Answer:No
Explanation:Because car use the energy produced by the car battery and humans get energy from the food they eat
A football is thrown down off of a building with a force of 70
Newtons. Gravity is pulling down the football with a force of
19 Newtons. What is the net force on the football?
Answer:
89 N downward.
Explanation:
Both forces are working in the same direction -- down.
Therefore you can add them together.
F = F1 + F2
F is the total force
F1 is the gravitational force
F2 is the force the person provides
F1 = 19
F2 = 70
F = ?
F = 19 + 70
F = 89 Newtons downward.
Answer:
wE dONt taLk aBt brunooooOoooOoOo ;')
Explanation:
A compact fluorescent bulb is 17.0% efficient. How much energy input would be required for the bulb to produce 252 J of light energy?
What time of energy is the waste output and how much would be created?
Help
Answer:
1482.35J
Explanation:
Efficency=(output energy/input energy) x100
17.0 =(252/input energy) x100
input energy =100x252/17=1482.35J
wasted energy=1482.35-252=1230.35J
helpppppppppppppppppppppppppppp............
an object is moving with initial velocity of 5 m/s. After 10 seconds final velocity is 10 m/s. Calculate its acceleration.
Answer:
0.5 m/s 2 is the acceleration
Explanation:
hope it helped!!!
The activity of a radioisotope is found to decrease 40% of its original value in 2.59 x 10 s.
Calculate the decay constant of the radioisotope.
Answer: [tex]0.0353\ s^{-1}[/tex]
Explanation:
Given
Radioactive material is found to decrease 40% of its original value in [tex]2.59\times 10\ s[/tex]
Sample at any time is given by
[tex]N=N_oe^{-\lambda t}[/tex]
where, [tex]\lambda=\text{decay constant}[/tex]
Put values
[tex]\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10[/tex]
Taking natural logarithm both side
[tex]\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}[/tex]
Jack has a weight of 300 N and sits 2.0 m from the pivot of see - saw. Jill has a weight of 450 N and sits 1.5 m from pivot. Who will move down?
Answer:
Jill will move down first
Explanation:
Scenario
In a game of football player theme running and each other A wider running out with lotyy. A du x ahead the wide receivers a comeback heading directly towards him at a different below . The wide role and comeback continue running toward each other ar constant spend The location where the cornerback tackle the wide receiver, la moured from the wide recer initial position
Draw a diagram of the situation described above.
Answer:
hello your question is poorly written and I have tried to understand it hence I will give a general diagram as related to your question
answer : attached below
Explanation:
Attached below is the required diagram of the situation you are trying to describe
let the wide receiver be ; W
Corner back = C
westward velocity = Vw
eastward velocity = Vc
at initial position X = Xw
x = 0 ( initial position )
Which of the following best describes Earth's crust, according to the theory of plate tectonics?
Answer:
the Earth's crust is broken into about 12 plates that float on hotter, softer rocks in the underlying mantle
Explanation:
Please Help NO LINKS
Part A
What is the magnitude of the momentum of a 2500 kg
k
g
car traveling at 11 m/s
m
/
s
?
Express your answer to two significant figures and include the appropriate units.
Part B
What is the magnitude of the momentum of a 200 g
g
baseball thrown at 32 m/s
m
/
s
?
Express your answer to two significant figures and include the appropriate units.
Answer:
A) 28000kgm/s
B) 6.40kgm/s
Explanation:
Momentum = mass * velocity of a body
A) Given
mass m = 2500kg
velocity v = 11m/s
Substitute
Momentum = 2500 * 11
Momentum = 27500kgm/s
Momentum = 28000kgm/s (to 2 sf)
B) Given
mass m = 200g = 0.2kg
velocity v = 32m/s
Substitute
Momentum = 0.2 * 32
Momentum = 6.4kgm/s
Momentum = 6.40kgm/s (to 2 sf)
please help with both questions I’m giving all my points :) it’s 23 and 24
Choose the correct statement regarding the sign conventions for lenses.
a. The focal length f is positive for diverging lenses.
b. Virtual images appear on same side of the lens as the object and have a negative value for the image distance.
c. Real images appear on the opposite side of the lens from the object and have a negative value for the image distance.
d. The focal length f is negative for converging lenses.
e. Virtual images appear on same side of the lens as the object and have a positive value for the image distance.
Answer:
a) false
b) True
c) True
d) False
e) False
Explanation:
a) False
For a diverging lens, the focal length is negative while it is positive for a converging lens
b) True
Image distances for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged
c) True
d) False
For a diverging lens, the focal length is negative while it is positive for a converging lens
e) False
Image distances for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged
Which of the following should a warm up NOT include?
Select one:
A. A few minutes of cardiovascular exercise
B. A light version of the exercise you will be performing
C. Intense stretching
D. All of the above should be included in a warm up
What is the answer to this?
Answer:
B
Explanation:
Answer:
B
Explanation:
The diagram shows four locations in the electric field of a positive point charge.
A positively charged particle has vectors perpendicular to the surface and pointing laterally outward. The vectors farther from the surface are shorter. There are 4 dots to the right of the positively charged particle. The closest to the particle is labeled W and the particles further away are labeled X, Y and Z.
At which point is the electric potential the greatest?
W
X
Y
Z
Answer:
W
Explanation:
The closer you get to the center of the positive point of the electrical feild the stronger the pull gets. Therfore W is the best option.
Answer:
W
Explanation:
The closer you get to the center of the positive point of the electrical feild the stronger the pull gets. Therfore W is the best option.
One way to probe the nucleus is to bombard a sample with high-energy electrons. To learn about the nuclear structures in a sample, the de Broglie wavelengths of these electrons would need to be a little smaller than a nuclear radius. Estimate the energy of such electrons. Give your answer in electron-volts. (Assume that the wavelength used is about 9.0 fm.) eV
Answer:
E = 1.38 x 10⁸ eV = 138 MeV
Explanation:
The energy associated with the given wavelength can be found from the following formula:
[tex]E = \frac{hc}{\lambda}[/tex]
where,
E = Energy of electron = ?
h = Plank's Constant = 6.625 x 10⁻³⁴ J.s
c = Speed of Light = 3 x 10⁸ m/s
λ = wavelength = 9 fm = 9 x 10⁻¹⁵ m
Therefore,
[tex]E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9\ x\ 10^{-15}\ m}\\\\E = (2.21\ x\ 10^{-11}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})[/tex]
E = 1.38 x 10⁸ eV = 138 MeV
A 10 kg migratory swan cruises at 20m/s. A calculation that takes into ac-count the necessary forces shows that this motion requires 200W of mechanical power. If we assume an efficiency similar to humans (say, 25%), a reasonableassumption, then the metabolic power of the swan is significantly higher thanthis. The swan does not stop to eat during a long day of flying; it get theenergy it needs from fat stores. Assuming an efficiency similar to humans, after12 hours of flight.
Required:
a. How far has the swan traveled?
b. How much metabolic energy has it used?
c. What fraction of its body mass does it lose?
Answer:
Part A:
Distance=864000 m=864 km
Part B:
Energy Used=ΔE=8638000 Joules
Part C:
[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]
Explanation:
Given Data:
v=20m/s
Time =t=12 hours
In Secs:
Time=12*60*60=43200 secs
Solution:
Part A:
Distance = Speed**Time
Distance=v*t
Distance= 20*43200
Distance=864000 m=864 km
Part B:
Energy Used=ΔE= Energy Required-Kinetic Energy of swans
Energy Required to move= Power Required*time
Energy Required to move=200*43200=8640000 Joules
Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]
[tex]K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules[/tex]
Energy Used=ΔE=8640000 -2000
Energy Used=ΔE=8638000 Joules
Part C:
Fraction of Mass used=Δm/m
For This first calculate fraction of energy used:
Fraction of energy=ΔE/Energy required to move
ΔE is calculated in part B
Fraction of energy=8638000/8640000
Fraction of energy=0.99977
Kinetic Energy=[tex]\frac{1}{2}mv^2[/tex]
Now, the relation between energies ratio and masses is:
[tex]\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2[/tex]
[tex]\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977[/tex]
[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]
In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being in- vestigated occurs when the external field Bext has magnitude 0.78 T. Assume that Bint and Bext are in the same direction and take the pro- ton magnetic moment component u, to be 1.41 X 10-26 J/T. What is the magnitude of Bint?
Answer:
[tex]B_{int}=-0.015T[/tex]
Explanation:
From the question we are told that:
RF source oscillation speed [tex]\sigma= 34 MHz[/tex]
The external field [tex]Bext =0.78 T[/tex].
Pro- ton magnetic moment component [tex]\mu=1.41 X 10-26 J/T[/tex]
Generally the equation for magnitude of [tex]B_{int}[/tex] is mathematically given by
[tex]B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}[/tex]
[tex]B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}[/tex]
[tex]B_{int}=0.78-0.7957[/tex]
[tex]B_{int}=-0.015T[/tex]
how does gravity affect objects of different mass close to earth, and how does that effect change as an object moves farther from earth?
Answer:
It makes it lighter when its closer and heavier when its farther way.
Explanation:
Energy may be measured in?
Answer:
Energy May be measured in joule
An experiment is performed on an unknown material and produces the given heat curve. The temperature of the material is shown as a function of heat added. Other experiments determine that the material has a temperature of fusion of fusion=235 °C and a temperature of vaporization of vapor=471 °C.
If the sample of material has a mass of =9.80 g, calculate the specific heat when this material is a solid, s, and when it is liquid, l.
The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.
The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.
In this case, we need to determine that specific heat for solid and liquid states of matter. By Heat Physics, we understand that specific heat is contained in the slopes of the two sensible phases in the following form:
[tex]\frac{\Delta T}{\Delta Q} = \frac{1}{m\cdot c}[/tex] (1)
Where:
[tex]\Delta T[/tex] - Temperature change, in degrees Celsius.[tex]\Delta Q[/tex] - Heat received, in joules.[tex]m[/tex] - Mass of the sample, in grams.[tex]c[/tex] - Specific heat of the sample, in joules per kilogram-degrees Celsius.Solid phase
If we know that [tex]m = 9.80\,g[/tex], [tex]T_{1} = 40\,^{\circ}C[/tex], [tex]T_{2} = 235\,^{\circ}C[/tex], [tex]Q_{1} = 183\,J[/tex] and [tex]Q_{2} = 819\,J[/tex], then the specific heat of the solid phase is:
[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]
[tex]c = \frac{819\,J-183\,J}{(9.80\,g)\cdot (235\,^{\circ}C - 40\,^{\circ}C)}[/tex]
[tex]c = 0.333\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.
Liquid phase
If we know that [tex]m = 9.80\,g[/tex], [tex]T_{3} = 230\,^{\circ}C[/tex], [tex]T_{4} = 471\,^{\circ}C[/tex], [tex]Q_{3} = 1470\,J[/tex] and [tex]Q_{4} = 2870\,J[/tex], then the specific heat of the liquid phase is:
[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]
[tex]c = \frac{2870\,J - 1470\,J}{(9.80\,g)\cdot (471\,^{\circ}C - 230\,^{\circ}C)}[/tex]
[tex]c = 0.593\,\frac{J}{g\cdot ^{\circ}C}[/tex]
The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.
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PLS HELP.
A rope breaks when the speed of a 0.309 kg mass moving in a circle of radius of 0.429 m reaches 12.9 m/s
how much tension is in the strings when it breaks?
unti=n
120 works for acellus
Answer: 119.9
Explanation:
F = (mv^2)/r
Here we know m (0.309kg), v (12.9m/s) and r (0.429m)
So F = (0.309*12.9^2)/0.429 = 119.861748
What makes astronomers think that impact rates for the Moon must have been higher earlier than 3.8 billion years ago?
Answer: See explanation
Explanation:
The reason why astronomers think that the rates of impact for the Moon must have been higher earlier than 3.8 billion years ago is because on the older highlands, there are ten times more craters than on the younger maria.
It is believed that the impact rate was higher earlier and thus can be seen when the numbers of the craters that can be seen on the lunar highlands is being compared to that on the maria. It should be noted that there are about 10 times more craters that can be found on the highlands than those on the maria.
If there was a constant rate of impact throughout the history of the Moon, then the highlands be about 10 times older and therefore will have been formed about 38 billion years ago.
No file link please
A positively charged particle travels horizontally northward and enters a region where a field may exist. This region may contain only a magnetic field, only an electric field, or both a magnetic field and an electric field. When the particle enters the region, the particle does not speed up or slow down, but it does deflect in the downward direction. What is correct about the field or fields within the region and provides correct reasoning?
Answer:
Only magnetic field present
Explanation:
Since, the positively charged particle does not speed up or slow down, but it does deflect in the downward direction. This means only magnetic field is present.
This is because electric field changes the velocity and magnetic field changes the direction.
The magnetic force F is given by
F = qvBsinθ
Where, q = charge magnitude, v = velocity of charge, B = strength of magnetic field. and θ =the angle between the directions of v and B.
Another word for kinetic energy
could be
energy.
A. Safe
B. Moving
C. Stored
D. Potential
Answer:
moving
Explanation:
hope it helped!!!
Answer:
B
Explanation:
Pi] A car's engine produces a useful output power of 6.5 10° W
The car of mass 950 kg is moving up a hill at a steady speed.
The slope of the hill is 12° to the horizontal. Resistive forces on the car are negligible.
1?
What is the steady speed of the car? rk
[1 mark]
A 7.0ms7 [e]
B 12ms7 [e]
Cc 34ms7 [=]
D 68ms7 [e]
Recall that
P = F v
where
P = power produced by the engine
F = magnitude of the component force supplied by the engine in the direction of the car's motion
v = speed of the car
Draw a free-body diagram for the car, and decompose all the forces acting on it into components that act parallel and perpendicular to the slope. The car moves parallel to the slope, so we only care about 2 forces: the parallel component of the car's weight, and the force provided by the engine.
By Newton's second law, since the car is moving at a constant speed,
∑ F = F - m g sin(12°) = 0
where m = 950 kg and g = 9.80 m/s². Solve for F :
F = m g sin(12°) ≈ 1935.658 N
The engine provides P = 6.5 × 10⁴ W, so the car's speed v is
v = P / F = (6.5 × 10⁴ W) / (1935.658 N) ≈ 33.6 m/s
which makes C : 34 ms⁻¹ the closest answer.
The Steady speed of a car moving up a hill, with a slope of 12° to the horizontal, whose output power is 6.5×10⁴ W and whose mass is 950 kg is A. 7.0 m/s
Speed: This can be defined as the ratio of distance to the time travelled by a body. The S.I unit of speed is m/s
The question above can be solved using the formula below[tex]P = (mgcosx)V[/tex].................... Equation 1
Where P = output Power of the engine, m = mass of the car, g = acceleration due to gravity, x = angle of the hill with the horizontal, V = steady speed of the car.
Make V the subject of the equation[tex]V = P/mgcosx[/tex]............... Equation 2
From the question,
Given: P = 6.5×10⁴ W, m = 950 kg, x = 12° Constant: g = 9.8 m/s²Substitute the values into equation 2[tex]V = (6.5*10^{4} )/(950*9.8*cos12)[/tex]
[tex]V = 65000/9106.55[/tex]
[tex]V = 7.13 m/s[/tex]
From the question, the right option is A. 7.0 m/s
Hence the steady speed of the car is 7.0 m/s
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A cell has an emf of 1.5 V and an internal resistance of 0.65 Q.
The cell is connected to a resistor R.
[0[8].[4] State what is meant by an emf of 1.5 V.
[2 marks]
I
[0] 8]-[2] The current in the circuit is 0.31 A.
Show that the total power output of the cell is approximately 0.47 W.
[1 mark]
[o[s).[3] Calcutate the energy dissipated per second in resistor R.
[2 marks]
energy dissipated per second = Ist
[o[si.[4] The cell stores 14 kJ of energy when it is fully charged. The cell's emf and internal resistance are constant as the cell f discharged.
Calculate the maximum time during which the fully-charged cell can deliver energy to
resistor R.
[2 marks)
maximum time = 5
1. A cell has an emf of 1.5 V means a battery with an Emf rating of 1.5v can supply a maximum of 1.5v of electrical energy in an open circuit. Once it is connected to a closed circuit, it will have a potential difference of 1.5v(considering it has negligible resistance).
2. Voltage(V) in the circuit is 1.5 V and The current (I) in the circuit is 0.31 A.
So we have to find out power (p) in the circuit.
So P = VI
P= 1.5(0.31)
P = 0.465
P= 0.47 W
the total power output of the cell is approximately 0.47 W.
3. We have to obtain the energy dissipated per second in resistor R.
P = 0.465 - I²R
P = 0.465 - 0.31²(0.65)
P = 0.402535 J/s
P = 0.40 J
Total 0.40 Joule per second energy has been dissipated from the resistor R.
4. The cell stores 14 kJ of energy when it is fully charged. The cell's emf and internal resistance are constant as the cell f discharged.
The energy dissipated per second(e) in resistor R is .40
Cell have total (E) 14 kJ (14000 J) of energy.
We have to calculate the maximum time (t) during which the fully-charged cell can deliver energy to resistor R.
[tex]\(E=\into{{e}_}}{t}\\\)[/tex]
14000 = 0.40 × t
t =14000 / 0.40
t = 35 000 s or about 9.75 hrs.
Fully-charged cell can deliver energy to resistor R up to 9.75 hrs.
What is Electromotive Force?
Electromotive force, abbreviation E or emf, energy per unit electric charge that is imparted by an energy source, such as an electric generator or a battery. Energy is converted from one form to another in the generator or battery as the device does work on the electric charge being transferred within itself.
Learn more about Electromotive force:
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Radioactive isotopes are used in medicine to create images of internal
structures of the human body. What process is used to create theses
isotopes?
A. Gamma decay
B. Fusion
C. Alpha decay
D. Fission
Answer:
The correct option is D) Fission
Explanation:
There are several methods through which Radioactive isotopes are created.
Using a nuclear reactor that has a field of neutrons, insert a stable sample such as Lutetium-176. When it gets bombarded with neutrons, it acquires some, and fission is said to have occurred. Note that when Lutetium-176 acquires a neutron, it becomes radioactive Lu-177. Fission is also used to create Fluorine-18. To obtain the same, you need to bombard pure or enriched [[tex]^{18} O[/tex]] water with ~18 MeV protons which are high energy in nature. The bombarding is achieved using a cyclotron or an accelerator.Cheers
A 90-gram wire that is 1.0 m long is under tension. When a transverse wave of frequency 890 Hz travels down the wire, its wavelength is 0.10 m and its amplitude is 6.5 mm. What is the tension in the wire
Answer:
The tension will be "713 N".
Explanation:
The given values are:
Frequency,
= 890 Hz
Mass,
= 90 gm
or,
= 0.09 kg
Length of wire,
= 1.0 m
Wavelength,
= 0.10 m
The velocity will be:
= [tex]Frequency\times Wavelength[/tex]
= [tex]890\times 0.10[/tex]
= [tex]89 \ m/s[/tex]
Now,
The tension will be:
⇒ [tex]T=v^2\times \frac{M}{L}[/tex]
On substituting the values, we get
⇒ [tex]=(89)^2\times \frac{0.09}{1.0}[/tex]
⇒ [tex]=7921\times \frac{0.09}{1.0}[/tex]
⇒ [tex]=7921\times 0.09[/tex]
⇒ [tex]=713 \ N[/tex]
A uniform ladder of length 10.8 m is leaning against a vertical frictionless wall. The weight of the ladder is 323 N, and it makes an angle of 1.16 radians with the floor. A man weighing 734 N climbs slowly up the ladder. When he has climbed to a point that is 7.46 m from the base of the ladder (measured along the ladder), the ladder starts to slip. Find the coefficient of static friction between the floor and the ladder.
Answer:
0.3625
Explanation:
From the given information:
Consider the equilibrium conditions;
On the ladder, net torque= 0
Thus,
[tex]\tau_{net} = 0[/tex]; and
[tex]-fL \s in \theta +m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta = 0[/tex]
However, by rearrangement;
[tex]fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta \\ \\ \mu(m_L + m) gL \ sin \theta = (323 \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0[/tex]
[tex]\mu= \dfrac{ (323 \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}[/tex]
[tex]\mathbf{\mu= 0.3625 }[/tex]
In this equation, what shows that transmutation has taken place?
242 Cm – 328 Pu + He
A. There is conservation of both nucleons and atoms.
B. The nucleus of an atom changes
C. It involves more than one element
D. The number of atoms is conserved,but the number of nucleons is not