A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat three people. The lift is operating at a steady speed of 10 km/h. Neglecting friction and air drag, and assuming that the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also, estimate the power required to accelerate this ski lift in 17 s to its operating speed when it is first turned on.

Answer:

yes bc they are falling :]

Explanation:

Answer:

d) electrical repulsion

Explanation:

When you stand on a floor, the two different bodies i.e your shoes and the floor on which you stand, are in a very close contact, the electrons that surround your shoes and that of the floor exert a repulsive force on each other, also known as Coulomb's force of repulsion or  electrical repulsive force.

This electrical repulsive force prevents you from falling through the floor.

Explanation:

(m2-m1)/t

25-0/5

25/5

5m/s

Answer:

The new water temperature is 26.4 °C

Explanation:

Given;

mass of copper, [tex]M_{cu}[/tex] = 26 g = 0.026 kg

temperature of copper, t = 300 °C

volume of water, V = 120 mL = 0.12 L

temperature of water, t = 21 °C

density of water, ρ = 1 kg/L

mass of water = density x volume

mass of water = (1 kg/L) x 0.12 L = 0.12 kg

heat lost by copper = heat gained by water

Both copper and water reach final temperature, T

Heat gained by water, [tex]Q_w[/tex] = [tex]m_w[/tex]cΔθ = [tex]m_w C(T - t)[/tex]

[tex]Q_w = m_w C(T - t)\\\\Q_w = 0.12*4200(T-21)\\\\Q_w = 504(T-21)[/tex]

Heat lost by copper is given by;

[tex]Q_{cu} = m_{cu}C(300-T)\\\\Q_{cu} = 0.026*385(300-T)\\\\Q_{cu} = 10.01(300 - T)[/tex]

[tex]Q_{cu} = Q_w[/tex]

504(T- 21) = 10.01(300 - T)

504 T - 10584 = 3003 - 10.01 T

504 T + 10.01 T= 3003 + 10584

514.01 T = 13587

T = (13587) / 514.01

T = 26.4 °C

Therefore, the new water temperature is 26.4 °C

The new water temperature using the given parameters is;

26.49°C

We are given;

Mass of copper; m_cu = 26 g = 0.026 kg

Temperature; T_cu = 300 °C

Volume of water; V = 120 mL = 0.12 L

Temperature of water, T_w = 21 °C

Density of water; ρ = 1 kg/L

Let us find the mass of water from the formula;

m_w = ρ × V

m_w = 1 × 0.12

m_w = 0.12 kg

Now, from the principle of conservation of energy, we can say that the total heat lost by a hot body is equal to the total heat gained by a cold body.

The hot body here is copper while the cold body is water. Thus;

Heat lost by copper = Heat gained by water

Formula for heat lost by copper is;

Q_cu = m_cu * c_cu * (T_f - T_cu)

Formula for heat gained by water is;

Q_w = m_w * c_w * (T_f - T_w)

Where;

T_f is final temperature reached by copper and water

c_cu is specific heat capacity of copper = 385 J/kg.°C

c_w is specific heat capacity of water = 4200 J/kg.°C

Thus;

Q_cu = 0.026 × 385 × (300 - T_f)

Q_cu = 3030 - 10.01T_f

similarly;

Q_w = 0.12 × 4200 × (T_f - 21)

Q_w = 504T_f - 10584

Thus;

3030 - 10.01T_f = 504T_f - 10584

3030 + 10584 = 504T_f + 10.01T_f

13614 = 514.01T_f

T_f = 13614/514.01

T_f = 26.49°C

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Answer:

1.    Kicking a soccer ball

2.      Playing tug of war  

3.         Bouncing a Ball

Explanation:

Answer:

B, and D

Explanation:

The two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.

The water cycle is defined as a cycle of events that involves precipitation as rain and snow, drainage in streams and rivers, and return to the atmosphere by evaporation and transpiration. Water moves through this cycle between the earth's oceans, atmosphere, and land.

It can also be defined as the route all water takes as it travels around Earth in various conditions.

There are basically six stages of water cycle.

Thus, the two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.

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Answer:

ANSWER

A = 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

plz give brainlist

hope this helped

Answer:

[tex]here \: amplitude = 5cm (convert \: to \: si \: unit) = .05m \\ t = .2sec \\ omega = \frac{2\pi}{t} \\ = \frac{2\pi}{.2 } = 10\pi \frac{rad}{s} \\ we \: want \: find \: a \: and \: v \\ we \: know \: that \: a = - {omega}^{2} x \\ v = omega \sqrt{ {r}^{2} - {x}^{2} } \\(1)x = .05m \\ a = - {10\pi}^{2} \times .05 = - 5 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.05}^{2} } = 0 \\ 2)x = 3cm = .03m \\ a = {(10\pi)}^{2} \times .03 = - 3 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.03}^{2} } = 10\pi \times .04 = .4\pi \frac{m}{s} \\ 3)x = 0 \\ a = - {(10\pi)}^{2} \times 0 = 0 \\ v = 10\pi \times \sqrt{ {.05}^{2} - {0}^{2} } = - 10\pi \times .05 = .5\pi \frac{m}{s} \\ thank \: you[/tex]

Use the formula,

[tex]\Delta x=v_it+\dfrac12at^2[/tex]

where [tex]\Delta x[/tex] is the cart's displacement (from the origin), [tex]v_i[/tex] is its initial speed, [tex]a[/tex] is its acceleration, and [tex]t[/tex] is time.

[tex]\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2[/tex]

[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]

Alternatively, since acceleration is constant, we have

[tex]\dfrac{v_f+v_i}2=\dfrac{\Delta x}t[/tex]

That is, we have these two equivalent expressions for average velocity, where [tex]v_f[/tex] is the cart's final velocity. Solve for [tex]\Delta x[/tex]:

[tex]\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}[/tex]

[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]

Answer:

Answer: potassium (K) germanium (Ge) selenium (Se)

Explanation:

I just took the test, the farther to the right on the periodic table you go the less conductive it gets.

Answer:

Answer: potassium (K) germanium (Ge) selenium (Se)

Explanation:

The most conductive to least conductive is from left to right

Answer:

The true statements are:

c) Temperature differing by 25 on the Fahrenheit scale must differ by 45 on the Celsius scale.

d) Temperatures which differ by 10 on the Celsius scale must differ by 18 on the Fahrenheit scale.

Explanation:

option a is false, 90°C is equal to 194°F which is not warmer than 202°F

option b is false, 40K is equal to -233°C

option e is false 0°F corresponds to -17.8°C.

Answer:

C An ideal isotropic antenna is a theoretical antenna that does not exist in practice, but is useful in explaining power density and unguided EM signal attenuation

Explanation:

Because practically speaking there is no ideal isotropic antenna it is only imaginary radiating in all directions and use as an arbitrary point for antenna gain

Answer:

Explanation:

Given the position of a particle  expressed by the equation x = (2t^3 - 6t^2 + 12)m, where t is in seconds, the acceleration function can be gotten by taking the second derivative of the function with respect to t as shown;

a = d/dt(dx/dt)

First let us get dx/dt

dx/dt = 3(2)t³⁻¹-2(6)t²⁻¹+0

dx/dt = 6t²-12t

a = d/dt(dx/dt)

a = d/dx(6t²-12t)

a = 2(6)t²⁻¹-12t¹⁻¹

a = 12t - 12t⁰

a = 12t-12

If the acceleration is zero, then;

12t-12 = 0

add 12 to both sides

12t-12+12 = 0+12

12t = 12

t = 12/12

t = 1sec

Hence the time when acceleration is zero is 1sec

Answer:

When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.

Answer:

SI unit is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion.

Explanation:

hope it helps

Si unit is used almost all around the world, so our use of it allows

Scientist from disparate regions to use a single standered in communicating scientific data without vocabulary confusion

Answer:

The value of m = 3.735 and the value of n = 7.

Explanation:

The equation for the momentum of the train is,

                                          P = mv  

Here, m is the mass of the train and v is the speed of the train.

Substitute [tex]4.5 \times {10^5}{\rm{ kg}} , 8.3 \times {10^1}{\rm{ m/s}}[/tex] for m and v respectively in above equation.

[tex]\begin{array}{c}\\P = \left( {4.5 \times {{10}^5}{\rm{ kg}}} \right)\left( {8.3 \times {{10}^1}{\rm{ m/s}}} \right)\\\\ = 37.35 \times {10^6}{\rm{ kg}} \cdot {\rm{m/s}}\left( {\frac{{{{10}^1}{\rm{kg}} \cdot {\rm{m/s}}}}{{10\,{\rm{kg}} \cdot {\rm{m/s}}}}} \right)\\\\ = 3.735 \times {10^7}{\rm{ kg}} \cdot {\rm{m/s}}\\\end{array}[/tex]  

According to the scientific notation, here the value of m is 3.735 and the value of n is 7 in the final answer of the momentum.

                        The value of m = 3.735 and the value of n = 7.

Answer:

23.5 degree North

Explanation:

The latitude of the vertical (direct) rays of the sun is 23.5 degree North. This is however due to the sun's vertical rays being located directly above 23.5 degree South which is the position of the Tropic of Capricorn.

The sun ray is highest at this pap tion and it occurs during Summer solstice which happens between June 21 / 22 of every year.

Answer:

Explanation:

check the attachment for the diagram of the set up for proper clarification.

From the diagram, the man's distance from the pole is labelled AB.

To get AB =, we need to get side CB first which is equal to ED i.e ED = CB.

From ΔCDE, opposite side = 45 feet and the adjacent is ED.

USing TOA, according to trig function SOH, CAH, TOA;

tan 16° = opp/adj = BD/ED

tan 16° = 45/ED

ED = 45/tan 16°

ED = 156.93 feet

Since ED = CB, hence CB = 156.93 feet

From ΔABC, adj = CB and hypotenuse = AB. According to CAH;

cos 12° = adj/hyp = CB/AB

cos 12° =  156.93/AB

AB =  156.93/cos 12°

AB = 160.44 feet

Hence the man's distance from the pole is 160.44 feet

Explanation:

Hey there!!

a = 40° { corresponding angles on a parallel lines are equal}.

As A and B are parallel, a and 40° are corresponding angles.

Hope it helps...

The value of angle "a" will be 40°

The angles which are on the same side of one of two lines cut by a transversal and are on the same side of the transversal is called corresponding angles.

since , angle a and angle B = 40° (given) are on the same relative position at the intersection where a straight line crosses two others and since both the lines are parallel to each other hence ,both the angles are said to be corresponding angles .This implies both the angles must be equal .

angle a = 40°

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Given :

The acceleration of a piston is given by :

[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]

To Find :

The amplitude of the acceleration of piston .

Solution :

Now , acceleration is :

[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]

Now , amplitude is maximum of any function .

So , maximum value of [tex]cos\ \theta[/tex] is equal to 1 .

Therefore , amplitude of the acceleration of piston is :

[tex]A=r\omega^2[/tex]

Hence , this is the required solution .

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

[tex] c = \frac{x}{t} [/tex]

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

[tex] x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km [/tex]

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

Answer:no

Explanation:the way to see air is by steaming something sometimes you might not see the air but you can see the shadow

Answer:

The new charge on the second object is q₂ - q₁

Explanation:

Given;

charge on the first object,  q₁

charge on the second object, q₂

Since there is no external force on isolated system, the total charge in the system is given by;

Q = q₁ + q₂

q₂ = Q - q₁

When the charge on object 1 is doubled, q₁' = 2q₁

let q₂' be the new charge on object 2

q₂' = Q - q₁'

q₂' = (q₁ + q₂) - 2q₁

q₂' = q₁ + q₂ - 2q₁

q₂' = q₂ - q₁

Therefore, the new charge on the second object is q₂ - q₁

The charge on object 2 will be "q₂ - q₁".

The charge's quantity on such an item represents the degree of imbalance between its electrons as well as protons.

Just to calculate the overall charge of a positively (+) charged item, divide the total no. of valence electrons by the overall number of protons.

According to the question,

Object 1's charge = q₁

Object 2's charge = q₂

Let,

New charge of object 2 be "q₂'".

The total charge in the system will be:

Q = q₁ + q₂

or,

q₂ = Q - q₁

hence,

The charge on object 2 be:

→ q₂' = Q - q₁'

By substituting the values,

        = (q₁ + q₂) - 2q₁

        = q₁ + q₂ - 2q₁

        = q₂ - q₁                

Thus the answer above is appropriate.

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Answer:

Explanation:

A. We know that 1microsecond= 10^-6s

So = 9.12*10^-6=9.12*10^-6s

B 1km= 1000m

So 3.42km= 3240m

C.1ms= 10^-3s

And 1cm= 10^-2m

So 44*10^-2m/10^-3s=440m/s

D. 80km/hr= 80*1000/3600= 22.2m/s

Because 1km= 1000m

1hr= 3600s

Answer:

Part A

[tex]x = 0.33 \ m[/tex]

Part B

[tex]y = 23.49 \ m/s[/tex]

Explanation:

From the question we are told to convert

    13 inches to meters

Now

     1 in   =   0.0254 \  m

      13 \  in  =  x  m

=>  [tex]x = \frac{13 * 0.0254 }{1}[/tex]

=>     [tex]x = 0.33 \ m[/tex]

For  part B  we are told to covert 77ft/s  to meters /srconds

   So  

         1 ft/s  =  0.305 \ m/s  

         77 ft/s  =  y

=>      [tex]y = \frac{77 * 0.305 }{1}[/tex]

=>     [tex]y = 23.49 \ m/s[/tex]

Answer: The amount of matter is higher in a bus than a ball.

Explanation:

Answer:

(a) 1, average velocity = -65.6 m/s

   2, average velocity = -64.8 m/s

   3, average velocity = -64.16 m/s

(b) The instantaneous velocity is -96 m/s

Explanation:

(a)

Average velocity is given  by;

[tex]y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}[/tex]

(1)

[tex]y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s[/tex]

(2)

[tex]y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s[/tex]

(3)

[tex]y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s[/tex]

b. y = 235 - 16t²

The instantaneous velocity is given by;

v = dy /dt

dy / dt = -32t

when t = 3 s

v = -32(3)

v = -96 m/s

(a)The average velocity for the 0.1 sec,0.05 sec,0.01 sec will be  -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.

The total displacement traveled by an object divided by the total time taken is the average velocity.

Average velocity is given by;

[tex]\rm y(t_2,t_1)=\frac{y(t_2)-y(t_1)}{t_2-t_1} \\\\[/tex]

The average velocity for case 1;

[tex]\rm y(2.1,2)=\frac{(235-16\times 2\times 1^2)-(235-16 \times 2^2)}{2.1-2} \\\\ \rm y(2.1,2)=\-65.6 \ m/sec[/tex]

The average velocity for case 2;

[tex]\rm y(2.015,2)=\frac{(235-16\times 2.05^2)-(235-16 \times 2^2)}{2.05-2} \\\\ \rm y(2.1,2)=\-64.8 \ m/sec[/tex]

The average velocity for case 3;

[tex]\rm y(2.01,2)=\frac{(235-16\times 2.01^2)-(235-16 \times 2^2)}{2.01-2} \\\\ \rm y(2.1,2)=\-64.16 \ m/sec[/tex]

Hence the average velocity for the 0.1 sec,0.05 sec,0.01 sec will be  -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.

(b) The instantaneous velocity will be -96 m/s.

The given equation in the problem is;

[tex]\rm y = 235 - 16t^2[/tex]

The instantaneous velocity is given as;

[tex]v = \frac{dy}{dt} \\\\ \frac{dy}{dt} = -32t\\\\ t = 3 s\\\\ v = -32\times 3\\\\ v = -96 m/s[/tex]

Hence the instantaneous velocity will be -96 m/s.

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