Answer:
The correct output will be "46.76 h.p.".
Explanation:
The given values are:
Gasoline's energy content,
= 46000 J/g
= 4.6 x10⁷ J/kg
Fuel usage,
= 13 kg/h
= [tex]\frac{13}{3600} \ kg/s[/tex]
Fuel efficiency,
= 21%
= 0.21
Now,
The average output will be:
= [tex](Energy \ content \ of \ gasoline)\times (fuel \ usage)\times (fuel \ efficiency)[/tex]
On substituting the values, we get
= [tex]4.6\times 10^7\times (\frac{13}{3600} )\times 0.21\times (\frac{1}{746}) h.p.[/tex]
= [tex]46.76 \ h.p.[/tex]
Answer:
the average horsepower output of the engine is 46.76 hp
Explanation:
The computation of the average horsepower output of the engine is shown below
The output of the engine is
= (Energy content of gasoline) × (usage of the fuel ) × (Fuel efficiency)
As we know that
1 hp = 746 W
where,
Energy content of gasoline = 46000 J / g = 4.6 × 107 J/kg
Usage of the fuel = 13 kg / h = 13 ÷ 3600 kg/s
Fuel efficiency = 21% = 0.21
Therefore, the output of the engine is
= 4.6 × 107 × (13 ÷ 3600) × 0.21 × (1 ÷ 746) h.p.
= 46.76 h.p.
hence, the average horsepower output of the engine is 46.76 hp
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 1.60 kg and 16.0 g whose centers are separated by about 3.30 cm. Calculate the gravitational force between these spheres, treating each as a particle located at the center of the sphere.
Answer:
The value is [tex]F = 1.568 *10^{-9} \ N[/tex]
Explanation:
From the question we are told that
The mass of the first lead sphere is [tex]m = 1.60 \ kg[/tex]
The mass of the second lead sphere is [tex]M = 16 \ g = 0.016 \ kg[/tex]
The separation between masses is [tex]r = 3.30 \ cm = 0.033 \ m[/tex]
Generally the gravitational force between each sphere is mathematically represented as
[tex]F = \frac{G * m * M }{r^2 }[/tex]
Here G is the gravitational constant with value [tex]G = 6.67 *10^{-11 } \ m^3 \cdot kg^{-1} \cdot s^{-2}[/tex]
[tex]F = \frac{6.67 *10^{-11 } * 1.60 * 0.016 }{0.033^2 }[/tex]
=> [tex]F = 1.568 *10^{-9} \ N[/tex]
Which of the following is an instantaneous speed?
A: All of the above
B: 80 ft/s
C. 80 yds./min
D. 80 km/hr
Answer:
A: All of the above
Explanation:
The instantaneous speed of an object is simply the current seed of the object at any given time. The SI unit is m/S and it is a vector quantity.
Therefore, according to the given options, they all have SI units that are consistent with distance and time which makes them all an example of instantaneous speed.
A 15.0kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 degrees above the horizontal. The block is displaced 5.0 m, and the coefficient of kinetic friction is 0.3.
a. Draw a free body diagram. Draw your coordinate system and label the axes.
b. Calculate the work done on the block by the 70 N force.
c. Calculate the work done on the block by the normal force.
d. Calculate the work done on the block by the gravitational force.
e. Calculate the work done on the block by the force of friction.
Answer:
[tex]W=70 * 5cos20 = 328.89 J[/tex]
[tex]W_n = 0[/tex]
[tex]W_g=0[/tex]
[tex]W_f= -184.59J[/tex]
Work done is 0
Explanation:
From the question we are told that
Weight of block =15.0kg
Force acting on the block = 70.0N
At an angle of 20 degree
Displacement of block is 5m
Coefficient of kinetic friction 0.3
b) Generally work done by force is give by [tex]W=fdcos \theta[/tex]
therefore
[tex]W=70 * 5cos20 = 328.89 J[/tex]
c) there is no work done by the normal force in this scenario because
normal force in this case is perpendicular to the displacement of the motion
[tex]W_n = 0[/tex]
d) The displacement in the vertical direction is 0
Therefore the gravitational work done is 0 [tex]W_g=0[/tex]
e)Generally in finding work done by friction we first find frictional force
Mathematically the equation for frictional force is given [tex]f = \alpha N[/tex]
Given that
[tex]N=mg-Fsin20[/tex]
[tex]N= 15.0*9.8 - 70 sin20[/tex]
[tex]N=123 N[/tex]
[tex]f=0.3* 123.06 = 36.92N[/tex]
Mathematically solving to get work done by frictional force [tex]W_f[/tex]
[tex]W_f= -fd\\W_f = -36.92 * 5[/tex]
[tex]W_f= -184.59J[/tex]
the frictional force work done is [tex]W_f= -184.59J[/tex]
The energy of a photon is ________ proportional to its wavelength.
Answer:
The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy.
Explanation:
Plz mark brainliest thanks
In a test run, a certain car accelerates uniformly from zero to 20.4 m/s in 2.60 s.
Required:
a. What is the magnitude of the cars acceleration?
b. How long does it take the car to change speed from 10.0 m/s to 20 m/s.
c. Will doubling the time always double the change in speed? why?
Answer:
(a) The acceleration is 7.85 m/s²
(b) It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Doubling the time will double the change in velocity if the acceleration is kept constant.
Explanation:
(a) Acceleration is the physical quantity that measures the rate of change of velocity with time. That is, acceleration relates changes in speed with the time in which they occur, that is, it measures how fast the changes in speed are.
The average acceleration is calculated using the following expression:
[tex]a=\frac{vf-vi}{t}[/tex]
where a is the acceleration, vf is the final velocity, vi is the initial velocity and t is the time.
In this case:
vf= 20.4 m/svi=0 m/st= 2.60 sReplacing:
[tex]a=\frac{20.4 \frac{m}{s} - 0\frac{m}{s} }{2.60 s}[/tex]
a= 7.85 m/s²
The acceleration is 7.85 m/s²
(b) In this case you know:
a= 7.85 m/s²vf= 20 m/svi= 10 m/sReplacing:
[tex]7.85 \frac{m}{s^{2} } =\frac{20 \frac{m}{s} - 10\frac{m}{s} }{t}[/tex]
and solving you get:
[tex]t=\frac{20 \frac{m}{s} - 10\frac{m}{s} }{7.85 \frac{m}{s^{2} } }[/tex]
t=1.27 s
It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.
(c) Being:
[tex]a=\frac{vf-vi}{t}[/tex]
Then:
a*t= vf - vi
vf - vi represents the change in velocity. You can see that, if a (acceleration) is constant, then (vf - vi) is directly proportional to the time t: therefore, if t doubles, the change in velocity doubles as well.
In other words, doubling the time will double the change in velocity if the acceleration is kept constant.
A hockey player strikes a puck that is initially at rest. The force exerted by the stick on the puck is 975 N, and the stick is in contact with the puck for 0.0049 s.
(a) What is the impulse imparted by the stick to the puck.
___________ kg m/s
(b) What is the speed of the puck (m= 1.67 kg)just after it leaves the hockey stick?
____________ m/s
Explanation:
Given that,
The force exerted by the stick on the puck is 975 N
The stick is in contact with the puck for 0.0049 s
Initial speed of the puck, u = 0 (at rest)
(a) We need to find the impulse imparted by the stick to the puck.
Impulse = Force × time
J = 4.7775 kg-m/s
(b) Mass of the puck, m = 1.76 kg
We need to find the speed of the puck just after it leaves the hockey stick.
Let the speed be v.
As impulse is equal to the change in momentum.
[tex]J=m(v-u)\\\\4.7775=1.67(v-0)\\\\v=\dfrac{4.7775}{1.67}\\\\v=2.86\ m/s[/tex]
So, when the puck leaves the hockey stick its speed is 2.86 m/s.
BERE
Which describes the positions on a horizontal number line?
0
O All points to the left of one are positive.
O All points to the right of one are positive.
O All points to the left of zero are negative.
O All points to the right of zero are negative.
Mark this and return
Save and Exit
Next
Submit
Answer:
All points to the left of zero are negative
Explanation:
Answer:
C
Explanation:
on edge
Do scientific theories have math in them?
Answer:
Yes of course science has maths
A truck covers 40.0 m in 9.50 s while uniformly slowing down to a final velocity of 2.75 m/s.
a. Find its original speed.
b. Find its acceleration.
Explanation:
Given that,
Distance covered, d = 40 m
Time, t = 9.5 s
Final velocity, v = 2.75 m/s
(a) Let u be the original speed of the truck. We can find it using first equation of motion.
[tex]v=u+at\\\\2.75=u+2.75\times 9.5\\\\2.75-26.125=u\\\\u=-23.375\ m/s[/tex]
(b) Acceleration = rate of change of velocity
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{2.75-(-23.375)}{9.5}\\\\=2.75\ m/s^2[/tex]
So, the original speed is -23.375 and acceleration is 2.75 m/s².
What is the strength of the electric field in a region where the electric potential is constant?
Answer:
Where the electric potential is constant, the strength of the electric field is zero.
Explanation:
As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e
Eₓ = - dV / dx ----------(i)
From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.
Therefore, a constant electric potential means that electric field is zero.
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the ydirection of ay = 7.30 m/s2. The engines turn off after firing for 675 s, at which point the spacecraft has velocity components of vx = 3630 m/s and vy = 4276 m/s.What was the magnitude and direction of the spacecraft's initial velocity, before the engines were turned on?Express the direction as an angle measured counterclockwise from the x -axis.
Answer:
v₀ = 677.94 m / s , θ = 286º
Explanation:
We can solve this exercise using the kinematic expressions, let's work on each axis separately.
X axis
has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
let's calculate
v₀ₓ = 3630 - 5.10 675
v₀ₓ = 187.5 m / s
Y Axis
[tex]v_{y}[/tex] = v_{oy} - a_{y} t
v_{oy} = v_{y} - a_{y} t
let's calculate
v_{oy} = 4276 - 7.30 675
v_{oy} = -651.5 m / s
we can give the speed starts in two ways
a) v₀ = (187.5 i ^ - 651.5 j ^) m / s
b) in the form of module and angle
Let's use the Pythagorean theorem
v₀ = [tex]\sqrt{v_{ox}^{2} + v_{oy}^{2} }[/tex]
v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]
v₀ = 677.94 m / s
we use trigonometry
tan θ = [tex]\frac{v_{oy} }{v_{ox} }[/tex]
θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }
θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])
θ = -73.94º
This angle measured from the positive side of the x-axis is
θ‘ = 360 - 73.94
θ = 286º
List the four outer planets from smallest to largest.
WILL MARK BRAINLIEST! ASAP
Answer: Mercury, Mars, Venus, Earth, Neptune, Uranus, Saturn, and Jupiter.
Explanation:
That's all of the planets if you need them. Hope this helps!
For an object like a planet, with a typical temperature of a few hundred kelvin, what kind of blackbody radiation would it principally emit
Answer:
Low-temperature blackbody
Explanation:
There are 3 types of blackbody temperatures.
Low-temperature blackbody
High temperature extended area blackbody
High-temperature cavity blackbody
A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.
Why do we perform stork stand test
Answer:
umm becuase it is a test and you need them
Explanation:
Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline. Which has the larger speed at the bottom?
The question incomplete, the complete question is;
Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline.
(A)the one falling vertically
(B)the one on the incline
(C)Both have the same speed.
(D)cannot be determined
Answer:
(C)Both have the same speed.
Explanation:
When we consider the question closely, we will discover that an object falling down a frictionless incline is comparable to an object falling freely under gravity.
In both instances, the acceleration of objects is just the same irrespective of mass.
Hence, the object falling vertically and the object sliding down a frictionless plane will have the same speed at the bottom.
You work at a garden store for the summer. You lift a bag of fertilizer with a force of 112 N, and it moves upward with an acceleration of 0.790 m/s^2.
a. What is the mass of the fertilizer bag?
b. How much does the fertilizer bag weigh?
Given :
Force provided, F = 112 N.
Acceleration of the bag, a = 0.79 m/s².
To Find :
a. What is the mass of the fertilizer bag?
b. How much does the fertilizer bag weigh?
Solution :
We know, force is given by :
F = ma
m = F/a
m = 112/0.79 kg
m = 141.77 kg
Now, weight is given by :
W = mg
W = 141.77 × 9.8 N
W = 1389.35 N
Therefore, the mass of fertilizer bag is 141.77 kg and weight us 1389.35 N.
According to the Law of Conservation of Energy, why does the first hill on a roller coaster always have to be the tallest of all the other hills?
30 points? I have no clue
Answer:
The second graph, B
Explanation
A 20-kg object sitting at rest is struck elastically in a head-on collision with a 10-kg object initially moving at 3.0 m/s. Find the final velocity
Answer:
1 m/s
Explanation:
Using law of conservation of momentum
m1v1 + m2v2 = (m1 + m2)vf , where
m1 = mass of object at rest, 20 kg
v1 = initial velocity of object at rest, 0 m/s
vf = final velocity of the bodies
m2 = mass of object in motion, 10 kg
v2 = initial velocity of object in motion, 3 m/s
On substituting, we have
(20 * 0) + (10 * 3) = (20 + 10) vf
0 + 30 = 30 vf
vf = 30 / 30
vf = 1 m/s
Therefore, the velocity of the bodies after hitting each other is 1 m/s
In schematic diagrams, currents are indicated using arrows. What do the arrows indicate? a) the direction of motion of the electrons b) the direction of the current vector c) the direction of motion of the charge carriers d) the direction that positive charge carriers would move e) nothing; they are just a convenient drawing tool
Answer:
D
Explanation:
The direction that positive charge would move
A tennis ball is hit with a vertical speed of 10 m/s and a horizontal speed of 30 m/s. How far will the ball travel horizontally before landing?
a. 10 m
b. 20 m
c. 40 m
d. 60 m
e. 80 m
Answer:
D) 60 m
Explanation:
We can use the constant acceleration equation that contains displacement, initial velocity, acceleration, and time. We want to solve for the time that the ball was in the air first.
Δx = v_i * t + 1/2at²Let's use this equation in terms of the y-direction.
Δx_y = (v_i)y * t + 1/2a_y * t²The vertical displacement will be 0 meters since the ball will be on the floor. The initial vertical velocity is 10 m/s, the vertical acceleration is g = 10 m/s², and we are going to solve for time t.
Let's set the upwards direction to be positive and the downwards direction to be negative. We must use -g to be consistent with our other values.
Plug the known values into the equation.
0 m = 10 m/s * t + 1/2(-10 m/s²) * t²Simplify the equation.
0 = -10t + 5t² 0 = 5t² - 10tFactor the equation.
0 = 5t(t - 2)Solve for t by setting both factors to 0.
5t = 0t - 2 = 0We get t = 0, t = 2. We must use t = 2 seconds because it is the only value for t that makes sense in the problem.
Now that we have the time that the ball was in the air, we can use the same constant acceleration equation to determine the horizontal displacement of the tennis ball. We will use this equation in terms of the x-direction:
Δx = v_i * t + 1/2at² Δx_x = (v_i)x * t + 1/2a_x * t²Plug the known values into the equation.
Δx_x = 30 m/s * 2 sec + 1/2(0 m/s²) * (2 sec)²We can eliminate the right side of the equation since anything multiplied by 0 outputs 0.
Δx_x = 30 * 2 Δx_x = 60The horizontal displacement of the ball is 60 meters. Therefore, the answer is D) 60 m.
A child blows a leaf from rest straight up in the air. the leaf has a constant upward acceleration of magnitude 1.0 m by s square. how much time does it take the leaf to displace 1.0m upwards?
Answer:
√2
Explanation:
From the question, we're given that the
Acceleration of the leaf is 1 m/s²
Change in displacement of the leaf is 1 m/s.
Again, from the question, we can tell that the initial velocity u = 0, since the object starts at rest
Now, to solve this, we don't the equation of motion to ur
S = ut + 1/2at², substituting the whole parameters, we then have
1 = 0 * t + 1/2 * 1 * t²
1 = 1/2 * t²
t²/2 = 1
t² = 2
t = √2 seconds
Therefore the time it takes the leaf to dislodge is 2 seconds
Two particles are separated by 0.38 m and have charges of -6.25 x 10-°C and 2.91 x 10-°C. Use Coulomb's law to predict the force between the particles if the distance is cut in half. The equation for Coulomb's law is F = kqi 42, and the constant, k, equals 9.00 x 109 Nm2/C2 2
Answer:
-4.35 × 10^-6 N
Explanation:
i just answered it on ap3x :)
derive an expression for torque experiend by an electric dipole placed in a uniform electric field
Answer:
The torque τ on an electric dipole with dipole moment p in a uniform electric field E is given by τ = p × E where the "X" refers to the vector cross product. Ref: Wikipedia article on electric dipole moment.
Explanation:
For the following types of electromagnetic radiation, how do the wavelength, frequency, and photon energy change as one goes from the top of the list to the bottom?
a. radio waves
b. infared radiation
c. visible light
d. ultraviolet radiation
e. gamma radiation
Answer:
Wavelength, frequency and the photon energy changes as the one goes across the ranges of the electro-magnetic radiations.
Explanation:
Electro-magnetic radiations may be defined as the form of energy that is radiated or given by the electro-magnetic radiations. The visible light that we can see is the one of the electro-magnetic radiations. Other forms are the radio waves, gamma waves, UV rays, infrared radiations, etc.
The wavelength of the radiations decreases as we go from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The frequency of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
The photon energy of the radiations increases when we move from a. radio waves -- b. infrared radiation -- c. visible light -- d. ultraviolet radiation -- e. gamma radiation.
A student lifts a box of books 2 meters with a force of 45 N. He then carries the box 10 meters to the living room. What is the total amount of work done in this situation?
PLEASE ANSWER FAST
Answer:
90J
Explanation:
The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.
A student is performing an experiment that involves the charge on a metal sphere that is attached to a charged electroscope. A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more. The rod is then removed, and the leaves return to their initial separated position. The student repeats the procedure, but this time the electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves again end up separated. What can be concluded about the charge on the separated leaves of the electroscope
Answer:
The leaves have a charge in each experiment, but the sign of the charge cannot be determined.
Explanation:
In the first experiment, A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more.
Thus indicates that there are charges involved. Now, like charges would repel like what is happening here but we don't know if they are both positive or negative because in both cases, they will still repel.
Now for the second experiment, electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves end up being separated again.
Similar to the first time, it's clear there are charges but the charges repel. Thus, they are the same sign charges but we don't know if they are both positive or negative.
Thus, in both cases we can conclude that the leaves have charges but we don't know their signs.
define alpha and beta
alpha is the excess return on an investment after adjusting for market related volatility and random fluctuations.
beta is a measure of volatility relative to a benchmark ,such as the S&P 500.
Explanation:
alpha and beta are two different parts of an equation used to explain the performance of stocks and investments funds. But in maths alpha and beta is the Greek alphabet
What was your train of thought as you navigated the picture of the candle?
Answer:
Where is the picture
Explanation:
WHERE IS THE PICTURE
a 150kg roller coaster SITTING ON THE TOP OF A 200M HILL HAS HOW MUCH POTETNTIAL ENERGY
Answer:
Epot = 294300 [J]
Explanation:
Potential energy is defined as the product of mass by height by gravitational acceleration. The height is measured with respect to the reference level. At this reference level the potential energy is equal to zero.
[tex]E_{pot}=m*g*h\\[/tex]
where:
m = mass = 150 [kg]
g = gravity acceleration [m/s²]
h = elevation = 200 [m]
[tex]E_{pot}=150*9.81*200\\E_{pot}=294300 [J][/tex]