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.

Answer:

false

Explanation:

it cant defined the messy and clean states

The force the rod exerts on the ball at the given angle is determined as 3.94 N.

The force exerted is calculated as follows;

F = ma

F = mv²/r

F = mω²r

where;

r = 2cosθ

ωf² = ωi² + 2αθ

where;

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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Answer:

D

Explanation:

15m/ s

i hope this helps

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.

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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The work done is positive and is equal to 20000 J

Work done is defined as the product of force and the distance moved by the force.

Mathematically:

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

The work done is positive and is equal to 20000 J

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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

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,

[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

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:

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.

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

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

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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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

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.

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:

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The specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

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 velocity of the air relative to the runner is 5 m/s.

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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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.

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

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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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.

Total 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

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]

Answer:

2kg brick

Explanation:

2kg brick at 25 degrees.

Answer: A 2kg brick at 20*C

Explanation:

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.

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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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]

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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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) 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.

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

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

E = 5.57 x 10⁻¹¹ N/C

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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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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Explanation:

This case is inertia.

Mass is directly proportional to inertia

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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The mass and tension due to the system are as follows:

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

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)  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.

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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