Answer:
E = 16581.6 J
Explanation:
Given that,
Current, I = 4.9 A
Time for which the current is set up, I = 4.7 min = 282 s
The voltage of the battery, V = 12 V
We need to find how much chemical energy of the battery reduced. Let It is E. We know that,
E = P t
Where
P is power of battery, P = VI
So,
[tex]E=VIt[/tex]
Put all the values,
[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]
So, 16581.6 J of chemical energy of the battery is reduced.
compare the time period of two pendulums of length 4m and 9m
area= length × length
area = 4m × 9m
ans 36
Isotopes of the same element always have the same
(2 points)
O atomic mass number
O A-number
O Z-number
O neutrinos
Answer:
Z-number
Explanation:
The Z number is the number of protons in an atom, and this does not change when an isotope is created. I got it right on the test.
A voltage of 75 V is placed across a 150 Ω resistor. What is the current through the resistor?
Answer:
0.5 A
Explanation:
Applying,
V = IR.................. Equation 1
Where V = Voltage, I = current, R = Resistance.
make I the subject of the equation
I = V/R............... Equation 2
From the question,
Given: V = 75 V, R = 150 Ω
Substitute these values into equation 2
I = 75/150
I = 0.5 A.
Hence the cuurent through the resistor is 0.5 A
What is sieving? Give an example where this method is used. (2)
Answer:
sieving is when you separate particles of different sizes.
Explanation:
separating sand mixtures
separating chaffs from local garri
You want to produce a magnetic field of magnitude 5.50 x 10¹ T at a distance of 0.0 6 m from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 cm from the wire's center?
Answer:
(a) I = 1650000 A
(b) 4.125 T
Explanation:
Magnetic field, B = 5.5 T
distance, r = 0.06 m
(a) Let the current is I.
The magnetic field due to a long wire is given by
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]
(b) Let the magnetic field is B' at distance r = 0.08 m.
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]
PLEASE HELPPPPPP ME PLS PLS
If two dogs are pulling a bone with force-20Newtons in opposite direction, then the resultant force is
Answer:
Newtons third law of motion: Balanced forces
Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs. Once more, tug-of-war is a prime illustration.
What force in opposite direction follow newton law?The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.
The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.
Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.
Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.
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Two people that have identical weight are holding onto a massless pole while standing on horizontal frictionless ice. 1)If the guy on the left starts to pull on the pole, where do they meet
Answer:
Explanation:
From the missing image attached below, it is obvious that there no external force. This implies that they cannot change their position by merely just pulling the ropes. As a result, there will be no movement and no net force will exist.
So, if there is no external force;
The center of mass of the two people is:
[tex]X_{cm}= \dfrac{m_1x_1+m_2x_2}{m_1+m_2} \\ \\ X_{cm}= \dfrac{m(-3m)+m(+3m)} {m+m}\\ \\ X_{cm}= \dfrac{0}{2m} \\ \\ X_{cm} =0[/tex]
Thus, In the system, no movement occurs and all forces remain the same.
Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of 25 cm. The focal length of the magnifying glass is +5.0 cm. Assume that Keesha's eye is close to the magnifying glass.
(a) What should be the distance between the magnifying glass and the beetle?
(b) What is the angular magnification?
Answer:
a) p = 4.167 cm, b) m = + 6
Explanation:
a) For this exercise we must use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
In this case the distance to the image q = 25 cm and the focal length is f = 5.0 cm
Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.
[tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]
[tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]
[tex]\frac{1}{ p}[/tex] = 024
p = 4.167 cm
b) angular magnification
m = h ’/ h = - q / p
m = - (-25) /4.167
m = + 6
the positive sign indicates that the image is straight and enlarged
Answer this
a) which ink is likely to be pure? Why?
b) What does the chromatography tell us about ink Y
c) Why are the three different spots separated out from ink Y found at different heights?
Answer:
a) Ink X is likely to be pure because it only contain 1 spot.
b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.
c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.
The different spots from Y are found at various heights because they represent different compounds.
What is chromatography?The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.
We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.
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Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)
Answer:
39.40 MeV
Explanation:
Determine the minimum possible Kinetic energy
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV
the boiling point of F2 much lower than the boiling point of NH3
Answer:yeah it A
Explanation:
A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that d is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.
Required:
a. What is the tension in the cable?
b. Compute the energy stored in the electric field after the top plate was raised.
Answer:
A) F = V²E_o•πr²/2d²
B) U = E_o•Aπr²V²/2d
Explanation:
A) Since we have two circular plates, the formula for the electric field is expressed as;
E = V/d
Where;
V is voltage
d is distance
However, the net electric field produced is given by;
E' = V/2d
The tension in the cable can then be expressed as;
F = qE'
Where q is charge
Thus;
F = qV/2d - - - (eq 1)
We also know that;
C = q/V = E_o•A/d
A is area = πr²
Thus;
q/V = E_o•πr²/d
q = VE_o•πr²/d
Let's put VE_o•πr²/d for q in eq 1 to get;
F = V²E_o•πr²/2d²
B) formula for the energy stored in the electric field is;
U = ½CV²
From earlier, we saw that; C = E_o•A/d
Thus;
U = ½E_o•AV²/d
A = πr²
Thus;
U = E_o•Aπr²V²/2d
what is the magnitude of an electric field (in 106 n/c) that balances the weight of a plastic sphere of mass 2.1 g that has been charged to 3.0 nc
Answer:
[tex]E=6.86\times 10^6\ N/C[/tex]
Explanation:
Given that,
Mass of the sphere, m = 2.1 g = 0.0021 kg
Charge, q = 3 nC
We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,
qE = mg
[tex]E=\dfrac{mg}{q}[/tex]
Put all the values,
[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]
So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].
Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.
The index of refraction of n-propyl alcohol is 1.39. Find the angle of refraction of light in that medium if light comes from air with an angle of incidence of 55 degrees.
Answer:
36.11 degrees
Explanation:
index of refraction n = sin i/sinr
i is the angle of incidence
r is the angle of refraction
Substitute into the expression
1.39 = sin55/sin(r)
1.39 = 0.8191/sin(r)
sin(r) = 0.8191/1.39
sin(r) = 0.5893
r = arcsin(0.5893)
r = 36.11
hence the angle of refraction of light is 36.11 degrees
A 3.50 kg basket of cookies sits on a 2.00 m high shelf. What is the gravitational potential energy of the basket?
pls help
Answer:
68.6 J
Explanation:
Applying,
P.E = mgh............... Equation 1
Where P.E = Potential Energy of the basket, m = mass of the basket, g = acceleration due to gravity of the basket, h = height of the basket
From the question,
Given: m = 3.5 kg, h = 2.00 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
P.E = 3.5×2×9.8
P.E = 68.6 J
Hence the potential energy of the basket is 68.6 J
In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon? A. Emission spectra of atoms B. Blackbody radiation C. Interference patterns of electrons D. Photoelectric effect
Answer:
In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon is:
C. Interference patterns of electrons.
Answer:
C. Interference patterns of electrons
5N
5 N
19 N
19 N
Pls help look at the pic
Answer:
b. is the correct answer ....
People who do very detailed work close up, such as jewellers, often can see objects clearly at much closer distance than the normal 25 cm. a. What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.5 cm
This question is incomplete, the complete question is;
People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.
a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.) __ mm
Answer:
1) the power in D of the eyes of a woman is 61.7647 D
2) the size in mm of an image of a 8.00 mm object is -1.882 mm
Explanation:
Given the data in the question;
a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,
so
u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m
v = 2.00 cm = ( 2 / 100 )m = 0.02 m
Now, we know that power of lens p = 1 / u + 1 / v
so we substitute
p = ( 1 / 0.085 ) + ( 1 / 0.02 )
p = 11.7647 + 50
p = 61.7647 D
Therefore, the power in D of the eyes of a woman is 61.7647 D
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)
we know that;
m = -v / u
we substitute
m = -0.02 / 0.085
m = -0.2353
since H₀ = 8.0 mm
H[tex]_i[/tex] = m × H₀
H[tex]_i[/tex] = -0.2353 × 8.0
H[tex]_i[/tex] = -1.882 mm
the size in mm of an image of a 8.00 mm object is -1.882 mm
By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects
Answer:
[tex]Weight\ loss=1.6321N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=85.9kg[/tex]
Altitude [tex]h= 6.33 km[/tex]
Let
Radius of Earth [tex]r=6380km[/tex]
Gravity [tex]g=9.8m/s^2[/tex]
Generally the equation for Gravity at altitude is mathematically given by
[tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]
[tex]g_s=9.781m/s^2[/tex]
Therefore
Weight at sea level
[tex]W_s=9.8*85.9[/tex]
[tex]W_s=841.82N[/tex]
Weight at 6.33 altitude
[tex]W_a=9.781*85.9[/tex]
[tex]W_a=840.2N[/tex]
Therefore
[tex]Weight loss=W_s-W_b[/tex]
[tex]Weight loss=841.82-840.2[/tex]
[tex]Weight loss=1.6321N[/tex]
Engineers are working on a design for a cylindrical space habitation with a diameter of 7.50 km and length of 29.0 km. The habitation will simulate gravity by rotating along its axis. With what speed (in rad/s) should the habitation rotate so that the acceleration on its inner curved walls equals 8 times Earth's gravity
Answer:
The speed will be "0.144 rad/s".
Explanation:
Given that,
Diameter,
d = 7.50 km
Radius,
R = [tex]\frac{7.5}{2} \ Km[/tex]
Acceleration on inner curve,
= 8 times
Now,
As we know,
⇒ [tex]\omega^2R=8g[/tex]
or,
⇒ [tex]\omega=\sqrt{\frac{8g}{R} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }[/tex]
⇒ [tex]=\sqrt{\frac{78.4}{3750} }[/tex]
⇒ [tex]=\sqrt{0.0209}[/tex]
⇒ [tex]=0.144 \ rad/s[/tex]
effieiency of simple machine is always less than 100% why
Answer:
efficiency of a machine is less than 100% because some part is energy is utilized to overcome some opposing forces like friction which is wasted as heat ,sound energy etc
Explanation:
The maximum amount of pulling force a truck can apply when driving on
concrete is 8760 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
O A. 8760 N
O B. 12,680 N
O C. 10,950 N
O D. 7240 N
Answer:
8760 N
Explanation:
think this is the right answer :)
Who is a socio-economically disadvantaged child? Explain any four ways forhelping such a child
Explanation:
A socio-economically disadvantaged child is the one who is disadvantaged in terms of social position and economic position. Such children have limited resources in terms of education, money and future options. Four ways of helping such children are as follows:
1) Help them in education - You can help such children by giving them free tuition. If you belong to a well off family, you can get them admitted in schools as well. Provide them with books and uniform.
2) Encourage them to do well in school and pursue their passions.
3) Provide them with meals, if they do not have access to regular meals.
4) encourage them to go school regularly.
Answer:
sorry i dont know the answer bit mark me as BRAINLISTIf the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N
The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.
What is Force?The force is the action of push or pull which makes an object to move or stop.
Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by
F = m(g-a)
F = 13.9 x (9.81 - (-0.898))
F = 148.84 N
Thus, the magnitude of the force is 148.84 N.
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Last stage of a medium mass star
White Dwarf Stage
This ring is called a planetary nebula. when the last of the helium atoms in the core are fused into carbon atoms, the medium size star begins to die. Gravity causes the last of the star's matter to collapse inward and compact. This is the white dwarf stage.Compared to its weight on Earth, a 5kg object on the moon will weigh
The same amount
Less
More
Answer:
Less
Explanation:
Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.
Dos cargas puntuales iguales y negativas, q1=q2=-24micro C se localizan en x=0 y y=38m y x=0 y y=-7m, respectivamente. Calcula la magnitud de la fuerza electrica total en N que ejercen estas dos cargas sobre una tercera, tambien puntual, Q=26micro C en y=0 y x=16m
Answer:
F_net = 9.87 10⁻⁴ N
Explanation:
Let's use that force is a vector magnitude
∑ F = F₁₃ + F₂₃
De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract
the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m
∑ F = F₁₃ - F₂₃
F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]
in this case q₁ = q₂ = q
F_net = k q q₃ ( )
let's look for the distance
r₂₃ = y₂ - y₃
r₂₃ = -7 -16
r₂₃ = - 23 m
r₁₃ = 38 - 16
r₁₃ = 22 m
let's calculate
F_net = 9 10⁹ 24 26 10⁻¹² ( )
F_net = 5.616 ( 1.758 10⁻⁴ )
F_net = 9.87 10⁻⁴ N
Air in a thundercloud expands as it rises. If its initial temperature is 292 K and no energy is lost by thermal conduction on expansion, what is its temperature when the initial volume has tripled
Answer:
Explanation:
It is a case of adiabatic expansion .
[tex]T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}[/tex]
T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.
Given ,
V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .
[tex]( 3 )^{\gamma-1}= \frac{292}{ T_2}[/tex]
[tex]( 3 )^{1.4-1}= \frac{292}{ T_2}[/tex]
1.552 = 292 / T₂
T₂ = 188 K .