Answer:
a. 2 m
b. 0.15 m
Explanation:
(a) By what deck does the ship sink in fresh.
water, when it loads a cargo of 4000 tonnes
We know that the upward force , U on the ship equal the weight of fresh water displaced, W = ρVg where ρ = density of fresh water = 1000 kg/m³, V = volume of water displaced and g = acceleration due to gravity.
So, U = W = ρVg
This upward force must equal the weight of the ship, W' = mg where m = mass of ship = 4000 tonnes = 4000 × 1000 kg = 4 × 10⁶ kg
So. U = W'
ρVg = mg
V = m/ρ = 4 × 10⁶ kg/10³ kg/m³ = 4 × 10³ m³
Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in water h.
So, V = Ah
Thus h = V/A = 4 × 10³ m³/2000 m³ = 2 m
So, the ship sinks to a depth of 2 m in fresh water.
(b) if the ship + Cargo has a displacement tonnage
of 12300 tonnes; by what amount will the ship
rise in the water when it sails from fresh water
into Seawater (density of Sea water - 1025kgm⁻³
We know that the upward force , U' on the ship in fresh water equals the weight of fresh water displaced, W" = ρV'g where ρ = density of fresh water = 1000 kg/m³, V' = volume of water displaced and g = acceleration due to gravity.
So, U = W" = ρV'g
This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg
So. U' = W₁
ρV'g = m₁g
V' = m₁/ρ = 1.23 × 10⁷ kg/10³ kg/m³ = 1.23 × 10⁴ m³
Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in fresh water h'.
So, V' = Ah'
Thus h = V'/A = 1.23 × 10⁴ m³/2000 m³ = 6.15 m
So, the ship sinks to a depth of 0.6 m in fresh water.
Also,
We know that the upward force , U" on the ship in sea water equals the weight of sea water displaced, W₂ = ρ'V"g where ρ' = density of sea water = 1025 kg/m³, V"= volume of water displaced and g = acceleration due to gravity.
So, U" = W₂ = ρ'V"g
This upward force must equal the weight of the ship + cargo, W₁ = m₁g where m₁ = mass of ship + cargo = 12300 tonnes = 12300 × 1000 kg = 1.23 × 10⁷ kg
So. U" = W₁
ρ'V"g = m₁g
V" = m₁/ρ' = 1.23 × 10⁷ kg/1.025 × 10³ kg/m³ = 1.2 × 10⁴ m³
Now the volume of water displaced equals the volume occupied by the ship floating ship of cross-sectional area, A = 2000 m³ and depth in sea water h".
So, V" = Ah"
Thus h" = V'/A = 1.2 × 10⁴ m³/2000 m³ = 6 m
So, the ship sinks to a depth of 6 m in fresh water.
So, the rise in height from fresh water to sea water is Δh = h' - h" = 6.15 m - 6 m = 0.15 m
you are stowing items and come across an aerosol bottle of hairspray.what should you do?
Answer:
below
Explanation:
Suppose the water at the top of Niagara Falls has a horizontal speed of 2.73 m/s just before it cascades over the edge of the falls. At what vertical distance below the edge does the velocity vector of the water point downward at a 52.9 ° angle below the horizontal?
Answer:
required vertical distance below the edge is 0.6648 m
Explanation:
Given the data in the question;
Horizontal speed of water falls v = 2.73 m/s
direction of water falls 52.9° below the horizontal
The vertical velocity must be such that;
tanθ = v[tex]_y[/tex] / v[tex]_x[/tex]
Now, vertical speed of water falls;
v[tex]_y[/tex] = v[tex]_x[/tex] × tanθ
we substitute
v[tex]_y[/tex] = 2.73 × tan(52.9°)
v[tex]_y[/tex] = 2.73 × 1.322237
v[tex]_y[/tex] = 3.6097
Now, at the top of falls, initial speed u = 0
v² - u² = 2as
s = ( v² - u² ) / 2as
we substitute
s = ( 0² - (3.6097)² ) / (2 × 9.8)
s = 13.029934 / 19.6
s = 0.6648 m
Therefore, required vertical distance below the edge is 0.6648 m
A pulley has a mechanical advantage of 1. What does this tell you about the size and direction of the input and output forces?
Answer:
The number of input force is the same as output. ... If it equals once, then both numbers are equal making it the same.Explanation:
Does this helps
Answer:
The number of input force is the same as output. Formula for MA (Mechanical Advantage) is Input Force/Output Force. If it equals once, then both numbers are equal making it the same. In order to raise MA, you must lower efficiency, something you learn around grade 8. Good luck!
P.S. Direction is the same for both, meaning if you pull something, the object you pull will come towards you.
An object that sinks in water has a mass in air of 0.0675 kg. Its apparent mass when submerged in water is 0.0424 kg. What is the specific gravity SG of the object? What material is the object probably made?
Answer:
1. SG
true
=2.689
2. The object is probably some sort of minerals and rocks such as Feldspar, Corals, Beryl, etc.
Explanation:
Given:
mass in the air= 0.0675 kg
mass in water= 0.0424 kg
The specific gravity of the object will be 2.6892. It is the ratio of the density of the given fluid and the standard fluid.
What is density?Density is specified as the mass divided by the volume. It is represented by the unit of measurement as kg/m³.
The mass of the object in air;
m=Vρ₀
m=0.0675 kg
Buoyant force on the object;
B= Vρₐg
For equilibrium;
N+B=m₀g
n=m₀g-Vρₓg
N/g=m₀-Vρₓ
N/g=0.0424 kg
[tex]\rm \frac{V\rho_0}{V\rho_x} =\frac{0.0675 }{m_0-0.0424 \ kg} \\\\ \frac{\rho_0}{\rho_x} =\frac{0.0675}{0.0675-0.0424} \\\\ \frac{\rho_0}{\rho_x} =2.6892[/tex]
Hence, the specific gravity of the object will be 2.6892.
To learn more about the density refers to the link;
brainly.com/question/952755
#SPJ2
The graph below shows the distance traveled by the skateboarder on each of the different road conditions. Using the graph, determine which of the roads was dry, wet, or muddy. Explain your answer using complete sentences.
Answer:
Road A- dry
Road B- mud
Road C- wet
Explanation:
Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.
The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.
The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.
A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What
is the magnitude of the impulse?
The magnitude of the impulse of the truck is equal to 4000 Kg.m/s.
What is impulse?Impulse can be described as the integral of a force over the time interval for which it acts. Impulse is also a vector quantity since force is a vector quantity. Impulse can be applied to an object that generates an equivalent vector change in its linear momentum.
The S.I. unit of impulse is N⋅s and the dimensionally equivalent unit of momentum is kg⋅m/s. A resultant force gives acceleration and changes the velocity of an object for as long as it acts.
Given the mass of the truck, m= 2000 Kg
The initial speed of the truck, u = 6 m/s
The final speed of the truck, v = 4 m/s
The change in the linear momentum is equal to the impulse.
I = ΔP = mv - mu
I = 2000 ×4 - 2000 × 6
I = 8000 - 12000
I = - 4000 Kg.m/s²
Therefore, the magnitude of the impulse is 4000 Kg.m/s².
Learn more about Impulse, here:
https://brainly.com/question/16980676
#SPJ2
Wavelength of blue photons 495 nm, what is the frequency? and what is the energy?
Answer:
1.F: About 6*10^14 Hz
2.E: About 4*10^ -19 J
Explanation:
Frequency: We knew that the speed of a wave is its wavelength(λ)* frequency(f, in Hz). By the wave-particle duality we know we can calculate the frequency of light in the same way. So, c=495nm *f, f=c/495nm=> (299,792,458 m/s) / (4.95*10^-7 m)
=6.05*10^14 /s
Energy: The energy photon contains can be calculate by this formula-- E=hf
f is the frequency and h is Planck's constant which is about 6.62 ×10^-34 *m^2*kg/s (after dimensional analysis ) =6.62*10^ -34 J*s.
So, the energy of a blue photon is (6.05*10^14)*(6.62*10^-34)=40.051*10^-20= 4.051*10^-19 J
Please, describe low-frequency vs. high-frequency waves.
Answer: High-frequency sound waves are perceived as high-pitched sounds, while low-frequency sound waves are perceived as low-pitched sounds. The audible range of sound frequencies is between 20 and 20000 Hz, with greatest sensitivity to those frequencies that fall in the middle of this range.
Explanation: Obviously explained in the answer
On topographic maps, contour lines that are farther apart indicate what ?
Answer:
if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.
Answer:
gentle slopes
Explanation:
A 3 5m container is filled with 900 kg of granite (density of 2400 3 kg m/ ). The rest of the volume is air, with density equal to 3 1.15 / kg m . Find the mass of air and the overall (average) specific volume
Complete question:
A 5-m³ container is filled with 900 kg of granite (density of 2400 kg/m3). The rest of the volume is air, with density equal to 1.15 kg/m³. Find the mass of air and the overall (average) specific volume.
Answer:
The mass of the air is 5.32 kg
The specific volume is 5.52 x 10⁻³ m³/kg
Explanation:
Given;
total volume of the container, [tex]V_t[/tex] = 5 m³
mass of granite, [tex]m_g[/tex] = 900 kg
density of granite, [tex]\rho _g[/tex] = 2,400 kg/m³
density of air, [tex]\rho_a[/tex] = 1.15 kg/m³
The volume of the granite is calculated as;
[tex]V_g = \frac{m_g}{ \rho_g}\\\\V_g = \frac{900 \ kg}{2,400 \ kg/m^3} \\\\V_g = 0.375 \ m^3[/tex]
The volume of air is calculated as;
[tex]V_a = V_t - V_g\\\\V_a = 5 \ m^3 \ - \ 0.375 \ m\\\\V_a = 4.625 \ m^3[/tex]
The mass of the air is calculated as;
[tex]m_a = \rho_a \times V_a\\\\m_a = 1.15 \ kg/m^3 \ \times \ 4.625 \ m^3\\\\m_a = 5.32 \ kg[/tex]
The specific volume is calculated as;
[tex]V_{specific} = \frac{V_t}{m_g \ + \ m_a} = \frac{5 \ m^3}{900 \ kg \ + \ 5.32\ kg} = 5.52 \times 10^{-3} \ m^3/kg[/tex]
what is measured by the ammeter
Answer:
amperes
Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.
Explanation:
hope it helps
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5
Answer:
A. Power generated by meteor = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Workdone = 981000 J
Power required = 19620 Watts
Note: The question is incomplete. A similar complete question is given below:
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?
Explanation:
A. Power = workdone / time taken
Workdone = Kinetic energy of the meteor
Kinetic energy = mass × velocity² / 2
Mass of meteor = 1.5 g = 0.0015 kg;
Velocity of meteor = 50 km/s = 50000 m/s
Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J
Power generated = 1875000/2.1 = 892857.14 Watts
Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.
B. Work done by elevator against gravity = mass × acceleration due to gravity × height
Work done = 1000 kg × 9.81 m/s² × 100 m
Workdone = 981000 J
Power required = workdone / time
Power = 981000 J / 50 s
Power required = 19620 Watts
Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.
g the total mechanical energy of the satellite-Earth system when the satellite is in its current orbit is E. In order for the satellite to orbit Earth in a new stable circular orbit at an altitude of 12RE, the energy of the satellite-Earth system must be
Answer:
The correct answer is "[tex]\frac{4E}{3}[/tex]".
Explanation:
According to the question,
Energy of satellite,
⇒ [tex]E_s=-\frac{GM_sM_E}{2r}[/tex]
For the very 1st case:
[tex]r = R_E+R_E[/tex]
[tex]=2R_E[/tex]
or,
⇒ [tex]E=-\frac{GM_sM_E}{4R_E}[/tex]...(1)
For the new case:
[tex]r = R_E+\frac{R_E}{2}[/tex]
[tex]=\frac{3R_E}{2}[/tex]
then,
⇒ [tex]E'=-\frac{GM_sM_E}{2 \frac{3R_E}{2} }[/tex]
[tex]=-\frac{GM_sM_E}{3R_E}[/tex]...(2)
From equation (1) and (2), we get
⇒ [tex]E'=\frac{1}{3}(4E)[/tex]
[tex]=\frac{4E}{3}[/tex]
A 2890-lb car is traveling with a speed of 58 mi/hr as it approaches point A. Beginning at A, it decelerates uniformly to a speed of 18 mi/hr as it passes point C of the horizontal and unbanked ramp. Determine the total horizontal force F exerted by the road on the car just after it passes point B.
Answer:
4592.57 lb
Explanation:
The missing diagram for this question is attached in the image below.
Given that:
the weight of the car = 2890 lb
At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr
At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr
To ft/s:
[tex](V_A)[/tex] = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_A)[/tex] = 85.07 ft/s
[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s
[tex](V_C)[/tex] = 26.4 ft/s
Between A to C, the total distance is;
[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]
Now, we need to determine the deceleration of the car using the formula:
[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]
[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]
[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]
[tex]696.96-7236.9049 = 2 a (654.154)[/tex]
[tex]-6539.9449 = 2 a (654.154)[/tex]
[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]
a = -4.99 ft/s²
The velocity of the car as it passes via B
[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]
[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]
[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]
[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]
[tex]v_B =\sqrt{ 10540.2849}[/tex]
[tex]v_B =102.67 \ ft/s[/tex]
Along B, the car's acceleration is:
[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]
[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]
[tex]a_B = 51.17 \ ft/s^2[/tex]
Finally, the total horizontal force F exerted = m[tex]a_B[/tex]
[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]
= 4592.57 lb
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 14.5 m away
Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:
[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of hockey player = 97 kg
m₂ = mass of puck = 0.15 kg
u₁ = u₂ = initial velocities of puck and player = 0 m/s
v₁ = velocity of player after collision = ?
v₂ = velocity of puck after hitting = 48 m/s
Therefore,
[tex](97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s[/tex]
negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:
[tex]s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s[/tex]
Now, the distance covered by the player in this time will be:
[tex]s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)[/tex]
s₁ = 0.022 m
Why does the moon appear dark from space?
But why does it appear bright when observed from earth, especially when it is full moon?
Answer:
The moon is actually quite dim.
Explanation:
compared to other astronomical bodies. The moon only seems bright in the night sky because it is so close to the earth and because the trees, houses, and fields around you are so dark at night. In fact, the moon is one of the least reflective objects in the solar system.
Answer:
It reflects the light send from the sun.
Explanation:
If the moon is between you and the sun, you will see the back of it which doesnt reflect light.
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
plz answer the question
Answer:
Ray A - incident ray
Ray B - reflected ray
if it takes a force of 20n to stretch a spring 0.1 meter how much energy does the spring have?
Answer:
The energy stored in the spring would be 1 joule.
Explanation:
hope that helps?
A 4-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ.
Answer:
12kWhr
Explanation:
Energy = Power * Time
Power = 4kW
Time = 3hrs
Substitute into the formula
Energy used up = 4kW * 3hrs
Energy used up = 12kWhr
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.
Answer:
The location of helicopter is behind the packet.
Explanation:
As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.
The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.
So, the helicopter is behind the packet.
An irregular shape object has a mass of 19 oz. A graduated cylinder with and initial volume of 33.9 mL. After the object was dropped in the graduated cylinder, it had a volume of 92.8 mL. What is the density of object( g/mL)
Explanation:
m = 19 oz × (28.3 g/1 oz) = 537.7 g
V = 92.8 mL
[tex]\rho = \dfrac{m}{V}= \dfrac{537.7\:g}{92.8\:mL} = 5.79\:\frac{g}{mL}[/tex]
On Ramesh’s13th birthday, his father invited all his friends and their relatives. It was a big party with lots of food and DJs. Ramesh didn’t like the loud sound of DJs and asked his father to play it in a low volume so that their neighbours do not get much disturbed and people at the party can also enjoy the music. Ramesh’s father felt good for his wisdom and did as he said.
→Do you think when loud music is played at a party is acceptable to all the people living in, neighbourhood? Give a reason for your answer
→How can you control noise pollution at your end?
Answer:
The first fact to understand which carries the most weight is we share this small world. We all have such similar characteristics that if you listed them all in one column and then list traits that make each person an absolute individual, one of the lists would be liken to an encyclopedia where as the other more akin to a doodle on a napkin in comparison. Now to the question.
Explanation:
We all know inherently the rules, so to say. yes its acceptable to do as people do normally. to go outside those bounds on purpose to be a nuisance intentionally is not. like 3am loud pounding music or downright disrespect is not acceptable. because when its done that way, since we all know the unspoken rules, its is being done on purpose to annoy.
Find the final velocity if the initial velocity of 8 m/s with an acceleration of 7 m/s2 over a 3 second interval?
I don't know about it your answer will give another people
Answer: Let the final velocity be v.
Given,
Initial velocity(u)=8m/s
Acceleration(a)=7m/s2
Time(t)=3 sec
Then,
v=u+at
=8+7*3 m/s
=29m/s
Therefore, the final velocity is 29m/s.
Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
A car accelerates uniformly from rest to a speed of 55.0 mi/h in 13.0 s. (a) Find the distance the car travels during this time. m (b) Find the constant acceleration of the car. m/s2
Answer:
(a) 159.84 m
(b) 1.89 m/s²
Explanation:
Applying,
(a)
s = (v+u)t/2.................. Equation 1
Where s = distance traveled by the car, u = initial velocity, v = final velocity, t = time.
From the question,
Given: u = 0 m/s ( from rest), v = 55 mi/h = (55/2.237) m/s = 24.59 m/s, t = 13 s
Substitute these values into equation 1
s = (24.59+0)13/2
s = 159.84 m
(b)
Also applying,
a = (v+u)/t................. Equation 2
Where a = acceleration of the car.
substituting into equation 2,
a = (24.59+0)/13
a = 1.89 m/s²
An airplane, starting from rest, moves down the runway at constant acceleration for 23 s and then takes off at a speed of 66 m/s. What is the average acceleration of the plane (in m/s2)?
Answer:
46
Explanation:
What is the efficiency of a machine that uses 102 kJ of energy to do 98 kJ of work?
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
