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2 years ago

11

A skateboarder rolls horizontally off the top of a staircase and lands at the bottom. The staircase has a horizontal length of 1

2.0 with a vertical drop of 2.00. We can ignore air resistance.

1 answer:

TiliK225 [7]2 years ago

3 0

Answer: 18.8 m/s

Explanation:

You might be interested in

How much positive electric charge is in 10 moles of carbon? NA=6.022×1023mol−1, e=1.60×10−19C.

levacccp [35]

Answer:

4.75 x 10^-25 kg

Explanation:

5 0

3 years ago

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

<u>Step 4:</u> calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

3 years ago

A car traveling at 27.4 m/s hits a bridge abutment. A passenger in the car, who has a mass of 65.0 kg, moves forward a distance

Minchanka [31]

Answer:

$F=43570.9N$

Explanation:

We can calculate the acceleration experimented by the passenger using the formula $v_f^2=v_i^2+2ad$ , taking the initial direction of movement as the positive direction and considering it comes to a rest:

$a=\frac{v_f^2-v_i^2}{2d}=\frac{-v_i^2}{2d}$

Then we use Newton's 2nd Law to calculate the force the passenger of mass m experimented to have this acceleration:

$F=ma=\frac{-mv_i^2}{2d}$

Which for our values is:

$F=\frac{-(65kg)(27.4m/s)^2}{2(0.56m)}=43570.9N$

6 0

3 years ago

If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th

devlian [24]

Answer:

1470kgm²

Explanation:

The formula for expressing the moment of inertial is expressed as;

I = 1/3mr²

m is the mass of the body

r is the radius

Since there are three rotor blades, the moment of inertia will be;

I = 3(1/3mr²)

I = mr²

Given

m = 120kg

r = 3.50m

Required

Moment of inertia

Substitute the given values and get I

I = 120(3.50)²

I = 120(12.25)

I = 1470kgm²

Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²

7 0

3 years ago

An automobile tire having a temperature of

EastWind [94]

Answer:

Psm = 30.66 [Psig]

Explanation:

To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.

P * v = R * T

where:

P = pressure [Pa]

v = specific volume [m^3/kg]

R = gas constant for air = 0.287 [kJ/kg*K]

T = temperature [K]

<u>For the initial state</u>

<u />

P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)

T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)

Therefore we can calculate the specific volume:

v1 = R*T1 / P1

v1 = (0.287 * 270.4) / 266.8

v1 = 0.29 [m^3/kg]

As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.

V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.

T2 = 43 + 273 = 316 [K]

P2 = R*T2 / V2

P2 = (0.287 * 316) / 0.29

P2 = 312.73 [kPa]

Now calculating the manometric pressure

Psm = 312.73 -101.325 = 211.4 [kPa]

And converting this value to Psig

Psm = 30.66 [Psig]

3 0

3 years ago