Answer:
(a) When the resistance R is doubled, I = 1 A
(b) When the peak emf εo is doubled, I = 4 A
(c) When the frequency ω is doubled, I = 2 A
Explanation:
Given;
peak current through the resistor, I = 2.0 A
According to ohms law the peak current through the circuit is given by;
[tex]I = \frac{V}{R}[/tex]
(a) When the resistance R is doubled;
[tex]I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 = \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A[/tex]
(b)When the peak emf εo is doubled
[tex]I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1} = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A[/tex]
(c) When the frequency ω is doubled
Peak current through resistor is independent of frequency
I₂ = 2.0 A
As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:___________.
A) approaches zero.
B) approaches infinity.
C) approaches unity.
D) none of the given answers
Answer:
B) approaches infinity
Explanation:
The capacitive reactance of an AC capacitor is given by;
[tex]X_C = \frac{1}{\omega C } = \frac{1}{2\pi f C}[/tex]
Where;
C is the capacitance
f is the frequency of the ac voltage
[tex]X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi (0) C} \\\\X_C = \frac{1}{0} \\\\X_C = \ infinity[/tex]
Therefore, as the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity.
The correct option is (B) approaches infinity
When making maps of the large-scale universe, astronomers estimate distances to the vast majority of galaxies by using:
Answer:
The comoving distance and the proper distance scale
Explanation:
The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster. Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.
If you are pushing on a crate on a frictionless surface in one direction, and your friend is pushing on the crate in the opposite direction with an equal amount of force. Which of the following statements is the most accurate? a. The crate will not move as the forces cancel each other out b. Because the surface is frictionless, the crate will always move regardless of who is pushing c. The crate can continue to move, but it will move at a constant velocity d. The net force is towards the direction that you are pushing, as you started the crate's motion
Answer:
Its not A..
Explanation:
I chose A - was incorrect
The Kelvin temperature of the hot reservoir of an engine is twice that of the cold reservoir, and work done by the engine per cycle is 50 J.
Calculate:
(a) the efficiency of the engine,
(b) the heat absorbed per cycle, and
(c) the heat rejected per cycle.
Answer:
a) 50%
b) 100 J
c) 50 J
Explanation:
The cold temperature of the reservoir = [tex]T_{c}[/tex]
according to the problem, it is stated that the hot reservoir of an engine is twice that of the cold reservoir, therefore,
the hot temperature of the reservoir [tex]T_{h}[/tex] = [tex]2T_{c}[/tex]
The work done by the engine = 50 J
a) The max efficiency obtainable from a heat engine η = [tex]1 - \frac{T_{c} }{T_{h} }[/tex]
since [tex]T_{h}[/tex] = [tex]2T_{c}[/tex], the equation becomes
η = [tex]1 - \frac{T_{c} }{2T_{c} }[/tex] =
η = [tex]1 - \frac{1 }{2 }[/tex] = 0.5 = 50%
b) The heat absorbed per cycle will be gotten from
η = [tex]\frac{W}{Q}[/tex]
η is the efficiency of the system = 0.5
where W is the work done = 50 J
Q is the heat absorbed = ?
substituting, we have
0.5 = [tex]\frac{50}{Q}[/tex]
Q = 50/0.5 = 100 J
c) The heat rejected per cycle = 50% of the absorbed heat
==> 0.5 x 100 J = 50 J
Calculate the density of a rod of metal in g/cm3, with a mass of 9.58g, a diameter of 8 mm and a height of 3.5cm
Answer:
5.448 g/cm³
Explanation:
Density: This is defined as the ratio of the mass of a body to its volume.
The unit of density is kg/m³ other sub units are g/cm³, mg/mm³.
From the question,
D = m/πr²h......................... Equation 1
Where D = density, m = mass, r = radius, h = height.
Given: m = 9.58 g, r = 8/2 mm = 4 mm = 0.4 cm, h = 3.5 cm
Substitute this values into equation 1
D = 9.58/(3.14×0.4²×3.5)
D = 5.448 g/cm³
Hence the density of the metal rod is 5.448 g/cm³
what is the net force on an object that is experiencing a force of 25 N north, a force of 25 N south, a force of 50 N to the east and a force of 45 N to th west?
Answer:
5 n
Explanation:
25 and 25 cancel each other out and 50-45 is 5
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
Answer:
R_total = 14.57 Ω , V_C = 1.176 V
Explanation:
To solve this circuit we are going to find the equivalent resistance of each branch, let's remember
* Serial resistance
[tex]R_{eq}[/tex] = ∑ [tex]R_{i}[/tex]
* For resistance in parallel
1 / R_{eq} = ∑ 1/R_{i}
We solve the two branches of the wheatstone bridge
Series resistors
Branch B
R_B = Rb + R4
R_B = 2 + 18
R_B = 20 Ω
Branch C
R_C5 = Rc + R5
R_C5 = 3 + 12
R_C5 = 15 Ω
Resistance in parallel R_B and R_C5
1 / R_BC = 1 / R_B + 1 / R_C5
1 / R_BC = 1/20 + 1/15 = 0.116666
R_BC = 8.57 Ω
Now we have a single branch, we solve the series resistance
R_total = R_A + R_BC
R_total = 6 + 8.57
R_total = 14.57 Ω
b) they ask us for the voltage in the resistance R_C
Let's remember that the voltage in a series circuit is the sum of the voltages
10 = V_a + V_BC
10 = i R_a + i R_BC = i (R_a + R_BC)
i = 10 / (R_a + R_BC)
i = 10 / (14.57)
i = 0.6863 A
The current in the series circuit is constant
V_BC = i R_BC
V_BC = 0.6863 8.57
V_BC = 5.8819 V
This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.
Branch C
V_BC = i R_C5
i = V_BC / R_C5
i = 5.8819 / 15
i = 0.39213 A
In this branch we have two resistors in series, let's remember that the current in a series circuit is constant
V_C = i R_C
V_C = 0.39213 3
V_C = 1.176 V
"Find the change in gravitational potential energy that the box undergoes as it rises to its final height."
Explanation:
Given that,
Mass of a box is 2.6 kg
(1) We need to find the work done on the box by this force as it is pushed up the 5.00-m ramp to a height of 3.00 m.
It means that the position of the object is 3 m i.e. h = 3 m
Work done = Fd
= mgh
So,
[tex]W=2.6\times 9.8\times 3\\\\W=76.44\ J[/tex]
(2) Now the gravitational potential energy that the box undergoes as it rises to its final height is equal to the work done by the box. So,
Change in potential energy = 76.44 J
A runner jumps off the ground at a speed of 16m/s .At what angle did he jumped from the ground if he lands 8m away?
Answer:
128 degrees
Explanation:
speed divided by distance travelled
the angle he jumped from the ground is 8.92°
The question above is a projectile motion problem, and we can solve it using the formula of range.
⇒ Formula:
R = u²sin2∅/g................... Equation 1⇒ Where:
R = Range of the runneru = initial velocityg = acceleration due to gravity∅ = angle to the horizontalFrom the question,
⇒ Given:
R = 8 mu = 16 m/sg = constant = 9.8 m/s²⇒Substitute these values into equation 1
8 = 16²sin(2∅)/9.8⇒ Solve for ∅
8×9.8 = 16²sin2∅78.4 = 256sin2∅sin2∅ = 78.4/256sin2∅ = 0.306252∅ = sin⁻¹(0.30625)2∅ = 17.83∅ = 17.83/2∅ = 8.92°Hence the angle he jumped from the ground is 8.92°
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An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce x rays. Nonrelativistically, what would be the maximum speed of these electrons
Answer:
1.187 x 10^8 m/s
Explanation:
the potential of the electric field V = 40 kV = 40000 V
the charge on an electron e = 1.6 x 10^-19 C
The energy of an accelerated electron in an electric field is given as
E = eV
E = 1.6 x 10^-19 x 40000 = 6.4 x 10^-15 J
This energy is equal to the kinetic energy with which the electron moves, according to the conservation of energy.
The kinetic energy = [tex]\frac{1}{2}mv^{2}[/tex]
where
m is the mass of the electron = 9.109 x 10^-31
v is the speed of the electron.
Equating the energy, we have
6.4 x 10^-15 = [tex]\frac{1}{2}*9.109*10^-31*v^{2}[/tex]
6.4 x 10^-15 = 4.55 x 10^-31 [tex]v^{2}[/tex]
[tex]v^{2}[/tex] = 1.41 x 10^16
[tex]x^{2} v = \sqrt{1.41*10^{16}}[/tex] = 1.187 x 10^8 m/s
While taking the stairs it takes you 10 seconds to reach the top. The next time you take the same stairs, it takes you 5 seconds to reach the top stair. During which of these trips up the stairs did you use more power to climb? Explain your answer in complete sentences with proper spelling, grammar, and other language mechanics.
Answer:
P₂ = 2 P₁
we conclude that in the second time the power used is double that in the first rise
Explanation:
In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.
Now we can analyze the required power,
P = W / t
From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,
therefore the first time the power is
P₁ = E / 10
P₁ = 0.1 E
for the second time the power is
P₂ = E / 5
P₂ = 0.2 E
we see that the power in the second case is
P₂ = 2 P₁
Therefore, we conclude that in the second time the power used is double that in the first rise.
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV. (in nm)
Answer: Wavelength of 424 nm can produce photoelectrons from silver
Explanation:
[tex]\phi=h\times \nu_o=\frac{hc}{\lambda}[/tex]
[tex]\phi[/tex] = work function = energy of photon
h = Planck's constant = [tex]6.63\times 10^{-34}Js[/tex]
[tex]\nu_0[/tex] = frequency of the metal
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] =longest wavelength of the radiation
Now put all the given values in the above formula, we get the wavelength of the photons.
[tex]\lambda=\frac{hc}{\phi}[/tex]
[tex]\lambda=\frac{6.63\times 10^{-34}\times 3\times 10^8m/s}{2.93\times 1.6\times 10^{-19}J}[/tex] ( as 1ev=[tex]1.6\times 10^{-19}J[/tex] )
[tex]\lambda=4.24\times 10^{-7}m[/tex]
[tex]1nm=10^{-9}m[/tex]
[tex]\lambda=424nm[/tex]
Here, wavelength of 424 nm can produce photoelectrons from silver
2. A 15 kg mass fastened to the end of a steel wire of un-
stretched length 0.5 m is whirled in a vertical circle with an
angular velocity of 2 rev/s at the bottom of the circle. The cross
section of the wire is 0.02 cm2. Calculate the elongation of the
wire when the weight is at the lowest point of the path. Steel
has Y.M.= 2.0 x 1011 Pa. [1.66mm]
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm
A catapult flings a stone at 16 m/s, giving it 1892 J of kinetic energy. What is the mass of the stone?
Calculate the mass of the stone.
Formula used:-Kinetic energy = 1/2 × m × v²
Solution:-We know that,
Kinetic energy = 1/2 × m × v²
⇒ 1892 = 1/2 × m × ( 16 )²
⇒ 1892 = 1/2 × m × 256
⇒ 1892 = 128m
⇒ m = 1892/128
⇒ m = 14.78 kg
what is the pressure of the gas inside the apparatus on teh right if the outside pressure P atm = 750mmHg
Complete question:
Check file uploaded for diagram of the apparatus
Answer:
The pressure of the gas inside the apparatus is 1000 mmHg
Explanation:
Given;
outside pressure, Patm = 750 mmHg
change in height of the two columns, Δh = 25 cm = 250 mm
The pressure of the gas inside the apparatus is given by;
P(gas) = P(atm) + P(Hg)
where;
P(atm) is the outside pressure
P(Hg) is the gauge pressure = height difference of the two columns.
P(gas) = 750 mmHg + 250 mmHg
P(gas) = 1000 mmHg
Therefore, the pressure of the gas inside the apparatus is 1000 mmHg
The pressure of the gas inside the apparatus is 1000 mmHg.
Calculation of the pressure:Since
outside pressure, Patm = 750 mmHg
change in height of the two columns, Δh = 25 cm = 250 mm
so we know that
The pressure of the gas inside the apparatus is
P(gas) = P(atm) + P(Hg)
here;
P(atm) is the outside pressure
P(Hg) is the gauge pressure = height difference of the two columns.
So,
P(gas) = 750 mmHg + 250 mmHg
P(gas) = 1000 mmHg
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A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comfortable internal pressure of 101 kilopascals. How deep can it dive in the ocean before it would risk collapsing from the pressure
Answer:
The depth will be equal to 6141.96 m
Explanation:
pressure on the submarine [tex]P_{sea}[/tex] = 62 MPa = 62 x 10^6 Pa
we also know that [tex]P_{sea}[/tex] = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
[tex]P_{sea}[/tex] = 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine [tex]P_{g}[/tex] = 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure [tex]P_{sea}[/tex], we have
62 x 10^6 = 10094.49h
depth h = 6141.96 m
A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm.
Required:
What is the gauge pressure at the bottom of the cylinder?
Answer:
Pressure, P = 3724 Pa
Explanation:
Given that,
Depth of water, [tex]h_w=20\ cm =0.2\ m[/tex]
Depth of oil, [tex]h_o=40-20=20\ cm=0.2\ m[/tex]
The density of water, [tex]d_w=1000\ kg/m^3[/tex]
The densinty of oil, [tex]d_o=900\ kg/m^3[/tex]
We need to find the gauge pressure at the bottom of the cylinder. So, total pressure is equal to :
[tex]P=d_wgh_w+d_ogh_o\\\\P=(d_wh_w+d_oh_o)g\\\\P=(1000\times 0.2+900\times 0.2)\times 9.8\\\\P=3724\ Pa[/tex]
So, the gauge pressure at the bottom of the cylinder is 3724 Pa.
HELP PLEEAAAASSSEEEEEEE What is the definition of net force?
Answer:
the sum of all force being applied to an object.
Explanation:
A 50g marble is moving at 2m/s when it strikes a 20g marble at rest. Immediately after the collision, the 50g ball is moving at 1m/s. Is this an elastic collision
Answer:
Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.
Explanation:
Given;
mass of first marble, m₁ = 50g = 0.05 kg
initial velocity of the first marble, u₁ = 2 m/s
mass of second marble, m₂ = 20 g = 0.02 kg
initial velocity of the second marble, u₂ = 0
final velocity of the first marble, v₁ = 1 m/s
Let the final velocity of the second marble, = v₂
Determine the final velocity of the second marble by applying principle of momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.05 x 2 + 0.02 x 0 = 0.05 x 1 + 0.02v₂
0.1 = 0.05 + 0.02v₂
0.02v₂ = 0.1 - 0.05
0.02v₂ = 0.05
v₂ = 0.05 / 0.02
v₂ = 2.5 m/s
During inelastic collision both objects will move at the same velocity after collision.
During elastic collision both objects will move at different velocities after collision.
Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.
An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 8.30 m before stopping. How far does the lighter fragment slide
Answer:
the distance d traveled by the lighter fragment is 58.1 m.
Explanation:
mass of the lighter fragment = m
the lighter fragment traveled a distance = ?
mass of the heavier fragment = 7m
the distance covered by the heavier fragment = 8.30 m
The two particles will be given the same amount of energy from the explosion. This energy is used to do work by the two fragments.
work done by heavier fragment w = mgd
where m is the mass
g is acceleration due to gravity
d is the distance traveled.
substituting, the work done by the heavier fragment is
w = 7m x g x 8.3 = 58.1mg
The same way, the lighter fragment does work of
w = mgd
equating the two work done since they are given the same amount of energy from the explosion, we have
58.1mg = mgd
mg cancels out, we have
the distance d traveled by the lighter fragment d = 58.1 m
What is the maximum distance allowed between the center of hole #2 and datum B as seen in the front view?
Answer:
4.003" (inches )
Explanation:
The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .
Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is
A 11.0kg bucket is lowered vertically by a rope in which there is 164N of tension at a given instant.
Determine the magnitude of the acceleration of the bucket.
Express your answer to two significant figures and include the appropriate units.
Answer:
The magnitude of the acceleration is 5.11 m/s²
Explanation:
Given;
Tension on the rope, T = 164 N
mass of the bucket, m = 11 kg
Weight of the rope is given by;
W = mg = 11 x 9.8 = 107.8 N
According to Newton's second law of motion, the tension on the rope is given by;
T = W - ma
ma = W - T
ma = 107.8N - 164N
ma = -56.2 N
a = -56.2 / m
a = -56.2 / 11
a = -5.11 m/s²
The magnitude of the acceleration is 5.11 m/s²
what is force??
...
Answer: force is a push or pull that results in the movement of an object ..
Explanation:
Hope this helps you!!!!
Suppose a big chunk of gold is submerged in water and its volume is found to be 12.5 cm?
Compute the mass of the chunk of gold in grams if you know the density is 19.3 g/cm2. Round
appropriately.
Answer:
The mass of the chunk of gold is 241.25 g
Explanation:
Since density is defined as mass divided by volume, we can solve for the mass (m) via its equation:
[tex]density=\frac{mass}{volume} \\19.3=\frac{m}{12.5} \\m=19.3\,(12.5)\, grams\\m = 241.25 \,\,g[/tex]
What is Physics to you? What do you know about it?
Answer:
Physics is the study of matter and the way living things behave everyday....It is related to maths in how we measure the way objects or people do specific things physics has many branches under it that can be helpful too.....Physics teaches us about the world,about the mechanical things we do about air,space,matter about the solar system and about simple machines and things that we do everyday in life.... What we do everyday is related to physics....
The coil has a radius of 8 cm and 110 turns. While moving the magnet closer towards the coil, at a certain point in time, the voltage meter shows -310 mV. What is the magnitude of the rate of change of the flux through the coil?
Answer:
The magnitude of the rate of change of the flux through the coil is 2.82 x 10⁻³ T.m²/s
Explanation:
Given;
radius of coil, r = 8 cm = 0.08 m
number of turns of the coil, N = 110 turns
the induced emf through the coil, E = -310mV
The induced emf through the coil is given by;
[tex]E = - N\frac{d \phi}{dt} \\\\\frac{d \phi}{dt} = \frac{E}{-N}[/tex]
Where;
dФ/dt is the magnitude of the rate of change of the flux through the coil
[tex]\frac{d\phi}{dt} = \frac{-0.31}{110} \\\\\frac{d\phi}{dt} =2.82*10^{-3} \ T.m^2/s[/tex]
Therefore, the magnitude of the rate of change of the flux through the coil is 2.82 x 10⁻³ T.m²/s
Define fluid flow. Write five difference between uniform and non uniform flow.
Answer:
Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.
Difference:
Whereas in real fluids velocity varies across the section. But when the size and shape of cross section are constant along the length of channels under consideration, the flow is said to be uniform. A non-uniform flow is one in which velocity is not constant at a given instant.
PLEASE HELP
It's the part of the Scientific Method that describes the steps to the experiment. What is is called?
A. materials B.procedure C.purpose D.hypothesis
Answer:
B
Explanation:
Because The Scientific Method is a way to identify a problem or Solve and in the choice above the best way to describe the scientific method will a procedure to achieve something
The procedure is the part of the Scientific Method that describes the steps of the experiment, therefore the correct answer is the option B
what is the scientific investigation?Scientific investigation is the process of looking for answers by doing extensive research and finding the answers through experimental results.
The scientific investigation very much relies on true experimental results that can be supported by evidence.
To establish facts or generate information, the scientific method employs a sequence of stages. Although the general procedure is generally known, the particulars of each stage may vary based on what is being examined and who is performing it. Only questions that can be verified or refuted by testing can be addressed using the scientific method.
Thus, The correct response is option B because the procedure is the component of the scientific method that describes the steps of the experiment.
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Plane-polarized light is incident on a single polarizing disk, withthe direction of E0 parallel to thedirection of the transmission axis. Through what angle should thedisk be rotated so that the intensity in the transmitted beam isreduced by a factor of each of the following?
(a) 2.20
(b) 5.20
(c) 12.0
Answer;
Cos²စ= I/Io
So
A. Cos²စ = 1/2.2
Cosစ= √1/2.2
စ = cos^-1 0.68
= 47.2°
B.
Cosစ = √1/5.2
စ = ,cos^-1 0.4385
= 64°
C.
Cosစ = √1/12
စ = cos^-1 0.2886
= 73.2°
A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T.the force on a proton is
Answer:
The force on the proton is 7.85 x 10⁻¹² N
Explanation:
Given;
kinetic energy of the proton, K.E = 2MeV = 2 x 10⁶ x 1.602 x 10⁻¹⁹ J
= 3.204 x 10⁻¹³ J
magnitude of the magnetic field, B = 2.5 T
The kinetic energy of the proton is given by;
[tex]K.E = \frac{1}{2} m v^2\\\\v^2 = \frac{2K.E}{m}\\\\v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2*3.204*10^{-13}}{1.67 *10^{-27}}} \\\\v = 1.959*10^7 \ m/s[/tex]
The force on the proton moving perpendicular to magnetic field is given by;
F = qvB
F = 1.602 x 10⁻¹⁹ x 1.959 x 10⁷ x 2.5
F = 7.85 x 10⁻¹² N
Therefore, the force on the proton is 7.85 x 10⁻¹² N