What is astatine: a non-metal, a metalloid, or a metal?

s=1/2(u+v)t check the dimensional consistency for this? where s is displacement, t is time, u is initial velocity, v is final ve

Anettt [7]

Answer:

See the explanation below.

Explanation:

The units of work are consistent since if we work in the international system of measures we have the following dimensional quantities for velocity, distance and time.

s = displacement [m]

v and u = velocity [m/s]

t = time [s]

Now using these units in the given equation.

$s = 0.5*([m/s]+[m/s])*[s]\\s=0.5*[m/s]*[s]\\s = 0.5*[m]$

So the expression is good, and dimensional has consistency.

8 0

2 years ago

A device that uses electricity and magnetism to create motion is called a

Lady bird [3.3K]

Answer:

electric motors is the answer

8 0

1 year ago

Read 2 more answers

Two objects, C & D, have the same momentum. Object C has ½ the mass of object D. Find the value of the ratio of velocity C t

Savatey [412]

These are two questions and two answers.

Part 1. Fin the value of the ration of velocity C to velocity D.

Answer: 2

Explanation:

1) Formula: momentum = mass * velocity

2) momentum C = mass C * velocity C

3) momentum D = mass D * velocity D.

4) C and D have the same momentum =>

mass C * velocity C = mass D * velocity D

5) mass C = (1/2) mass D => mass C / mass C = 1/2

6) use in the equation stated in the point 4)

velocit C / velocity D = mass D / mass C

using the equation stated in point 5:

mass D / mass C = 1 / [ mass C / mass D] = 1 / [1/2] = 2

=>

7) velocity C / velocity D = mass D / mass C = 2

Part 2: <span>ratio of kinetic energy C to kinetic energy D.
</span>
Answer: 2

Explanation:

1) formula: kinetic energy KE = (1/2) mass * (velocity)^2

2) KE C = (1/2) mass C * (velocity C)^2

3) KE D = (1/2) mass D * (velocity D)^2

4) KE C / KE D =

(1/2) mass C * (velocity C)^2 mass C (velocity C)^2
--------------------------------------- = --------------- * ---------------------- = (1/2) * (2)^2
(1/2) mass D *( velocity D)^2 mass D v(velocity D)^2

= 4 / 2 = 2

3 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

1.A body of certain mass is kept at a height of 15m from the ground, taking the value of g as 10m/s find out the mass of the bod

ehidna [41]

Answer:

1.#potential energy = PE, m = mass in kg, g = force of gravity, h= vertical height above the ground. ** means to the power of ie exponent. * means multiply.

PE = mgh

300 = m(10)(15)

m = 300/(10)(15)

m= 2kg

2. KE = 1/2 mv**2

= 1/2(50)(50)**2

= 2500 joules

Explanation

Is as in solution

4 0

3 years ago