All categories

4 years ago

What is the fundamental source of all energy in the universe

Physics

1 answer:

seropon [69]4 years ago

5 0

The Sun is the world's main source of energy. Without it we would not be able to live.

You might be interested in

A bicyclist is initially traveling at 3 m/s. The bicyclist accelerates at 1 m/s2 for 5 seconds.

leonid [27]

The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s

Distance = (3*5) + (1/2*1*5^2)= 15+12.5= 27.5m

7 0

3 years ago

At what position or positions on the x-axis is the electric field zero?

ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

$E=\dfrac{kq}{r^2}$

The electric field vector due to charge one

$\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})$

The electric field vector due to charge second

$\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})$

We need to calculate the electric field

Using formula of net electric field

$\vec{E}=\vec{E_{1}}+\vec{E_{2}}$

$\vec{E_{1}}+\vec{E_{2}}=0$

Put the value into the formula

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0$

$\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})$

$(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}$

$\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}$

Put the value into the formula

$\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}$

$2.0+x=x$

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0

4 years ago

Which are the three types of mutations?

WITCHER [35]

Replication, Multiplication, and Substitution.

7 0

4 years ago

What is net force?

kramer

Answer:

A. The sum of all the forces acting on an object.

3 0

3 years ago

what was the initial temperature is 250 calories reply .1 kg of gold the final temperature of the gold was 175°c ? the specific

weqwewe [10]

Answer:

$115.2^{\circ}C$

Explanation:

When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by $\Delta T$ , according to the equation

$Q=mC_s \Delta T$

where $C_s$ is the specific heat capacity of the substance.

In this problem, we have:

$Q=250 \cdot 4.184 =1046 J$ is the amount of heat supplied to the sample of gold

m = 0.1 kg = 100 g is the mass of the sample

$C_s = 0.175 J/gC$ is the specific heat capacity of gold

Solving for $\Delta T$ , we find the change in temperature

$\Delta T = \frac{Q}{m C_s}=\frac{1046}{(100)(0.175)}=59.8^{\circ}$

And since the final temperature was

$T_f = 175^{\circ}$

The initial temperature was

$T_i = T_f - \Delta T= 175 -59.8=115.2^{\circ}C$

3 0

3 years ago