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3 years ago

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An Olympic diver is on a diving platform 5.40 m above the water. To start her dive, she runs off of the platform with a speed of

1.17 m/s in the horizontal direction. What is the diver's speed, in m/s, just before she enters the water

1 answer:

dybincka [34]3 years ago

4 0

Answer:

10.36m/s

Explanation:

From the question we are given;

Distance S = 5.40m

Initial velocity v= 1.17m/s

Required

Final velocity of the diver

Using the equation of motion

v² = u²+2gS

v²= 1.17²+2(9.8)(5.40)

v² = 1.3689+105.84

v² = 107.2089

v = √107.2089

v= 10.36m/s

Hence the required velocity is 10.36m/s

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When designing a user interface, the most important information should be placed in the ______ of the screen.

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4 years ago

A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe

pentagon [3]

Answer:

(A) l = 0.39 m

(B) l =0.38 m

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

potential energy of the object = mg(1.05 + l)

potential energy of the spring = 0.5 x k x l^{2} (k= spring constant)

therefore we now have

mg(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) = 0.5 x 300 x l^{2}

15.68 (1.05 + l) = 150 x l^{2}

16.5 + 15.68l = 150l^{2}

l = 0.39 m

(B) with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l) = 0.5 x 300 x l^{2}

(16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}

16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

15.71 + 14.93l = 150^{2}

l =0.38 m

(C) where g = 1.63 m/s^{2} and neglecting air resistance

the equation mg(1.05 + l) = 0.5 x k x l^{2} now becomes

1.6 x 1.63 x (1.05 + l) = 0.5 x 300 x l^{2}

2.608 (1.05 +l) = 0.5 x 300 x l^{2}

2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0

3 years ago

A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago

Two iron bolts of equal Mass one at a hundred see another at 55 Sierra place in the insulated cylinder assuming the heat capacit

malfutka [58]

Answer:

$T_2 = 77.5c$

Explanation:

From the question we are told that

Temp of first bolts $T_1=100$

Temp of 2nd bolt $T_2=55$

Generally the equation showing the relationship between heat & temperature is given by

$q=cm \triangle T$

Generally heat released by the iron bolt = heat gained by the iron bolt

Generally solving mathematically

$-(0.45*m* (T_2-100 \textdegree c)) = 0.45*m*(T_2 -55\textdegree c)$

$-(T_2-100 \textdegree c)) = (T_2 -55 \textdegree c)$

$T_2 +T_2= 100 \textdegree c+55 \textdegree c$

$T_2=\frac{155 \textdegree c}{2}$

$T_2 = 77.5 \textdegree c$

Therefore $T_2 = 77.5 \textdegree c$ is the final temperature inside the container

5 0

3 years ago