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3 years ago

THE NUMERICAL RATIO OF DISPLACEMENT TO DISTANCE IS:

Physics

1 answer:

Tamiku [17]3 years ago

7 0

There's no general rule.

Displacement is the length of a straight line from start to finish, and distance is how far you actually traveled from start to finish.

The only thing we really know is that distance can never be shorter than displacement. So I guess the answer is:

<em>The numerical ratio of displacement to distance is always 1 or less</em>.

(But it has to be written ALL IN CAPITALS.)

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A horizontal force of 50 N is applied to the object.

USPshnik [31]

Answer:

Mass of object (m) = 5.102 kg

Explanation:

Given:

Horizontal Force (F) = 50 N

Find:

Mass of object (m) = ?

Computation:

We know that, acceleration due to gravity (g) = 9.8 m/s²

⇒ Horizontal Force (F) = mg

⇒ 50 N = m (9.8 m/s²)

⇒ Mass of object (m) = 50 / 9.8

⇒ Mass of object (m) = 5.102 kg

Mass of object (m) is 5.1 kg (Approx)

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4 years ago

The electrons that are gained or lost in an ionic bond are called...?

Gala2k [10]

Answer:anion

Explanation:

4 0

4 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

The greenhouse effect presentation summarized?

emmainna [20.7K]

Answer:

What is the greenhouse effect?

The greenhouse effect is the way in which heat is trapped close to Earth's surface by “greenhouse gases.”

Explanation:

These heat-trapping gases can be thought of as a blanket wrapped around Earth, keeping the planet toastier than it would be without them. Greenhouse gases include carbon dioxide, methane, nitrous oxides, and water vapor. (Water vapor, which responds physically or chemically to changes in temperature, is called a "feedback.") Scientists have determined that carbon dioxide's warming effect helps stabilize Earth's atmosphere. Remove carbon dioxide, and the terrestrial greenhouse effect would collapse. Without carbon dioxide, Earth's surface would be some 33°C (59°F) cooler.

Credit: NASA

I hope this helps, if it doesn't then just message me and ill be more than happy to help :)

8 0

2 years ago

3. How does the mass of a proton compare to the mass of an electron?

o-na [289]

Answer:

Proton, stable subatomic particle that has a positive charge equal in magnitude to a unit of electron charge and a rest mass of 1.67262 × 10−27 kg, which is 1,836 times the mass of an electron.

7 0

3 years ago

THE NUMERICAL RATIO OF DISPLACEMENT TO DISTANCE IS:​

THE NUMERICAL RATIO OF DISPLACEMENT TO DISTANCE IS: