Answer:
Explanation:
The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .
The weight component acting on box parallel to incline plane
= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N
This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .
Force exerted by person = 98 N
distance travelled in 5 s
= velocity x time
= 2 x 5 = 10 m
Work done by person
= 98 x 10
= 980 J .
kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
0.2 mph
4.8 mph
5.5 mph
144.1 mph
it is actually science i couldn't find the word science
Answer:
4.8mph
Explanation:
Speed= Distance/time
Speed= 26.2/5.5
= 4.76mph
( To the nearest tenth ) = 4.8mph
Answer:
38.7 mph
Explanation:
I just add all the numbers together then divided them by 4, which is the amount of numbers you gave.
Electric power is to be generated by installing a hydraulic turbine-generator at a site 120 m below the free surface of a large water reservoir that can supply water at a rate of 2300 kg/s steadily. Determine the power generation potential.
Answer:
the power generation potential is 2.705 x 10⁶ J/s.
Explanation:
Given;
height below the free surface of a large water reservoir, h = 120 m
mass flow rate of the water, m' = 2300 kg/s
The power generation potential is calculated as;
[tex]Power = \frac{Energy}{time} = \frac{F\times h}{t} = \frac{(mg) \times h}{t} = \frac{m}{t}\times gh = m' \times gh\\\\Power = m' \times gh\\\\Power = 2300 \ kg/s \ \times \ 9.8 \ m/s^2 \ \times \ 120 \ m\\\\Power = 2.705 \times 10^6 \ J/s[/tex]
Therefore, the power generation potential is 2.705 x 10⁶ J/s.
You are given three pieces of wire that have different shapes (dimensions). You connect each piece of wire separately to a battery. The first piece has a length L and cross-sectional area A. The second is twice as long as the first, but has the same thickness. The third is the same length as the first, but has twice the cross-sectional area. Rank the wires in order of which carries the most current (has the lowest resistance) when connected to batteries with the same voltage difference.
Rank the wires from most current (least resistance) to least current (most resistance).
a. Wire of Lenght L and area A
b. Wire of Lenght 2L and area A
c. Wire of Lenght L and area 2A
Answer:
The answer is below
Explanation:
The resistance of a wire is directly proportional to the length of the wire and inversely proportional to its area. The resistance (R) is given by:
[tex]R=\frac{\rho L}{A}\\\\where\ L=length \ of\ wire,A=cross\ sectional\ area, \rho=resistivity\ of\ wire.[/tex]
Let us assume that all the wires have the same resistivity.
a) Wire of Length L and area A
[tex]R_1=\frac{\rho L}{A}[/tex]
b) Wire of Length 2L and area A
[tex]R_2=\frac{\rho *2L}{A}=2R_1[/tex]
C) Wire of Length L and area 2A
[tex]R_3=\frac{\rho L}{2A}=\frac{1}{2}R_1[/tex]
Therefore the wire of least resistance is R3 and R2 has the highest resistivity.
R₃ < R₁ < R₂
Therefore, the ranking of the wires from most current (least resistance) to least current (most resistance) is:
R₃ < R₁ < R₂
Starting from the front door of your ranch house, you walk 55.0 m due east to your windmill, turn around, and then slowly walk 35.0 m west to a bench, where you sit and watch the sunrise. It takes you 30.0 s to walk from your house to the windmill and then 36.0 s to walk from the windmill to the bench.
Required:
a. For the entire trip from your front door to the bench, what is your average velocity?
b. For the entire trip from your front door to the bench, what is your average speed?
Answer:
Explanation:
Average velocity = Total displacement / total time
Average speed = total distance covered / total time
a )
For the entire trip from your front door to the bench
Total displacement = 55 - 35 = 20 m [ first displacement is positive and second displacement is negative , because second displacement is in opposite direction ]
Total displacement = 20 m
Total time = 30 + 36 = 66 s
Average velocity = 20 / 66
= .303 m / s
b )
For the entire trip from your front door to the bench
Total distance covered = 55 + 35 = 90 m
Total time = 30 + 36 = 66 s
Average speed = 90 / 66
= 1.36 m / s
Anyone know this question?
A car of mass 150 kg is riding down at constant velocity. What is the normal force?
The formula is ( Fg=mg ; Fnet =ma ; Fx = Fcos theta ; Fy = Fsin theta)
Answer and I will give you brainiliest
Answer:
1500N
Explanation:
Normal force = mg - F sin theta
constant velocity means acceleration = 0
F= ma = 150× 0 = 0N
thus;
normal force = mg = 150 × 10 = 1500N
Answer:
864N
Explanation:
True or false it is impossible to determine weather you are moving unless you can touch another object
Answer: false
Explanation:
Answer:
false
Explanation:
According to Newton's First Law of motion, an object remains in the same state of motion unless a resultant force acts on it.
The cart is given an initial push up the ramp. After this push, as the car moves up the ramp, the direction of the acceleration of the cart is ________ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _______.
Answer:
Explanation:
The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .
Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .
When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .
After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.
a device that spreads light into different wavelengths is a what?
maybe a spectrograph ?
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 520 lines/mm , and the light is observed on a screen 1.4 m behind the grating.
Required:
What is the distance between the first-order red and blue fringes?
Answer:
0.143 m
Explanation:
Since
d = 1/N = 1/520 = 1.92 * 10^-3 mm
For red light;
θ = sin^-1 (1 * λred/d) = sin^-1 (1 * 656 * 10^-9/1.92 * 10^-6) = 19.98°
L = 1.4 * (tan 19.98) = 0.509 m
For blue light;
θ = sin^-1 (1 * λblue/d) = sin^-1 (1 * 486 * 10^-9/1.92 * 10^-6) = 14.66°
L = 1.4 * (tan 14.66°) = 0.366 m
Distance between the first-order red and blue fringes= 0.509 m - 0.366 m = 0.143 m
A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car slows down, and its average acceleration is . For the next 4.03 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 5.82-second period is +18.4 m/s. The ratio of the average acceleration values is = 1.53. Find the velocity of the car at the end of the initial 1.79-second interval.
Answer:
first value+2nd +3rd
Explanation:
thug life and there
A circuit has 12 Amps and 220 Volts. What is the Resistance of the circuit?
Answer:
:To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)
To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)
To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)
To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)
I need help please will mark brainliest
Answer:
c
Explanation:
A person is holding a bucket by applying a force of 10N. He moves a horizontal distance of 5m and then climbs up a vertical distance of 10m. Find the total work done by him?
Answer:
dgfggddhdbxbxjxddhsnsxnc
PHYSICS QUESTION PLS HELP
The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is
mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J
The total energy is the same, 970,200 J.
Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.
At point B, the coaster has dropped to a height of 10 m, so it has PE
mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J
which means it must have KE
970,200 J = KE + 294,000 J → KE = 676,200 J
which gives the coast a speed v at point B of
1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J → v ≈ 21.2 m/s
At point C, the coaster has a speed of 16.0 m/s, so it has KE
1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J
and hence PE
970,200 J = 384,000 J + PE → PE = 586,200 J
This lets us determine the height h at C:
mgh = (3000 kg) (9.80 m/s²) h = 586,200 J → h ≈ 19.939 m
which means the loop has diameter h - 10 m ≈ 9.94 m.
At point D, the coaster is 15 m above the ground so its PE at D is
mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J
and so its KE is
970,200 J = KE + 441,000 J → KE = 529,200 J
and hence has speed v at D
1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J → v ≈ 18.9 m/s
Lolliguncula brevis squid use a form of jet propulsion to swim—they eject water out of jets that can point in different directions, allowing them to change direction quickly. When swimming at a speed of 0.15m/s0.15m/s or greater, they can accelerate at 1.2m/s21.2m/s 2 .
(a) Determine the time interval needed for a squid to increase its speed from 0.15m/s0.15m/s to 0.45m/s0.45m/s.
(b) What other questions can you answer using the data?
Answer:
a) t = 0.25 s, b) x = 0.075 m
Explanation:
a) For this exercise we will use kinematic relationships in one dimension
v = v₀ + a t
in the problem they indicate the initial velocity v₀ = 0.15 m / s, the final velocity v = 0.45 m / s and the acceleration of the squid a = 1.2 m / s²
t = [tex]\frac{v -v_o}{a}[/tex]
we calculate
t = [tex]\frac{0.45 - 0.15}{1.2}[/tex]
t = 0.25 s
b) We can also find the distance traveled during this acceleration
v² = v₀² + 2a x
x = [tex]\frac{v^2 -v_o^2 }{2a}[/tex]
let's calculate
x = [tex]\frac{0.45^2 - 0.15^2 }{2 \ 1.2}[/tex]
x = 0.075 m
Which three statements are true of all matter?
A.
It is filled with air.
B.
It takes up space.
C.
It contains aluminum.
D.
It has mass.
E.
It is made up of atoms
Answer:
B, D and E, not all matter can be filled with air
A Car is moving at a speed of 20 m/s. How Much Distance it will cover in 1 min? Express the answer in km.
Answer:
d=20m/sx60s=1200m=1200/1000Km=1.2km
Explanation:
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating. For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force is needed on a 25 kg stone (a typical mass) to maintain the stone's motion once a gust has started it moving
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
HELP ASAP PLEASE
Which 2 factors must be present for chemical vapor deposition to be successful:
A.) The size of the diamond is larger than the most.
B.) The conditions during cooling are controlled.
C.) The heat in a vacuum forms a gas of single atoms.
D.) The heat in a vacuum is decreased to freezing.
E.) The pressure of a reaction vessel is negative.
2. Who was the first to show ancient hieroglyphs of swimming?
Answer:
Rafi Bahalul
Explanation:
Israeli Haaretz newspaper reported Feb. 4 that Rafi Bahalul stumbled upon a 3,400-year-old Egyptian stone anchor engraved with hieroglyphs during a morning swim along Israel's northern shores last year.
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).
(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
Nikki was walking around a department store shopping one day, and did not realize that the shirt she was wearing looked just like the shirts worn by employees. When a stranger asked, "do you work here," she thought it was funny. The other customers' assumption that Nikki was a store employee demonstrates the Gestalt principle of _______.
Answer: Similarity
By.
when air mass is caught between two cold fronts the result is a _______ front.
Answer choices
A.occluded
B.warm
C.cold
D.stationary
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water
Answer:
See explanation below
Explanation:
This is a typical exercise of free falling. In this case a rock thrown straight down from the bridge, and we are asked to determine the final velocity of the rock and it's displacement at those given times.
First, just for this problem, as the rock is going straight down, we'll say that the downward direction is positive, therefore, the following expressions to calculate velocity and speed will be:
V = V₀ + gt (1)
X = V₀t + gt²/2 (2)
In this case, g = 9.8 m/s²
Now, let's see the displacement and velocity for each given time:
a) For t = 0.5 s
V = 14 + (9.8)*0.5
V = 18.9 m/s
X = (14*0.5) + (9.8)(0.5)²/2
X = 7 + 1.225
X = 8.225 m
b) For t = 1.00 s
V = 14 + (9.8)*1
V = 23.8 m/s
X = (14*1) + (9.8)(1)²/2
X = 14 + 4.9
X = 18.9 m
c) For t = 1.5 s
V = 14 + (9.8)*1.5
V = 28.7 m/s
X = (14*1.5) + (9.8)(1.5)²/2
X = 21 + 11.025
X = 32.025 m
d) For t = 2 s
V = 14 + (9.8)*2
V = 33.6 m/s
X = (14*2) + (9.8)(2)²/2
X = 28 + 19.6
X = 47.6 m
e) For t = 2.50 s
V = 14 + (9.8)*2.5
V = 38.5 m/s
X = (14*2.5) + (9.8)(2.5)²/2
X = 35 + 30.625
X = 65.625 m
Hope this helps
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.
Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?
Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)
Answer:
30.4 s
Explanation:
A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g
a = 4 x 9.8 = 39.2 m /s²
initial velocity u = 0
Final velocity = 3.60 times speed of sound
= 3.6 x 331 = 1191.6 m /s
v = u + at
Putting the values
1191.6 = 0 + 39.2 t
t = 30.4 s .
I need help will mark brainliest
Answer: ITS 1 TRUST ME MAN BYE K
Explanation: OK BYE TRUST YEAH
A bartender slides a beer mug at 1.3 m/s towards a customer at the end of a frictionless bar that is 1.3 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar. (a) How far away from the end of the bar does the mug hit the floor
Answer:
x = 0.67 m
Explanation:
For this problem, let's use the projectile launch equations, as the jug goes through the bar, it comes out with horizontal speed vx = 1.3 m / s, which does not decrease as there is no friction.
Let's find the time or it takes to get to the floor
y = y₀ + v_{oy} - ½ g t²
in this case I go = 0 and when I get to the floor y = 0
0 = y₀ + 0 - ½ g t²
t² = 2y₀ / g
t² = 2 1.3 / 9.8 = 0.2653
t = 0.515 s
now let's find the distance traveled in this time
x = vx t
x = 1.3 0.515
x = 0.6696 m
x = 0.67 m
Explain what is happening in this picture
Answer:
in this video waves are coming up for the BOTTOM to the top of the sandbar