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2 years ago

Mechanical energy is the sum of energy and potential energy.

Physics

1 answer:

MAVERICK [17]2 years ago

7 0

Answer:

True, the total amount of mechanical energy is merely the sum of the potential energy and the kinetic energy. This sum is simply referred to as the total mechanical energy.

Explanation: Hope it helps you:))))

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2. For a rotating rigid body, which of the following statements is NOT correct?

AfilCa [17]

Answer:

dasgfwe

Explanation:

6 0

3 years ago

A 1.5m wire carries a 6 A current when a potential difference of 68 V is applied. What is the resistance of the wire?

ANTONII [103]

Answer:

$11.3 \Omega$

Explanation:

We can find the resistance of the wire by using Ohm's law:

$V=RI$

where

V is the voltage applied

R is the resistance

I is the current

In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:

$R=\frac{V}{I}=\frac{68 V}{6 A}=11.3 \Omega$

6 0

3 years ago

The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average r

Dvinal [7]

Answer:

The lifetime of the particle is $\Delta t = 2.6*10^{-25} \ s$

Explanation:

From the question we are told that

The average rest energy is $E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J$

The intrinsic width is $\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J$

The lifetime is mathematically represented as

$\Delta t = \frac{h}{\Delta E}$

Where h is the Planck's constant with a value of $1.055*10^{-34} \ J\cdot s$

substituting values

$\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}$

$\Delta t = 2.6*10^{-25} \ s$

6 0

3 years ago

If a 12 kg cat is sitting 5 m up in a tree, how much PE does it have?

Helen [10]

Answer:

588 J

Explanation:

PE (potential energy) = (mass) x (gravity) x (height)

mass = 12 kg

gravity = 9.8m/s^2

height = 5 m

PE = (12) x (9.8) x (5) = 588 J (Joules)

5 0

2 years ago

Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23

Sloan [31]

Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

The radius of Phobos orbit is R_2 = 9380 km

The radius of Deimos orbit is $R_1 = 23500 \ km$

Generally from Kepler's third law

$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

G is the gravitational constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

$T^2 = R^3 * constant$

=> $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and $T_1$ is the period of Phobos

$[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=> $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

=> $\frac{T_1}{T_2} = 3.965$

8 0

3 years ago