What needed to be present in marine mud to form fossils fuels?

One consequence of Einstein's theory of special relativity is that mass is a form of energy. This mass-energy relationship is pe

Zarrin [17]

Answer:

Unit of energy is kg-m/s or Joules.

Explanation:

The mass- energy relationship in physics is given by :

$E=mc^2$

Where

m is the mass of the object

c is the speed of light

The SI unit of m is kilogram while the SI unit of c is as same as speed of light i.e. m/s. So, the unit of energy is kg-m/s. It is also equivalent to Joules.

So, the SI unit of energy is kg-m/s or Joules. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally

Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$ , downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0

3 years ago

A uniformly charged, thin ring has radius 15.0 cm and total charge +24.0 nC. An electron is placed on the ring’s axis a distance

djyliett [7]

Answer:

A. The electron will begin to move along the axis, towards the centre and the instantaneous velocity because the force acting on it depends largely on acceleration and x until it reaches maximum velocity at centre.

B. Veloctiy (Vb) = 1.66m/s

Explanation:

Given the following data

x(a) = 0.3m

x(b) = 0

q = 1.6×10^-19

Q = 24nc

r = 0.15m

Required: the motion of the electron and the velocity (Vb)

1. At point A the electron will begin to move along the axis from point A to point B, the magnitude of the electric field will change while moving which depends on that and this will produce instantaneous force which will later change and the acceleration will change too while moving, the velocity would reach maximum value at point B

2. Potential energy and kinetic energy are given by

U(a) + K(a) = U(b) + K(b). . .1

Initial P.E and K.E are given as

U(a) = kQ/√x²(a) + a2

By substitution, we have

U(a) = 9×10^9 × (-1.9×10^-19)×24×10^-9/√(0.15)²+(0.3²)

U(a) = -1.03×10^-16

Final P.E and K.E are given as

U(b) = KQ/√x²(b) + a2

By substitution, we have

U(b) = 9×10^9×(-1.9×10^-19)×24×10^-9/√(0.15)²+(0)²

U(b) = -2.3×10^16

3. By substitution into equation 1 becomes

-1.03×10^-6 - 2.3×10^-16 + MV²(b)/2

V(b) = √2×1.27×10^-16/9.1×10^31

V(b) = 1.66×10^7m/s

4 0

4 years ago

The enthalpy of fusion of mercury is 2.292 kJ mol-1 at its normal freezing point of 234.3 K; the change in molar volume on melti

aev [14]

Answer:

T_2= 234.37 K

Explanation:

According to Claperyon, we know that

$P_2-P_1= \frac{\Delta H_{fus}}{\Delta V}\times\frac{T_1}{T_2}$

P_1= Atmospheric pressure 760 mm Hg

P_2 = pressure at the bottom of the column

= 10×10^3 mm of Hg+ 760 mm of Hg

= 10760 mm of Hg

now,

P_2-P_1= 10760-760= 10^4 mm

P_2-P_1 ( in pascals) = 10^4× 133.322= 1333220 mm

the enthalpy of fusion (ΔH-fus) of mercury is 2.292 KJ/mol

use the above equation to calculate ΔT as follows

$1333220= \frac{2292}{0.517\times10^{-6}}\times ln\frac{T_2}{234.3K}$

therefore, T_2= 234.37 K

6 0

3 years ago

a 5.5 kg box is pushed across the lunch table.the net force applied to the box is 9.7 N.what is the acceleration of the box?

skelet666 [1.2K]

Answer:

1.76m/s²

Explanation:

Acceleration is the time rate of change in velocity of a body. It is a vector quantity that is it has both magnitude and direction

From newton's second law of motion which states that the rate of change of momentum of a body is proportional to the applied force which takes place in the direction of force applied.

This law gives a formula which relate force, mass and acceleration.

Force = mass x acceleration

Given that force = 9.7N , mass = 5.5kg

Since force(F)= mass(m) x acceleration(a)

Therefore F = ma

Divide both sides by m

F/m = ma/m

Therefore,

Acceleration (a) = F/m

Acceleration = 9.7N ➗ 5.5kg

Acceleration = 1.76m/s²

The S. I unit of acceleration is m/s²

I hope this was helpful, Please mark as brainliest

5 0

3 years ago