Why is energy transferred from the substance to the surroundings when a substance freezes

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizo

Andreyy89

Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here

A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.

What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?

Answer:

$W_{work}=2.67*10^{-3}J$

Explanation:

Given data

Charge q=28 nC

Electric field E=5.00×10⁴ V/m.

Distance d=2.70 m

Angle α=45°

To find

Work done by electric force

Solution

$W_{work}=F_{force}*D_{distance}Cos\alpha \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J$

8 0

3 years ago

Un avión tarda en llegar a su destino 12 horas. Si recorrió una distancia de 10.700 kilómetros. Calcular su velocidad y expesarl

elixir [45]

Given that,

Distance, d = 10700 km

Time taken by the airplane to complete the destination, t = 12 hours

We need to find the speed of the airplane. It is equal to the total distance covered divided by total time taken. So,

$v=\dfrac{d}{t}\\\\v=\dfrac{10700\ km}{12\ h}\\\\v=891.66\ km/h$

We know that,

1 km = 1000 m

1 h = 3600 s

So,

$v=891.66\ km/h =\dfrac{891.66\times 1000\ m}{3600\ s}\\\\v=247.68\ m/s$

So, the speed of the airplane is 891.66 km/h or 247.68 m/s.

5 0

3 years ago

Which statement best describes the type of magnetism generated by attaching a wire to a battery and wrapping the wire around an

Mazyrski [523]

It only occurs when there is an electric current

4 0

3 years ago

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

So

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

8 0

3 years ago

Which type of energy is released when a bond between atoms is broken

scoray [572]

Answer:

D

Explanation:

7 0

4 years ago