The center of the Milky Way galaxy lies in the direction of the _constellation, aboutlight-years away.

Answer:

Answer:196 Joules

Explanation:

Hello

Note:  I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work  is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

[tex]F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules[/tex]

Answer:196 Joules

I hope it helps

Answer:

so specific heat capacity of unknown metal is 656.8 J/degree C kg

Explanation:

Here by energy balance we can say that energy given by metal alloy at 100 degree = heat absorbed by water at 18.7 degree C

now we have

[tex]Q_{in} = Q_{out}[/tex]

[tex]390 s (100 - 26.8) = 553 (4186 ) ( 26.8 - 18.7)[/tex]

[tex] 28548 s = 18750349.8[/tex]

now we have

[tex]s = \frac{18750349.8}{28548}[/tex]

[tex]s = 656.8 J/kg ^o C[/tex]

so specific heat capacity of unknown metal is 656.8 J/degree C kg

Answer:

Impulse, J = 9.50 kg-m/s

Explanation:

Mass of the volleyball, m = 0.35 kg

Incoming velocity, u = +3.27 m/s

Outgoing velocity, v = -23.9 m/s

We need to find the magnitude of the impulse that the player applies to the ball. It is equal to the change in momentum. It is given by :

[tex]J=m(v-u)[/tex]

[tex]J=0.35\ kg(-23.9\ m/s-3.27\ m/s)[/tex]

J = -9.50 kg-m/s

So, the impulse that the player applies to the ball is 9.5 m/s but in opposite direction.

Impulse is the product of the mass of an object and the change in its velocity. Here, the magnitude of the impulse applied to the volleyball is determined to be 9.5095 kg*m/s.

The question involves the concept of impulse, which in physics is the change in momentum of an object. The momentum of an object is calculated as its mass times its velocity. If velocity changes, then momentum changes and this change in momentum is what we call 'impulse'.

So in this context, we need to find the change in the volleyball's velocity, which is the final velocity minus the initial velocity. The final velocity is -23.9 m/s and the initial velocity is +3.27 m/s. Subtracting the initial velocity from the final velocity gives us a change in velocity of -27.17 m/s.

Now, the impulse can be calculated by multiplying the mass of the volleyball (0.350 kg) by the change in velocity (-27.17 m/s). Therefore, the impulse applied to the volleyball is -9.5095 kg*m/s. As we are looking for the magnitude of the impulse, we take the absolute value, which is 9.5095 kg*m/s.

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Answer:

Explanation:

m1= 3.65kg

v1= 5 m/s

m2= 1.3 kg

v2= 2.5 m/s

V=?

m1*v1 + m2*v2 = (m1+m2) * V

V= (m1*v1 + m2*v2) / (m1+m2)

V= 4.34 m/s

Answer:

6900 m/s

Explanation:

The mass of the rocket is:

m = 330000 − 280000 (t / 250)

m = 330000 − 1120 t

Force is mass times acceleration:

F = ma

a = F / m

a = F / (330000 − 1120 t)

Acceleration is the derivative of velocity:

dv/dt = F / (330000 − 1120 t)

dv = F dt / (330000 − 1120 t)

Multiply both sides by -1120:

-1120 dv = -1120 F dt / (330000 − 1120 t)

Integrate both sides.  Assuming the rocket starts at rest:

-1120 (v − 0) = F [ ln(330000 − 1120 t) − ln(330000 − 0) ]

-1120 v = F [ ln(330000 − 1120 t) − ln(330000) ]

1120 v = F [ ln(330000) − ln(330000 − 1120 t) ]

1120 v = F ln(330000 / (330000 − 1120 t))

v = (F / 1120) ln(330000 / (330000 − 1120 t))

Given t = 250 s and F = 4.1×10⁶ N:

v = (4.1×10⁶ / 1120) ln(330000 / (330000 − 1120×250))

v = 6900 m/s

Answer:

The difference in the length of the bridge is 0.42 m.

Explanation:

Given that,

Length = 1000 m

Winter temperature = 0°C

Summer temperature = 40°C

Coefficient of thermal expansion [tex]\alpha= 10.5\times10^{-6}\ K^{-1}[/tex]

We need to calculate the difference in the length of the bridge

Using formula of the difference in the length

[tex]\Delta L=L\alpha\Delta T[/tex]

Where, [tex]\Delta T [/tex]= temperature difference

[tex]\alpha[/tex]=Coefficient of thermal expansion

L= length

Put the value into the formula

[tex]\Delta L=1000\times10.5\times10^{-6}(40^{\circ}-0^{\circ})[/tex]

[tex]\Delta L=0.42\ m[/tex]

Hence, The difference in the length of the bridge is 0.42 m.

Answer:

376.8 Volt

Explanation:

N = 100, A = 400 cm^2 = 400 x 10^-4 m^2, B = 0.25 T, f = 60 rps

Maximum value of induced emf is given by

e = N x B x A x w

e = 100 x 0.25 x 400 x 10^-4 x 2 x 3.14 x 60

e = 376.8 Volt

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.27 s

c) Total distance traveled = 402.43 m

Explanation:

a) Maximum speed is obtained after the end of acceleration

        v = u + at

        v = 13.5 + 1.9 x 6.2 = 25.28 m/s

    Maximum speed = 25.28 m/s

b) We have maximum speed = 25.28 m/s, then it decelerates 1.2 m/s² until it stops.

         v = u + at  

         0 = 25.28 - 1.2 t

         t = 21.07 s

    Total time = 6.2 + 21.07 = 27.27 s

c) Distance traveled for the first 6.2 s

          s = ut + 0.5 at²

          s = 13.5 x 6.2 + 0.5 x 1.9 x 6.2² = 120.22 m

   Distance traveled for the second 21.07 s

          s = ut + 0.5 at²

          s = 25.28 x 21.07 - 0.5 x 1.2 x 21.07² = 282.21 m

   Total distance traveled = 120.22 + 282.21 = 402.43 m

Answer:

a) Maximum speed = 25.28 m/s

b) Total time = 27.26 s

c) Total distance traveled = 390,5537

Explanation:

In order to solve the first proble we just have to use the next formula:

[tex]Vf= Vo+ Acc-t\\Vf= 13,5 + 6,2*1,9\\Vf=25,28 m/s\\[/tex]

So the maximum speed would be 25,28 m/s.

THe total time of the trip will be given by adding the inital time plus the velocity divided by the acceleration rate:

[tex]Time= 6,2 s+ \frac{25,28}{-1,2} \\Time= 6,2 +21,06\\Time= 27,26[/tex]

Remember that when dealing with time in physics you will always use positive numbers since there is no negative time.

To calculate the total distance covered we use the next formula:

[tex]D=Vo*t+ 1/2a*t^2\\D= 13,5*6,2 + 1/2(1,9)(6,2)^2\\D=87,75+36,518\\D=124,268m[/tex]

This is the first part now we calculate it with the stopping of the car:

[tex]D=Vo*t+ 1/2a*t^2\\D=25,28*21,06+ 1/2(-1,2)(21,06)^2\\D=532,3968-266,1141\\D= 266,2857meters[/tex]

No we just add the two distances to discover the whole distance:

Total distance= 124,268+266,2857= 390,5537

Answer:

a)

v = 31.15 m/s

t = 0.393 sec

h = 62.5 m

Explanation:

a)

v₀ = initially speed of the stone = 35.0 m/s

v = final speed of the stone = ?

y = vertical displacement of the stone = 13.0 m

a = acceleration due to gravity = - 9.8 m/s²

using the equation

v² = v₀² + 2 a y

v² = 35² + 2 (- 9.8) (13.0)

v = 31.15 m/s

t = time taken

using the equation

v = v₀ + a t

31.15 = 35 + (- 9.8) t

t = 0.393 sec

h = maximum height

v' = final speed at the maximum height = 0 m/s

using the equation

v'² = v₀² + 2 a h

0² = 35² + 2 (- 9.8) h

h = 62.5 m

Answer:

A large elliptical

Explanation:

A S0 type galaxy is a large elliptical.

A S0 type galaxy is a large elliptical.

Answer is C.

Answer:

8.85 x 10⁴ Nm²/C

Explanation:

d = diameter of the conducting sphere = 0.10 m

r = radius of the conducting sphere = (0.5) d = (0.5) (0.10) = 0.05 m

Area of the sphere is given as

A = 4πr²

A = 4 (3.14) (0.05)²

A = 0.0314 m²

σ = Surface charge density = 150 x 10⁻⁶ C/m²

Q = total charge enclosed

Total charge enclosed is given as

Q = σA

Q = (150 x 10⁻⁶) (0.0314)

Q = 4.7 x 10⁻⁶ C

Electric flux through one of the side is given as

[tex]\phi = \frac{Q}{6\epsilon _{o}}[/tex]

[tex]\phi = \frac{4.7\times 10^{-6}}{6(8.85\times 10^{-12})}[/tex]

[tex]\phi [/tex] = 8.85 x 10⁴ Nm²/C

The density of the hydrogen nucleus is approximately 1.49 × 1014 times greater than that of the complete hydrogen atom.

To determine the ratio of the density of the hydrogen nucleus to the hydrogen atom, we need to first find the volumes and masses involved and then calculate the densities.

Volume of proton = (4/3)π(1.0 × 10-15 m)3 = 4.19 × 10-45 m3

Volume of hydrogen atom = (4/3)π(5.3 × 10-11 m)3 = 6.23 × 10-31 m3

Mass of hydrogen atom ≈ 1.67 × 10-27 kg

Density of proton = (mass of proton) / (volume of proton) = 1.67 × 10-27 kg / 4.19 × 10-45 m3 = 3.99 × 1017 kg/m3

Density of hydrogen atom = (mass of hydrogen atom) / (volume of hydrogen atom) = 1.67 × 10-27 kg / 6.23 × 10-31 m3 = 2.68 × 103 kg/m3

Ratio = Density of proton / Density of hydrogen atom = 3.99 × 1017 kg/m3 / 2.68 × 103 kg/m3 = 1.49 × 1014

Therefore, the density of the hydrogen nucleus is approximately 1.49 × 1014 times greater than the density of the complete hydrogen atom.

Answer:

The period of the resulting oscillatory motion is 0.20 s.

Explanation:

Given that,

Spring constant [tex]k= 3\ N/m^2[/tex]

We need to calculate the time period

The object is at rest and has no elastic potential but it does has gravitational potential.

If the object falls then the the gravitational potential change in to the elastic potential.

So,

[tex]mgh=\dfrac{1}{2}kh^2[/tex]

[tex]m=\dfrac{1}{2}\times\dfrac{kh}{g}[/tex]

Where,h = distance

k = spring constant

Put the value into the formula

[tex]m=\dfrac{1\times3\times3.42\times10^{-2}}{2\times9.8}[/tex]

[tex]m=5.235\times10^{-3}\ kg[/tex]

Using formula of time period

[tex]T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{m}{k}}[/tex]

Put the value into the formula

[tex]T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{5.235\times10^{-2}}{3}}[/tex]

[tex]T=0.20\ sec[/tex]

Hence, The period of the resulting oscillatory motion is 0.20 s.

Explanation:

The weight of the car is equal to, [tex]W_c=m\times g[/tex]...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, [tex]F_N=mg\ cos\theta[/tex]

or

[tex]F_N=mg\ cos(13)[/tex].............(2)

The horizontal component of the force is, [tex]F_H=mg\ sin\theta[/tex]

Taking ratio of equation (1) and (2) as :

[tex]\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}[/tex]

[tex]\dfrac{F_N}{W_c}=cos(13)[/tex]

[tex]\dfrac{F_N}{W_c}=0.97[/tex]

or

[tex]\dfrac{F_N}{W_c}=\dfrac{97}{100}[/tex]

Hence, this is the required solution.

The ratio of the magnitude of the normal force to the weight of the car can be determined by taking the cosine of the angle of inclination.

When a car is traveling up a hill inclined at an angle θ above the horizontal, the ratio of the magnitude of the normal force to the weight of the car can be determined. The vertical component of the normal force is N cos θ, while the weight of the car is mg. Since the net vertical force must be zero, these two forces must be equal in magnitude:

N cos θ = mg

To find the ratio, divide both sides of the equation by mg:

N/mg = cos θ

So, the ratio of the magnitude of the normal force to the weight of the car is cos θ.

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Answer:

The power required to move each bale is 30 lbf.ft/sec.

Explanation:

F= 40lbf * sin (30º)

F=20lbf

P= F*v

P=20 lbf * 1.5 ft/sec

P= 30 lbf.ft/sec

Answer:

2.17 A

Explanation:

The magnetic field due to a circular current carrying coil is given by

B = k x 2i / r

For i = 7 A, r = 0.097 m, clockwise

B = k x 2 x 7 / 0.097 = 144.33 k (inwards)

The direction of magnetic field is given by the Maxwell's right hand thumb rule.

The magnetic field is same but in outwards direction as the current is in counter clockwise direction. Let the current be i.

Now, r = 0.03 m, B = 144.33 K, i = ?

B = k x 2i / r

144.33 K = K x 2 x i / 0.03

i = 2.17 A

Final answer:

To create a zero magnetic field at the center of concentric loops, the smaller loop needs to carry approximately 2.15 A in a counterclockwise direction, based on the proportionality of currents and radii.

Explanation:

The question asks about the conditions to achieve a zero magnetic field at the center of concentric circular loops of wire when a conventional current runs through them. For the two loops described, the larger one with a radius of 0.097 m has a current of 7 A flowing clockwise. To counteract the larger loop's magnetic field at the origin and ensure the total magnetic field is zero, the smaller loop with a radius of 0.03 m must have a counterclockwise current. According to Ampere's Law, the magnetic field at the center of a loop due to current is proportional to the current times the number of turns, divided by two times the radius. The larger loop already present creates a magnetic field, and to cancel this magnetic field, the second smaller loop should create an equal and opposite magnetic field. This implies the smaller loop must carry a current I such that (I/0.03 m) = (7 A/0.097 m). Solving for I gives I = (0.03 m / 0.097 m) * 7 A, which equals approximately 2.15 A going counterclockwise.

Answer:

1.84 x 10^7 N/C

Explanation:

Let q 1 = - 3 C, q2 = 2 C, q3 = 1 C

Electric field at P due to q1 is E1, due to q2 is E2 and due to q3 is E3.

E1 = k (3) / (50)^2 = 9 x 10^9 x 3  / 2500 = 108 x 10^5 N/C

E2 = k (2) / (50)^2 = 9 x 10^9 x 2 / 2500 = 72 x 10^5 N/C

E3 = k (1) / (150)^2 = 9 x 10^9 x 1 / 22500 = 4 x 10^5 N/C

Resultant electric field at P is given by

E = E1 + E2 + E3 = (108 + 72 + 4) x 10^5 = 184 x 10^5 = 1.84 x 10^7 N/C

Answer:

The momentum of the two-block system is 18 kg-m/s

Explanation:

It is given that,

Mass of block A, [tex]m_A=4\ kg[/tex]  

Mass of block B, [tex]m_B=6\ kg[/tex]  

Velocity of block A, [tex]v_A=3\ m/s[/tex]  

Velocity of block B, [tex]v_B=-5\ m/s[/tex] (it is moving in opposite direction)

We need to find the momentum of two block system. It is given by the product of mass and velocity for both blocks i.e.

[tex]p=m_Av_A+m_Bv_B[/tex]

[tex]p=4\ kg\times 3\ m/s+6\ kg\times (-5\ m/s)[/tex]

p = -18 kg-m/s

So, the momentum of two block system is 18 kg-m/s. Hence, this is the required solution.                                            

Final answer:

The total momentum of the two-block system, involving block A with a mass of 4.0 kg moving at 3.0 m/s and block B with a mass of 6.0 kg moving at 5.0 m/s in the opposite direction, is computed as -18 kg·m/s, which is in the -x-direction.

Explanation:

The question involves finding the momentum of a two-block system where the blocks are moving in opposite directions. To compute the total momentum of the system, we apply the principle of conservation of momentum, which states that the momentum of a system remains constant if no external forces act on it.

Momentum is defined as the product of an object's mass and its velocity (p = mv). For block A, which has a mass of 4.0 kg and is moving at 3.0 m/s, its momentum is calculated as:
pA = (4.0 kg) × (3.0 m/s) = 12 kg·m/s in the +x-direction.
For block B, which has a mass of 6.0 kg and is moving at 5.0 m/s in the opposite direction, its momentum is calculated as:
pB = (6.0 kg) × (-5.0 m/s) = -30 kg·m/s in the -x-direction (note the negative sign because it is moving in the opposite direction).

To find the total momentum of the system, we sum the momenta of both blocks:
ptotal = pA + pB = 12 kg·m/s + (-30 kg·m/s) = -18 kg·m/s.

The total momentum of the two-block system is -18 kg·m/s, indicating that there is a net momentum in the -x-direction.

Answer:

Partial pressure of nitrogen gas,[tex]p^o_{N_2] =0.44 atm[/tex]

Explanation:

Pressure of the mixture of gases before adding nitrogen gas = 0.85 atm

Pressure of the mixture of gases after adding nitrogen gas = 988 mmHg

1 mmHg = 0.001315 atm

988 mmHg=[tex]988 mmHg\times 0.001315 atm = 1.29 atm[/tex]

Partial pressure of nitrogen gas,[tex]p^o_{N_2] = 1.29 atm - 0.85 atm = 0.44 atm[/tex]

According to Dalton's law of partial pressure , the total pressure of the mixture of gases is equal to sum of all the partial pressures of each gas present in the mixture.

[tex]P_{total}=\sum p^o_{i}[/tex]

So, on addition of nitrogen gas to the mixture the pressure of the mixture increases.

Answer:

[tex]r = 5.13 \times 10^6[/tex]

Explanation:

As we know that the centripetal force for the space shuttle is due to gravitational force of earth due to which it will rotate in circular path with constant speed

so here we will have

[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]

here we know that

v = 19800 mi/h

[tex]v = 19800 \frac{1602 m}{3600 s}[/tex]

[tex]v = 8811 m/s[/tex]

also we know that

[tex]M = 5.97 \times 10^{24} kg[/tex]

now we will have

[tex]r = \frac{GM}{v^2}[/tex]

[tex]r = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(8811)^2}[/tex]

[tex]r = 5.13 \times 10^6[/tex]

Answer:

Speed of air = 1106.38 ft/s

Explanation:

Speed of sound in air with temperature

[tex]v_{air}=331.3\sqrt{1+\frac{T}{273.15}} \\ [/tex]

Here speed is in m/s and T is in celcius scale.

T = 50°F

[tex]T=(50-32)\times \frac{5}{9}=10^0C \\ [/tex]

Substituting

[tex]v_{air}=331.3\sqrt{1+\frac{10}{273.15}}=

337.31m/s \\ [/tex]

Now we need to convert m/s in to ft/s.

1 m = 3.28 ft

Substituting

[tex]v_{air}=337.31\times 3.28=1106.38ft/s \\ [/tex]

Speed of air = 1106.38 ft/s

Answer:

The distance the ball moves up the incline before reversing its direction is 3.2653 m.

The total time required for the ball to return to the child’s hand is 3.2654 s.

Explanation:

When the girl is moving up:

The final velocity (v) = 0 m/s

Initial velocity (u) = 4 m/s

a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).

Let time be t  to reach the top.

Using

v = u + a×t

0 = 4 - 2.45*t

t = 1.6327 s

Since, this is the same time the ball will come back. So,

Total time to go and come back = 2* 1.6327 = 3.2654 s

To find the distance, using:

v² = u² + 2×a×s

0² = 4² + 2×(-2.45)×s

s = 3.2653 m

Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.

To determine the distance and the total time for the ball to return to the girl, we apply kinematic equations with the initial velocity of 4 m/s and a constant acceleration of 0.25g down the incline. The ball travels 3.27 meters up the incline before stopping, and the total time for the round trip is 3.26 seconds.

To determine the distance s that a ball moves up an incline before reversing its direction and the total time t required for it to return to the girl, we can use kinematics equations. The acceleration a is given as 0.25g, which means the acceleration is 0.25 times the acceleration due to gravity, with g being 9.8 m/s². The initial velocity u of the ball is 4 m/s up the incline.

The final velocity v when the ball stops at the highest point is 0 m/s. By using the kinematic equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can solve for s:

The ball travels a distance of s = 3.27 m up the incline before stopping and reversing its direction.

To find the total time t, we know that the time taken to go up and down the incline is the same. We use the equation v = u + at, and by rearranging for t, we get t = (v - u) / a.

The ball takes 1.63 seconds to reach the highest point, so the total time for the round trip is t = 1.63 s * 2 = 3.26 s. Thus, it takes 3.26 seconds for the ball to return to the girl's hand.

Answer:

[tex]L = 20 + 37 = 57 cm[/tex]

Explanation:

As we know that two charges connected with spring is at equilibrium

so here force due to repulsion between two charges is counter balanced by the spring force between them

so here we have

[tex]F_e = F_{spring}[/tex]

here we have

[tex]\frac{kq_1q_2}{r^2} = kx[/tex]

[tex]\frac{(9 \times 10^9)(40 \mu C)(40 \mu C)}{(0.20 + x)^2} = 120 x[/tex]

[tex]14.4 = (0.20 + x)^2 ( 120 x)[/tex]

by solving above equation we have

[tex]x = 0.37 m[/tex]

so the distance between two charges is

[tex]L = 20 + 37 = 57 cm[/tex]

Answer:

Is dissipated E) P= 48 W in the 12 Ω resistor.

Explanation:

V= 36v

Req= (12 + 6)Ω

Req= 18Ω

I = V/ Req

I= 2 A

P= I² * R(12Ω)

P= 48 W

Answer:

The final speed of the rocket once the engine has fired is 29.16 m/s.

Explanation:

By the principle of momentum and amount of movement, we match the momentum data to the amount of movement and find the value of speed.

I=3.5 kg*m/s

m=120g= 0.12kg

V=?

I=F*t

F*t=m*V

I/m=V

(3.5 kg*m/s) / 0.12kg = V

V=29.16 m/s

Explanation:

The given data is as follows.

   J = 3.5 kg-m/s,    m = 120 g = 0.120 kg      (as 1 g = 0.001 kg)

It is known that formula to calculate impulse is as follows.

        J = [tex]m(v_{2} - v_{1})[/tex]

          [tex]0.120(v_{2} - v_{1})[/tex] = 3.5

      [tex]v_2 - v_1[/tex] = 29 m/s

So, [tex]v_{2} = 29 + v_{1}[/tex]

                = 29 + 0 = 29 m/s

Thus, we can conclude that final speed ([tex]v_{2}[/tex]) is 29 m/s.

Completing the square gives the answer right away.

[tex]-16t^2+64t+23=-16(t^2-4t)+23=-16(t^2-4t+4-4)+23=-16((t-2)^2-4)+23[/tex]

[tex]\implies h(t)=-16(t-2)^2+87[/tex]

which indicates a maximum height of 87 when [tex]t=2[/tex].

Answer:

The maximum height of the arrow is 87 meters.

Explanation:

If we look at the height function of the arrow

                [tex]h(t)=-16t^{2} +64t+23[/tex]

we see that its a parabola whose principal coefficient is negative, that means is inverted or upside down.

When the arrow reaches maximum height its velocity will be zero. The velocity of an object is the derivative of the position function, in this case the so called height function.

So we we derivate the height function to get that

                [tex]h'(t)=-32t+64[/tex]

we must find the t that makes this equation equal to zero:

                [tex]-32t+64=0[/tex]

                          [tex]32t=64[/tex]

                             [tex]t=2s[/tex]

we replace this value of t in the height function:

                [tex]h(2 s)=-16.(2s)^{2} +64.(2s)+23[/tex]

we get that

                 [tex]h(2s)=87m[/tex]

The maximum height of the arrow is 87 meters.

We have used the MKS system which uses the meter, kilogram and second as base units.

Answer:

The elevator has moved 54 meters upwards.

Explanation:

m= 50kg

F= 600 N

t= 3 sec

a= [F-(m*g)]/m

a= (600N-490N)/50kg

a= 2.2 m/s²

h= (a*t²)/2

h= (2.2m/s²*(3 sec)²)/2

h= 9.9 m

Final answer:

The elevator has moved upward a distance of 7 meters at the end of 3 seconds by considering the motion and acceleration of the elevator and the student.

Explanation:

The elevator has moved upward a distance of 7 meters at the end of 3 seconds.

To calculate this, we need to consider the motion of the elevator and the student. The apparent weight of the student is the normal force exerted by the elevator floor on the student, which is 600 N when moving up. This value is different from the student's actual weight due to the acceleration of the elevator.

Using the equation Net Force = Mass x Acceleration, and knowing that the student's weight (mg) - normal force = ma, we can find the acceleration, and subsequently, the distance the elevator has moved 7 meters after 3 seconds.

Answer:

a)1.13×10³

b)1.6×10³

Explanation:

Given:

Boltzmann's constant (K)=1.38×10^-23 J/K

atmoic mass of helium = 4 AMU or 4×1.66×10^-27kg

a)The formula for RMS speed (Vrms) is given as

[tex]Vrms=\sqrt{\frac{3KT}{m} }[/tex]

where

K= Boltzmann's constant

T= temperature

m=mass of the gas

[tex]Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 206}{6.64\times 10^{-27}}}[/tex]

[tex]Vrms=1.13\times 10^{3}m/s[/tex]

b) RMS speed of helium when the temperature is doubled

[tex]Vrms=\sqrt{\frac{3\times 1.38\times 10^{-23}\times 2\times 206}{6.64\times 10^{-27}}}[/tex]

[tex]Vrms=1.598\times 10^{3}m/s[/tex]

Answer: a)1.13×10³ b)1.6×10³

Explanation:

Answer:

The actual size is

69.33 nm.

Explanation:

It means that

3 cm is equivalent to 40 nm

So, 1 cm is equivalent to 40 / 3 nm

Thus, 5.2 cm is equivalent to

40 × 5.2 / 3 = 69.33 nm