A skier (m=55.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally.
a. If h = 6.50 m and D = 11.4 m, find H.
b. Find her total kinetic energy as she reaches the ground.

A Skier (m=55.0 Kg) Starts Sliding Down From The Top Of A Ski Jump With Negligible Friction And Takes

Answers

Answer 1

a. The value of H is 11.51 m

b. The total kinetic energy as she reaches the ground is 6.21 kJ.

What is mechanical energy?

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a skier (m=55.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. Value of h = 6.50 m and D = 11.4 m.

Time taken to reach the ground after take off is

t = √2h/g

t = √(2x6.50) /9.81

t =1.15 s

The speed at that time is

speed = distance /time

v = 11.4 /1.15

v = 9.913 m/s

a. From energy conservation principle,

v = √2g(H-h)

Plug the values, we get

9.913 = √(2 x 9.81 (H - 6.5)

Squaring both sides and solving, we have

H = 11.51 m

Thus, the value of H is 11.51 m.

b. The energy at the ground will be only potential energy

Total kinetic energy at the ground = mgH

T.K.E = 55 x 9.81 x 11.51

T.K.E =6.21 kJ

Thus, the total kinetic energy is 6.21 kJ.

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Related Questions


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.

Answers

C. Increase to 4 times its original magnitude

5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.
(a) How much charge is stored in the 5.00-μF capacitor?
(b) What is the potential difference across the 10.0-μF capacitor?

Answers

(a)  The charge stored in the 5.00-μF capacitor is 37.2  μC.

(b) The potential difference across the 10.0-μF capacitor is 3.72 V.

What is capacitor?

The capacitance of a capacitor is defined as the ratio of the charge stored and the potential difference between the capacitor.

The capacitance of a capacitor is denoted by C and expressed as

C = Q/V

Given, 5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.

(a) The equivalent capacitance is

1 / Ceq = 1 / C₁ +1 / C₂ + 1/ C₃

Substitute the values, we get

Ceq = 3.1  μF

The charge stored in 5.00-μF capacitor is

Q  = Ceq x V

Q = 3.1  μF x 12 V

Q = 37.2  μC

Thus, the charge stored in the 5.00-μF capacitor is 37.2  μC

(b) The potential difference across the 10.0-μF capacitor is given by

V = Q/C₂

Put the values, we get

V = 37.2 / 10

V = 3.72 V

Thus, the potential difference across the 10.0-μF capacitor is 3.72 V.

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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?

Answers

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.

A bumblebee
is flying towards a flower in a
straight line at 4.09 m/s when it begins to
accelerate at 1.01 m/s².
How long does it take the bee to reach the
flower if it is 23.4 m away?

Answers

Answer:

given -

initial velocity = 4.09 m/s

acceleration = 1.01 m/s²

distance = 23.4 m

time = ?

using second formula of motion,

s = ut + 1/2 at².

where, s = distance

u = initial velocity

t = time

a = acceleration

23.4 = 4.09(t) + 1/2(1.01)(t) ²

23.4 = 4.09t + 2.02t²

2.02t² + 4.09t - 23.4 = 0

solving the equation by using quadratic formula

Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :

a = 2.02

b = 4.09

c = -23.4

we get t=2.539 or t= -4.563

time cannot be negative so

t= 2.539 sec = 2.6 Sec is the answer

Which object has the least thermal energy?

Answers

Answer:

2kg brick

Explanation:

2kg brick at 25 degrees.

Answer: A 2kg brick at 20*C

Explanation:

Determine the amount of power
used in holding a 25 kg box, 1.5
meters above the floor, for 60
seconds.

[?] W

(answer is not 6.13)

Thank you in advance!

Answers

Here is your answer mate,

Question,

[tex]Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds[/tex]

Answer,

Power is equal to work done per unit time

Work is force × displacement

SI UNIT OF WORK Newton meter

SI UNIT OF POWER Watt

[tex][/tex]

Solution,

[tex][/tex]

Given,

[tex]MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: [/tex]

[tex][/tex]

WORK DONE (done against gravity) =

mass×acceleration due to gravity ×height

WORK = 25× 10× 1.5

[tex]\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: [/tex]= 375 Nm

[tex][/tex]

Now

POWER =

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} [/tex]

POWER

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt [/tex]

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25[/tex]

[tex]Therfore\: your \: answer\: is\: 6.25[/tex]

[tex][/tex]

Check this,

[tex]Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10[/tex]

[tex][/tex]

Have a good day

Which of the following causes a car's passenger to lean in the direction
opposite the direction in which the car is turning?
OA. Friction
OB. Inertia
OC. Centripetal acceleration
OD. Centripetal force

Answers

Explanation:

This case is inertia.

Mass is directly proportional to inertia

Entropy is how quickly things get messy.
O A. True
OB. False



Answer : False

Answers

Answer:

false

Explanation:

it cant defined the messy and clean states

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 ___.​

Answers

For objects like insulators (plastics), they can get charged by Friction.For metals, where there are more loosely bound electrons, they can get charged without contact by inductionFor metals, they can also be charged by direct contact by Electrostatic Induction.

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?

Answers

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

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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?

Answers

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 * distance

The work done by the force = 20 * 1000 = 20000J

The work done is positive and is equal to 20000 J

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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

Answers

The steel would expand by 4. 8 * 10^-3 cm

How to determine the linear expansion

The 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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Need help with this question!!

Answers

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₋₁

⁵⁶Fe₂₆ → ⁵⁶Mn₂₅ + ⁰e₋₁

²¹⁰Pb₈₂ → ²¹⁰Tl₈₁ + ⁰e₋₁

²⁴¹Pu₉₄ → ²⁴¹Am₉₅ + ⁰e₋₁

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Which picture correctly shows the path of refracted light rays given an object outside the focal point? Select one: a. A b. B c. C d. D

Answers

Answer:

Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor

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.

Answers

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 ms

Explanation:

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.

Units

When 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

Frequency

The frequency of the wave is 1/(0.460 ms) ≈ 2.17 kHz.

Period

Because 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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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

Answers

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 grocery cart is pushed with a force of 21.4 N. If 1,974 J of work is done in pushing the grocery cart through the store, what is the total distance that the grocery cart has traveled?

Answers

Answer:

Given - Force = 21.4 N

Work done = 1974 J

To find - Distance

Solution -

Work done = Force * displacement

1974 j = 21.4 N * displacement

1974/21.4 = displacement

92.24

[tex]\\ \rm\Rrightarrow Work=Force\times Displacement [/tex]

[tex]\\ \rm\Rrightarrow 1974=21.4×Displacement[/tex]

[tex]\\ \rm\Rrightarrow Displacement=92.24m[/tex]


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.​

Answers

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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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.)

Answers

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 N

What is tension?

Tension is a type of pulling force due transmitted by means of a string or cable.

Force = mass * acceleration due to gravity

a) 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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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​

Answers

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 path

r = 2cosθ

Angular speed when the ball moves 15 degrees

ωf² = ωi² + 2αθ

where;

θ is angular displacement in radians, 15⁰ = 15 x π/180 rad

ωf² = (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 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?

Answers

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 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

Answers

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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The answer is not B.

Answers

Answer:

D

Explanation:

15m/ s

i hope this helps

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?

Answers

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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HELP!! ASAP!!
2. Describe the difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl. Which molecule contains a polar covalent bond and which molecule contains a nonpolar covalent bond? Explain your reasoning alongside describing the differences between the types of bonds.

3. How is the metallic bonding different than ionic or covalent bonding? What are some properties of metals that result from this type of bonding? Explain/connect how the nature of the bonding leads to the properties of metallic substances.
Answer:

Answers

The difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl are HCl is a polar covalent compound because the chloride ion is extra electronegative than the hydrogen ion.

Why HCl is polar and H2 is now no longer?

HCl is a polar molecule. This is due to the fact the Chlorine (Cl) atom withinside the HCl molecule is extra electronegative and does now no longer proportion the bonding electrons similarly with Hydrogen (H). But H2 And Cl2 are nonpolar because of comparable electronegativity of each the atoms withinside the molecule H2 And Cl2 .

Hydrogen chloride is a diatomic molecule, such as a hydrogen atom H and a chlorine atom Cl related with the aid of using a polar covalent bond. The chlorine atom is an awful lot extra electronegative than the hydrogen atom, which makes this bond polar.

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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)

Answers

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]

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.

Answers

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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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?

Answers

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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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

Answers

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 ÷ time

Total Work done = Potential energy (P.E) = mgh

Total work done = mgh

Total work done = 1kg × 10m/ × 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

. 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.

Answers

(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 field

The 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 electron

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

For proton

E = (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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