Answer:
Fritional force
Explanation:
Fritional force is a force that always opposes the motion of an object and is frequently ignored
C
Explanation:
Applied force, air resistance
A person lifts a ball of mass 1 kg through the
height of 1 m in 10 s. The average power
supplied by him is [Take g = 10 m/s²]
(2) 1 W
(4) 2 W
(1) 10 W
(3) 0.1 W
Answer:
1 W
Explanation:
Power is the ratio of the total work done by the body to the time it will takes to do the work.
Therefore Power = total work done ÷ timeTotal Work done = Potential energy (P.E) = mgh
Total work done = mgh
Total work done = 1kg × 10m/s² × 1m
Total work done = 10 J
Therefore Total Work done is 10J.
And then time is 10 s.
Power = Total Work done divided by time
Power = 10 J÷ 10 s = 1 W
Therefore Power is 1W
If the distance between two objects is cut in half, what happens to the
gravitational force between them?
A. It decreases to 1/2 its original magnitude.
B. It decreases to 1/4 its original magnitude.
O
C. It increases to 4 times its original magnitude.
D. It increases to 2 times its original magnitude.
An object is released from an aeroplane which is diving at an angle of 30° from the horizontal with a speed of 50m/s. If the plane is at a height of
4000m from the ground when the object is released, find
(a) the velocity of the object when it hits the ground.
(b) the time taken for the object to reach the ground.
please I need the solution urgently.
a)The velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec
b) The time taken for the object to reach the ground will be 11.4 × 10³ sec.
What is projectile motion?The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.
The velocity in the x-direction;
[tex]\rm v_x = (vcos \theta)^2 +2gh \\\\ v_x = (50 cos 30)^2+ 2 \times 9.81 \times 4000 \\\\ v_x = 80355 m/sec[/tex]
Velocity in y-direction;
[tex]\rm v_ y = (vsin \theta)^2 + 2gh \\\\ v_ y = (25)^2+2 \times 9.81 \times 4000 \\\\ v_y =79105 \ m/sec[/tex]
The resultant velocity is found as 1.12 × 10⁵ m/sec.
The time taken to reach the ground is found as;
[tex]\rm v = u+gt \\\\ 1.12 \times 10^5 \ m/sec = 50 \ m/sec + 9.81 \times t \\\\ t = 11.4 \times 10^3 \ sec[/tex]
Hence, the velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec, and the time taken for the object to reach the ground will be 11.4 × 10³ sec.
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. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required magnitude and direction of the field when the point charge is: (a) an electron? (b) a proton? Comment on the obtained values.
(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
Magnitude of electric fieldThe magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
E is the electric fieldm is mass of the particleg is acceleration due to gravityq is charge of the particleFor an electronE = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
For protonE = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
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What work do you think is done when you carry a 20 N weight backpack for a 1000 m walk? will the work be positive, negative, or potentially zero?
The work done is positive and is equal to 20000 J
What is work done?Work done is defined as the product of force and the distance moved by the force.
Mathematically:
Work done = force * distanceThe work done by the force = 20 * 1000 = 20000J
The work done is positive and is equal to 20000 J
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a concave lens creates a virtual image at -47.0 cm and a magnification of +1.75. what is the focal length?
The focal length of given concave lens will be -26.85 cm
The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.
Given concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.
We have to find focal length
The focal length can be found out by following way:
Magnification = m = +1.75
m = hi/h
hi = -47 cm
1.75 = -47/h
h = -26.85 cm
So the focal length of given concave lens will be -26.85 cm
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Answer:
26.85
Explanation: I got it right on Acellus.
if you were to draw a 3rd harmonic of a tube open at both ends, what would you draw at the ends of the tube?
A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.
In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.
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The answer is not B.
Answer:
D
Explanation:
15m/ s
i hope this helps
What can a line graph be used to do?
A. Make predictions
B. Make comparisons
C. Identify positive correlation
D. Identify negative correlation
Answer:
B
Explanation:
i's 100% paset bro.....
A car and a truck, starting from rest, have the same acceleration, but the truck moves twice the length of time. Compared with the car, the truck will travel
Compared with the car, the truck will travel four times as far.
The initial velocity of car and truck, u = 0
The acceleration of both the truck and car = a
The length of time for the acceleration = t
Let the time the truck accelerated be 2t
How to calculate the distance traveled by car and truck?The distance traveled by car is calculated as;
s = ut + ¹/₂at²
s₁ = 0(t) + ¹/₂at²
s₁ = ¹/₂at²
The distance traveled by truck
s = ut + ¹/₂at²
s₂ = 0(2t) + ¹/₂a (2t)²
s₂ = ¹/₂a x 4t²
s₂ = 4 (¹/₂at²)
s₂ = 4(s₁)
We can conclude that the Truck distance is four times the car distance.
Therefore, Compared with the car, the truck will travel four times as far.
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Please help me!!
1. For objects like insulators (plastics), they can get charged by ___.
2. For metals, where there are more loosely bound electrons, they can get charged without contact by ___.
3. For metals, they can also be charged by direct contact by ___.
Consider the baby being weighed in Figure 4.25.
Figure 4.25
(a) What is the mass of the child and basket if a scale reading of 104 N is observed?
kg
(b) What is the tension T in the cord attaching the child to the scale?
N
(c) What is the tension T' in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?
N
(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. (Do this on paper. Your instructor may ask you to turn in this work.)
The mass and tension due to the system are as follows:
The mass of the child and scale = 10.6 kgThe tension T, in the cord attaching the child to the scale = 104N The tension T', in the cord attaching the scale to the ceiling T' = 108.9 NWhat is tension?Tension is a type of pulling force due transmitted by means of a string or cable.
Force = mass * acceleration due to gravitya) The mass of the child and scale = 104/9.81 = 10.6 kg
b) The tension T, in the cord attaching the child to the scale = scale reading = 104N
c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale
T' = 104 + (0.5 * 9.81)
T' = 108.9 N
d) The sketch is attached in the picture
In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.
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Which object has the least thermal energy?
Answer:
2kg brick
Explanation:
2kg brick at 25 degrees.
Answer: A 2kg brick at 20*C
Explanation:
A radioactive contaminant gives an unfortunate lab rat a dose of 1500 rem in just 1 minute. Assuming that the half life of the radioactive isotope in the contaminant is much longer than 1 minute.What would the activity of the contaminant be if the contaminant is 1.1MeV Beta Emitter?
The activity of the contaminant will be 3.5 ×10⁻¹⁰ Bq if the contaminant is 1.1MeV Beta Emitter.
Disclaimer, here the m is taken as 0.5kg.
Given, REM = 1500
Time, t= 1 minute =60 seconds
For beta particles,
RBE (Relative Biological Effectiveness) =2.
Absorbed dose, D =REM/RBE= 1500/2 = 750rad
As 100 rad=1 J/kg,
750 rad=7.5 J/kg
So, D=7.5 J/kg
D= total energy absorbed (E)/mass (m)
7.5 J/kg=E/m
E =7.5 ×0.5 = 3.75J
Therefore, the number of beta particles absorbed is 3.75 J.
The number of beta particles absorbed,
As 1 MeV= 1.6 ×10⁻¹³J
So, converting 1.1 MeV in J, 1.1×1.6 ×10⁻¹³J=1.76×10⁻¹³J
Numbers of beta particles,
n=3.75J/1.76×10⁻¹³J
n=2.13×10⁻¹²
Hence, the number of beta particles is 2.13×10⁻¹².
Activity, A= n/t
where n is the number of beta particles and t is the time.
A= n/t=(2.13×10⁻¹²)/60
A= 3.5 ×10⁻¹⁰ Bq
So, the activity is 3.5 ×10⁻¹⁰ Bq.
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At a given temperature, lighter gas molecules:
A) have an average speed less than heavier gas molecules.
B) have an average speed greater than heavier gas molecules.
C) ll move more slowly than heavier gas molecules.
At a given temperature, lighter gas molecules, have an average speed greater than heavier gas molecules.
What is temperature?Temperature is the measure of degree of hotness or coldness of a body. It is also the measure of average kinetic energy of a body.
As temperature increases, the average kinetic energy of a body increases.
K.E = ¹/₂mv²
v² = 2K.E/m
v = √( 2K.E/m)
where;
m is massv is velocityK.E is kinetic energyFrom the equation above, at a constant temperature, the speed of a gas molecule will decrease with increase in the mass of the molecules and vice versa.
Thus, at a given temperature, lighter gas molecules, have an average speed greater than heavier gas molecules.
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a 10.00mf parallel-plate capacitor is connected to a 24.0v battery. after the capacitor is fully charged ,the
battery is disconnected without loss of any of charge on the plates.
a) a voltmeter is connected across the two plates without discharging them.what does it read
b)what would the voltmeter read if the plate separationwere doubled?
The answer to the first part of the question is 24 V and the answer to the second part of the question is 48 V.
The formula which relates Charge, Capacitance and Voltage is
Charge = Capacitance × Voltage
Q = CV
where Q denotes the charge, C denotes the capacitance and V denote the voltage.
Using the above formula the charge on the capacitance will be
Q = CV
C = 10 mF
V = 24 V
Q = 240 mC
Charge on capacitance is 240 mC.
Now after we disconnect the battery
C = 10 mF (will remain same)
Q = 240 mC
V = Q/C
V = 240/10 V
V = 24 V
So after we removed the battery, The voltage will remain same.
Now we know that parallel plate capacitor formula is
C = ε(A/d)
from here
C ∝ 1/d, and as there is no loss of charge we can say that V ∝ 1/d
Form C ∝ 1/d and V ∝ 1/C we can state that
V ∝ d {where d is the measurement of separation between the plates}
so if we double the distance between the plates then, the voltage will also get double.
Previously our voltage was 24 V, now if we double the distance between the plates the voltage will also get double, and it will become 48 V.
So, the voltmeter will take a reading of 24 V if voltmeter is connected across the two plates without discharging them, and if we double the plate separation then it will take 48 V as reading.
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A metallic spoon is placed in a hot cup of coffee. If the coffee gives away 190 calories to the spoon to cool down by 0.75°C, what is the mass of the coffee? (Assume that ccoffee = 1.0 cal/gC°.)
According to the definition of calorimetry, the mass of the coffee is 253.33 g.
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
so, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
Q is the heat exchanged by a body of mass m.
c is the specific heat substance.
ΔT is the temperature variation.
Mass of coffee
In this case, Given is :
Q= 190 calories
c= 1
m= ?
ΔT= 0.75 C
Replacing in the equation that allows to calculate heat exchanges is:
190 cal = 1 × m× 0.75 C
Solving:
m= 190 cal÷ (1 × 0.75 C)
m=253.33 g
Finally, the mass of the coffee is 253.33 g.
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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 100-kilogram weight 2-decimeters above the ground with an energy efficiency of 25%. How many repetitions can she do with the energy supplied from a single Oreo cookie? What happens to the number of repetitions that can be done if the efficiency increases?
Answer:
Approximately [tex]325[/tex] (rounded down,) assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
The number of repetitions would increase if efficiency increases.
Explanation:
Ensure that all quantities involved are in standard units:
Energy from the cookie (should be in joules, [tex]{\rm J}[/tex]):
[tex]\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}[/tex].
Height of the weight (should be in meters, [tex]{\rm m}[/tex]):
[tex]\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}[/tex].
Energy required to lift the weight by [tex]\Delta h = 0.2\; {\rm m}[/tex] without acceleration:
[tex]\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}[/tex].
At an efficiency of [tex]0.25[/tex], the actual amount of energy required to raise this weight to that height would be:
[tex]\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}[/tex].
Divide [tex]2.551 \times 10^{5}\; {\rm J}[/tex] by [tex]784\; {\rm J}[/tex] to find the number of times this weight could be lifted up within that energy budget:
[tex]\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}[/tex].
Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.
which of the following statements is true about the suns energy and earth
The correct statement about the sun and the earth is that only about half of the suns energy hits the earths surface.
What is the sun?The sun is the source through which energy reaches the earth. The sun is a large star that produces energy by the process of nuclear fusion.
Hence, the correct statement about the sun and the earth is that only about half of the suns energy hits the earths surface.
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a 1kg ball is bening pushed by the rod to move in horizontal groved smooth slot if it startes from angle teta = zero degree . determaine the force the rod exertes on the ball at teta is =15 dgree if ai this instant the rod moves at angular speed of teta = 1 rad per sec end with angular acceleration theta = 2 rad persec and square the ball is only in contact with the outer side of the slot
The force the rod exerts on the ball at the given angle is determined as 3.94 N.
Force exerted on the rod by the ball
The force exerted is calculated as follows;
F = ma
F = mv²/r
F = mω²r
where;
m is mass of the ballω is angular speed of the ballr is radius of the pathr = 2cosθ
Angular speed when the ball moves 15 degreesωf² = ωi² + 2αθ
where;
θ is angular displacement in radians, 15⁰ = 15 x π/180 radωf² = (1)² + 2(2)(15 x π/180)
ωf² = 2.04
ωf = √2.04
ωf = 1.428 rad/s
F = mω²(2cosθ)
F = (1)(1.428)²(2 x cos15)
F = 3.94 N
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A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrofoam container. When thermal equilibrium is reached, the temperature of the alcohol-water solution is 12°C. What is the specific heat capacity of the alcohol? Assume the sealed container is an isolated system. The specific heat capacity of water is 4.19 kJ/kg · °C. 3.14 kJ/kg \xe2\x88\x99 °C 4.14 kJ/kg \xe2\x88\x99 °C 3.72 kJ/kg \xe2\x88\x99 °C 4.88 kJ/kg \xe2\x88\x99 °C
The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
What is the specific heat capacity?The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."
Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.
Given data;
Mass of liquid sample of Alcohol m₁ = 200-gram
The temperature of alcohol, T₁ = -6°C.
Mass of liquid sample of water m₂ = 400-gram
The temperature of the water, T₂= 20°C.
The specific heat capacity of the alcohol, S₁=?
The specific heat capacity of water is, S₂=4.19 kJ/kg.°C
As we know that;
[tex]\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C[/tex]
Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
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You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, it accelerates because the engines are at full throttle and because there is a catapult that propels the jet forward. You begin to wonder how much force is supplied by the catapult. You look on the Web and find that the flight deck of an aircraft carrier is about 90.0 m long, that an F-14 has a mass of 24800 kg, that each of the two engines supplies 27000 lb of thrust, and that the takeoff speed of such a plane is about 158 mi/h. Estimate the average force on the jet due to the catapult.
You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, the average force on the jet is due to the catapult is mathematically given as
What is the average force on the jet is due to the catapult?
Generally, the equation for acceleration is mathematically given as
[tex]a=\frac{vf^2-vi^2}{2s}\\\\\Therefore\\\\a=\frac{69.29^2-0^2}{2(90}\\\\a=26.673m/s^2[/tex]
In conclusion, The force
F=m*a
F=15100*26.673
F=40272.3N
F 402 KN
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Ivan exerts 20 N of force to lift a basket 2 m. He then carries the basket 5 m across a room to place it on a shelf.
Ivan does work when he
.
The amount of work he does is
The amount of the work done will be 100 J. Work done is denoted by W.
What is work done?Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.
Work may be zero, positive and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force it will be positive.
The given data in the problem is;
F is the force applied = 20 N
d is the displacement = 5.0 m
The work done is found as;
W = F × d
W = 20 N × 5
W= 100 J
Hence, the amount of work done will be 100 J.
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Answer:
lifts the basket, 40J
Explanation:
A piece of steel expands 10 cm when ur is heated from 20 to 50 degrees Celsius. How much would it expand if it was heated from 20 to 60 degrees Celsius
The steel would expand by 4. 8 * 10^-3 cm
How to determine the linear expansionThe change in length ΔL is proportional to length L. It is dependent on the temperature, substance, and length.
Using the formula:
ΔL= α LΔT
where ΔL is the change in length L = 10cm
ΔT is the change in temperature = 60° - 20° = 40° C
α is the coefficient of linear expansion = 1.2 x 10^-5 °C
Substitute into the formula
ΔL = [tex]1.2 * 10^-5 * 10 * 40[/tex]
ΔL = [tex]4.8 * 10 ^-3[/tex] cm
Therefore, the steel would expand by 4. 8 * 10^-3 cm
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Calculate the estimated density of each ball. Use the formula
D = m/V
where D is the density, m is the mass, and V is the volume. Record your calculations in Table A of your Student Guide.
Given that the density of water is 1.0 g/cm3, make a prediction about whether each ball will float in water. Record your prediction in Table A.
What is the estimated density of the table tennis ball? Record your answer to the nearest hundredth.
0.07
g/cm3
What is the estimated density of the golf ball? Record your answer to the nearest hundreth.
The estimated density of the golf ball is 700 kg/m³
What is density?Density is defined as Mass per unit Volume.
In displacement method,
First , we measuring the volume of water displaced by an object which tell us the volume of the object then we will use the physical balance to determine its mass.
Then calculate the density by dividing the mass by the volume.
i.e. D = m/V
Given, Density of water is 1.0 g/cm³
Using displacement method , The estimated volume of golf ball is 100 cm³ and estimated mass is 7g
Then ,
Density = 10 cm³ / 7 g= 0.07 g/ cm³ = 700 kg/m³
So the estimated density of golf ball is 700 kg/m³
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What is the weight of a 0.145kg baseball on the moon if the gravitational acceleration on the moon is approximately 1.625 m/s2
Answer:I would say that it would be about 2/3rds as fast as the moon and i have seen a similar question but it asked different info about it. I also believe that the basketball would have a lower weight due to it being in space. I feel bad saying this but i know that i got some info off this website when working with this and i hope it helps.
Explanation:
Note- I am new to Brainly so sorry if this isn't as helpful as you expected this to be.
A sinewave has a period (duration of one cycle) of 460 μs. What is the corresponding frequency of this sinewave in kHz, expressed to three significant figures (see Study Note 7.1 in the Using numbers (maths) booklet)?
If the frequency of the sinewave is now reduced by a factor of 4.25, what will be the new period value? Express your answer in ms to three significant figures.
Note: in the second part of this question, to avoid rounding errors in your calculations, use the full frequency value from the first part, not the rounded value to three significant figures.
Answer:
See below
Explanation:
Frequency = 1 / period
= 1 / 460 X 10^-6
2173.91 Hz = ~ 2.17 Khz
2.17391 kHz / 4.25 = .511 kHz <====new frequency
period = 1/frequency
= 1955 microseconds =1.96 ms
Answer:
2.17 kHz1.96 msExplanation:
You want to know the frequency in kHz of a sine wave with a period of 460 μs, and its period if the frequency is reduced by a factor of 4.25.
UnitsWhen you are interested in the frequency in kHz, it is convenient to use milliseconds (ms) to express the period. Milliseconds and kilohertz are inverse units.
The period of the given sine wave can be expressed in milliseconds as ...
460 μs = 460×(0.001 ms) = 0.460 ms
FrequencyThe frequency of the wave is 1/(0.460 ms) ≈ 2.17 kHz.
PeriodBecause period and frequency are reciprocals of each other, reducing the frequency by a factor of 4.25 increases the period by the same factor.
The new period is 4.25×0.460 ms = 1.96 ms.
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Need help with this question!!
A beta particle or an electron is released during beta decay. The charges can be positive or negative. Co changes to Ni, Fe to Mn, Pb to Tl, and Pu to Am.
What is beta decay?Beta-decay is the radioactive decay that involves the release of the beta particle or the positron or the electron. The larger nucleus splits during nuclear fission and releases smaller nuclei.
The nuclear reactions are shown as:
⁶⁰Co₂₇ → ⁶⁰Ni₂₈ + ⁰e₋₁⁵⁶Fe₂₆ → ⁵⁶Mn₂₅ + ⁰e₋₁²¹⁰Pb₈₂ → ²¹⁰Tl₈₁ + ⁰e₋₁²⁴¹Pu₉₄ → ²⁴¹Am₉₅ + ⁰e₋₁Learn more about beta decay here:
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What is the magnetic force on a proton that is moving at 5.2 x 107 m/s to the
right through a magnetic field that is 1.4 T and pointing away from you? The
charge on a proton is 1.6 × 10-19 C. Use F = qvx B sin(e)
Hello!
We can use the following equation for magnetic force on a charged particle:
[tex]F_B = qv \times B[/tex]
[tex]F_B[/tex] = Magnetic force (N)
q = Charge of particle (1.6 × 10⁻¹⁹ C)
v = velocity of particle (5.2 × 10⁷ m/s)
B = Magnetic field strength (1.4 T)
This is a cross-product, so the equation can be rewritten to F = qvBsinφ where φ is the angle between the magnetic field and particle velocity vectors.
Since the proton's velocity vector and the magnetic field vector are perpendicular, sin(90) = 1. We can reduce the equation to:
[tex]F_B = qvB[/tex]
Plug in the known values.
[tex]F_B = (1.6*10^{-19})(5.2*10^7)(1.4) = \boxed{1.1648 *10^{-11} N}[/tex]
22. On a day the wind is blowing toward the south at 3m/s, a runner jogs west at 4m/s. What is the velocity of the air relative to the runner?
The velocity of the air relative to the runner is 5 m/s.
What is the relative velocity?We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;
Velocity of the runner = 4m/s. due west
Velocity of the wind = 3m/s due south
The relative velocity is;
Vr = √(4)^2 + (3)^2
Vr = 5 m/s
Learn more about relative velocity:https://brainly.com/question/20813206
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