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3 years ago

Calculate the energy that 1.5 kg of mass contains

Physics

1 answer:

Vsevolod [243]3 years ago

4 0

Answer:

$4.5(10)^8$ joules

Explanation:

Mass–energy equivalence is the principle that says that anything that has mass has an equivalent amount of energy. Converse is also true.

The relation between energy and mass is given by $E=mc^2$

Here, E denotes energy, m denotes mass and c denotes speed of light

Mass $(m)=1.5\,\,kg$

Speed of light (c) = 3 × $10^8$ m/s

Therefore,

$E=1.5(3)(10)^8=4.5(10)^8$ joules.

The S.I. unit of energy is joules which is equal to $1\,\,kg\,\,m^2/s^2$

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Mila [183]

Answer:

F = 385.56 N

Explanation:

Given that,

Mass, m = 37.8 kg

He applies a horizontal force and cross the wooden floor.

We need to find the force that must be applied to move a dog with a constant speed of 1 m/s.

The coefficient of kinetic friction between the dog in the floor is 1.02.

The net force acting on it is given by :

F = f- μmg

f is force due to constant speed, f = 0 (since, a = 0)

F = μmg

= 1.02 × 37.8 × 10

= 385.56 N

Hence, the required force is 385.56 N.

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2 years ago

Can anyone solve this?

love history [14]

Answer:

F = 39.2 N

Explanation:

Since, the object is in uniform motion. Therefore, the frictional force on object will be:

Frictional Force = μk N = μk mg

where,

μk = coefficient of kinetic friction = 0.2

m = mass of crate = 10 kg

g = 9.8 m/s²

Therefore,

Frictional Force = (0.2)(10 kg)(9.8 m/s²)

Frictional Force = 19.6 N

The horizontal component of force must be equal to this frictional force to continue the uniform motion:

F Sin 30° = 19.6 N

F = 19.6 N/Sin 30°

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3 years ago

Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is abo

hoa [83]

To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.

Our values are given as

$f = 4.11 *10^{12} Hz$

$A = 1.23 * 10^{-11}m$

The angular velocity of a body can be described as a function of frequency as

$\omega = 2\pi f$

$\omega = 2\pi 4.11 *10^{12}$

$\omega=2.582*10^{13} rad/s$

PART A) The expression for the maximum angular velocity is given by the amplitude so that

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$a = A \omega^2$

$a =( 1.23 * 10^{-11})(2.582*10^{13})^2$

$a= 8.2*10^{15}m/s^2$

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3 years ago

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3 years ago

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Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted out

blondinia [14]

Answer:

$\theta = 20.98 degree$

Explanation:

As we know that the speed of the sound is given as

$v = 332 + 0.6 t$

now at t = 273 k = 0 degree

$v = 332 m/s$

so we have

$a sin\theta = N\lambda$

$a sin\theta = N(\frac{v_1}{f})$

now when temperature is changed to 313 K we have

$t = 313 - 273 = 40 degree$

now we have

$v = 332 + (0.6)(40)$

$v_2 = 356 m/s$

$a sin\theta' = N(\frac{v_2}{f})$

now from two equations we have

$\frac{sin19.5}{sin\theta} = \frac{332}{356}$

so we have

$sin\theta = 0.358$

$\theta = 20.98 degree$

7 0

3 years ago