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3 years ago

If you are in an elevator that speed up then the apparent weight is

Physics

1 answer:

Sholpan [36]3 years ago

3 0

Answer:

Fnet - Fg

Explanation:

When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.

The weight, W, of an object can be expressed as;

W = mg

where m is the object's mass, and g is the acceleration due gravity.

If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

$W_{net}$ = mg + ma

where: mg is the weight of the object, and ma is the apparent weight.

Apparent weight (ma) = $W_{net}$ - mg

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A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a

Colt1911 [192]

Answer:

The intensity of sound wave at the surface of the sphere $I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

Explanation:

B = Bulk modulus

Intensity, $I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }$

The amplitude of oscillation of the sphere is given by:

$P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\$

$A = \triangle R\\$

Substitute v and A into Pmax

$P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}$

$I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }$

$P_{total} = 4\pi R^{2} I$

$P_{total} =4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }$

The intensity of the sound wave at a distance is given by:

$I = \frac{P_{total} }{4\pi d^{2} }$

$I = 4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

5 0

4 years ago

How much energy is required to raise the temperature of one kilogram (liter) of water 1°c?

Gnoma [55]

Heat required in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
Heat = 1 kg (4.18 kJ / kg C)( 1 C)
<span>Heat = 4.18 kJ energy needed</span></span>

4 0

4 years ago

If a driver is tired, the thinking distance will be less. True or false.why?

Evgesh-ka [11]

Answer:

False

Explanation:

When a driver is impaired the brain will need an extra second or two to wake up, after it does wake up it tends to drift off regularly.

8 0

3 years ago

Select the correct answer.

Juliette [100K]

Answer: the answer is C

Explanation:

4 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