2=kg m^2/s^2 the fraction 1/2 won't be able to make any changes to to the dimensional expression of energy i.e half of energy is still energy therefore you can neglect the number .
>>kg m^2/s^2=kg m^2/s^2
>>J= J
Select the correct answer.
According to the Universal Law of Gravitation, every object attracts every other object in the universe. Why can’t you feel the force of attraction between you and Mars?
A.
There is no force of attraction between you and Mars.
B.
Your mass is too low.
C.
Mars is a larger planet than Earth.
D.
Mars is a long distance away.
Answer:
D. Mars is a long distance away
Allocate birr 5000 among the three workers in the ratio 1/3 :1/6 and 5/12.
Answer:
1666.7 ETB (birr)
833.3 ETB (birr)
2083.3 ETB (birr)
Explanation:
The first worker
5000*1/3=1666.7
The second worker
5000*1/6=833.3
The third worker
5000*5/12=2083.3
Hope this helps :) ❤❤❤