A skydiver has a mass of 140 kg. At what speed will she have a momentum
of 10,000 kg•m/s?

The coaster starts at rest, so the kinetic energy (KE) at point A is 0. It is situated 33 m above ground, so its potential energy (PE) at A is

mgh = (3000 kg) (9.80 m/s²) (33 m) = 970,200 J

The total energy is the same, 970,200 J.

Assuming no energy is lost to friction or sound etc, energy is conserved throughout the coaster's motion, so the total energy should be the same at each point.

At point B, the coaster has dropped to a height of 10 m, so it has PE

mgh = (3000 kg) (9.80 m/s²) (10 m) = 294,000 J

which means it must have KE

970,200 J = KE + 294,000 J   →   KE = 676,200 J

which gives the coast a speed v at point B of

1/2 mv ² = 1/2 (3000 kg) v ² = 676,200 J   →   v ≈ 21.2 m/s

At point C, the coaster has a speed of 16.0 m/s, so it has KE

1/2 mv ² = 1/2 (3000 kg) (16.0 m/s)² = 384,000 J

and hence PE

970,200 J = 384,000 J + PE   →   PE = 586,200 J

This lets us determine the height h at C:

mgh = (3000 kg) (9.80 m/s²) h = 586,200 J   →   h ≈ 19.939 m

which means the loop has diameter h - 10 m ≈ 9.94 m.

At point D, the coaster is 15 m above the ground so its PE at D is

mgh = (3000 kg) (9.80 m/s²) (15 m) = 441,000 J

and so its KE is

970,200 J = KE + 441,000 J   →   KE = 529,200 J

and hence has speed v at D

1/2 mv ² = 1/2 (3000 kg) v ² = 529,200 J   →   v ≈ 18.9 m/s

Answer:

the force your friend applied on the box is 40 N.

Explanation:

Given;

speed of the box, v₁ = 1 m/s

force applied to the box, F₁ = 20 N

the speed of the box when your friend pushes it, v₂ = 2 m/s

then your friends applied force, F₂ = ?

Assuming the time, t, through which both forces were applied and mass of the box, m, to be constant;

[tex]F_1 = \frac{mv_1}{t} \\\\\frac{m}{t} = \frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]

[tex]F_2 = \frac{F_1v_2}{v_1} \\\\F_2 = \frac{20\times 2}{1} \\\\F_2 = 40 \ N[/tex]

Therefore, the force your friend applied on the box is 40 N.

Answer:

zero

Explanation:

For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = [tex]\frac{1}{2}[/tex] mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             [tex]K_e = K_p[/tex]

              K_p = ½ m v_e²

              K_p = [tex]\frac{1}{2}[/tex]  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]

        v/c= 2.33 10⁻⁴

Answer: [tex]26.460\times 10^3\ J[/tex]

Explanation:

Given the mass of skier m=72 kg

distance traveled d=75 m

constant speed v=3.4 m/s

If speed is constant  then there must no force acting in the direction of motion

i.e. tension force must be equal to the component of weight

[tex]T=mg\sin 30^{\circ}[/tex]

Work done is given by

[tex]\Rightarrow W=F\cdot d=Td\cos 0^{\circ}\\\Rightarrow W=mg\sin 30^{\circ} d=72\times 9.8\times 0.5\times 75\\\Rightarrow W=26.460\times 10^3\ J[/tex]

Answer:

   F = 196 N

Explanation:

For this exercise we will use Newton's second law,  we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically

Y axis  

       N- W = 0

       N = mg

X axis

       F -fr = ma

In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.

       F- fr = 0

       F = fr

the friction force has the equation

       fr = μ N

       fr = μ mg

we substitute

        F = μ mg

let's calculate

         F = 0.80 9.8 25

         F = 196 N

Answer: 30 to 40 s

Explanation:

OK please your picture not perfect please try again

Answer:

The centripetal acceleration of the girl is 2.468 m/s²

Explanation:

Given;

number of turns, = ¹/₄ Revolution

distance traveled by the girl, d = 25 m

time of motion, t = 5.0 s

The linear speed of the of the girl is calculated as;

[tex]v = \omega \ r\\\\v =(\frac{1}{4}rev \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1}{5 \ s} ) (25 \ m)\\\\v = (0.3142 \ \frac{rad}{s} )(25 \ m)\\\\v = 7.855 \ m/s[/tex]

The centripetal acceleration of the girl is calculated as;

[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(7.855)^2}{25} \\\\a_c = 2.468 \ m/s^2[/tex]

Therefore, the centripetal acceleration of the girl is 2.468 m/s²

Answer: IT B TRUST ME MANN OK

Explanation: OK BYE TRUST

Answer:

The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.

Explanation:

Given;

first net force, F₁ = 100 N

second net force, F₂ = 50 N

If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.

Let a₁ be the acceleration produced by the first net force

then, a₂ be the acceleration produced by the second net force

Thus, a₁ = 2a₂

Answer:

A. estrogen

Explanation:

This is released in the female reproductive organ.

Answer:

432.78 Kg

Explanation:

From the question given above, the following data were obtained:

Distance apart (r) = 6.8 m

Force of attraction (F) = 5.4×10¯⁸ N

Mass of Daffy Duck (M₁) = 86.5 kg

Mass of Minnie Duck (M₂) =?

NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

The mass of Minnie Duck can be obtained as follow:

F = GM₁M₂ / r²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24

Cross multiply

6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24

Divide both side by 6.67×10¯¹¹ × 86.5

M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5

M₂ = 432.78 Kg

Therefore, the mass of Minnie Duck is 432.78 Kg

Explanation:

There are two answers

m/(s/s)=m

or

(m/s)/s=m/s²

Answer:

58.469 kg.m/s are moving in the xy plane

Answer:

popu

Explanation:

2u2uwju2i2je82jei

Answer: B

Explanation:

The vertical component of a vector such as velocity is the magnitude of the vector multiplied by the sine of the angle.

[tex]V_y=48*sin(53)=38.3m/s[/tex]

Answer: the at which the bar conducts now is 5 Js⁻¹

Explanation:

Given the data in the question;

we know that; Heat transfer by conduction is given by;

Q ∝ [tex]A / l[/tex]

such that,

[tex]Q_{1}/Q_{2}[/tex] = [tex]A_{1}l_{2} / A_{2}l_{1}[/tex]

[tex]Q_{2} = Q_{1} A_{2}l_{1}/A_{1}l_{2}[/tex]

so

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r}{2}[/tex])² × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])  

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r^{2} }{4}[/tex]) × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × πr² × 1/4 × [tex]l[/tex] ) / (πr² × 1/2 × [tex]l[/tex] )

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)

[tex]Q_{2} =[/tex]  ( 10 Js⁻¹ × 1/4) / ( 1/2)

[tex]Q_{2} =[/tex] 5 Js⁻¹

Therefore, the at which the bar conducts now is 5 Js⁻¹

Answer:

Ff = F₀ *(576/196)

Explanation:

       [tex]F_{o} = \frac{k*q^{2} }{d^{2}} (1)[/tex]

       [tex]F_{f} = \frac{k*(q/14)^{2} }{(d/24)^{2}} = \frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} (2)[/tex]

       ⇒[tex]F_{f} =\frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} = F_{o} * \frac{576}{196} (3)[/tex]

Answer:

using W=1/2kW2

k=22N/m w=2.9

w=1/2×22×2.9×2.9

w=92.51Joules

Approximately 93J answer is C

maybe a spectrograph ?

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns  to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

                                               = 1/5 ∫ 3*10^-3  d(t)  + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

 The reading of the voltmeter ( V )

V = IR

   = 0.96 mA * 21 k Ω  = 20.16 v

Answer:

the force exerted on the foot by the tibia would be 2975 N

Explanation:

Given the data in the question;

To maintain equilibrium between the foot and the ball vertically, the addition normal normal force [tex]N^>[/tex] (750 N)  and the tension in the Achilles tendon [tex]F^>_{Achilles}[/tex] (2225 N) must be equal to the force exerted on the foot by the tibia;

so

| [tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] | = | [tex]F^>_{Tibia}[/tex] |

so force exerted on the foot by the tibia will be;

| [tex]F^>_{Tibia}[/tex] | = |[tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] |

so we substitute IN OUR VALUES

| [tex]F^>_{Tibia}[/tex] | = 750 N + 2225 N

| [tex]F^>_{Tibia}[/tex] |  = 2975 N

Therefore, the force exerted on the foot by the tibia would be 2975 N

Answer:

v = 10.46 m/s

x =  30.134 m from house

Explanation:

given data

speed = 3.05 m/s

time = 7 s

height = 40.8 mm

solution

we get here first time required to fall that is t

t = [tex]\sqrt{\frac{2\times 40.8}{9.8}}[/tex]       ..................1

t = 2.88 s

now we take here initial speed that is v

so v × t = 3.05 × ( t+ 7)

v = [tex]\frac{3.05\times (2.88 + 7)}{2.88}[/tex]

v = 10.46 m/s

and

when she catch the bagel henrietta was at

x = 3.05 × ( 2.88 + 7)

x =  30.134 m from house

Answer:

h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]

Explanation:

Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )

Determine the ratio of heights  h1/h2

mass of tissues = same

radius of tissues = same

h1 = height of tissue 1

h2 = height of tissue 2

For the first tissue ( Tissue that dropped manually )

potential energy = kinetic energy

mgh = 1/2 mv^2  

therefore the final velocity ( v^2 ) = 2gH  ----- ( 1 )

second tissue ( Tissue that dropped while rotating )

gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )

To determine the ratio of heights we will equate equations 1 and 2

hence :

gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )

∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]

Answer: ITS 1 TRUST ME MAN BYE K

Explanation: OK BYE TRUST YEAH

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so [tex]V_{max1}[/tex] = A × ω

[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s

[tex]V_{max1}[/tex] =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

[tex]V_{max2}[/tex] = A' × ω

[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69

[tex]V_{max2}[/tex] = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s