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3 years ago

What are two factors that determine object's gravitational potential energy? an

Physics

1 answer:

Artyom0805 [142]3 years ago

6 0

Answer:

The factors that affect an object's gravitational potential energy are its height relative to some reference point, its mass, and the strength of the gravitational field it is in.

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David is driving a steady 26.0 m/s when he passes tina, who is sitting in her car at rest. tina begins to accelerate at a steady

nata0808 [166]

Relative speed of Tina with respect to David is given by

$v_r = v_t - v_d$

$v_r = 0 - 26$

$v_r = - 26 m/s$

now the acceleration of Tina with respect to David

$a_r = a_t - a_d$

$a_r = 2.80 - 0$

$a_r = 2.80 m/s^2$

now the relative displacement would be zero when Tina cross David

so now we have

$\deta x = 0 = v_r * t + \frac{1}{2} a_r t^2$

$0 = -26 * t + \frac{1}{2}*2.8*t^2$

$t = 18.6 s$

now the speed of Tina at this moment is given as

$v_f = v_i + a * t$

$v_f = 0 + 2.8 * 18.6$

$v_f = 52 m/s$

<em>so the speed will be 52 m/s</em>

5 0

3 years ago

Efficiency of electrons causes negative charge why?

Paraphin [41]

Answer:

Electron is negatively charged because you don't get work or in simple words charge out of it.

Explanation:

4 0

3 years ago

List the characteristics properties of all waves. at what speed do electromagnetic waves travel in a vacuum

ad-work [718]

All electromagnetic waves travel at

299,792,458 meters per second

in vacuum.

7 0

4 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

EleoNora [17]

Answer:

63.8 V

Explanation:

We are given that

$A_1=1.6 cm^2=1.6\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

$A_2=1.2 cm^2=1.2\times 10^{-4} m^2$

$A_3=4.4 cm^2=4.4\times 10^{-4} m^2$

$A_4=7 cm^2=7\times 10^{-4} m^2$

Potential difference,V=140 V

We know that

$R=\frac{\rho l}{A}$

According to question

$l_1=l_2=l_3=l_4=l$

In series

$R=R_1+R_2+R_3+R_4$

$R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})$

$R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})$

$R=\rho l(18284.6)$

$I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}$

Potential across 1.2 square cm= $V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V$

Hence, the voltage across the 1.2 square cm wire=63.8 V

3 0

3 years ago

What happens if both the maximum voltage and current are exceeded and 10W of power must be dissipated?

Aleksandr [31]

Answer:

If both the maximum voltage and current are exceeded in a resistor, that the point will lie in the thermal breakdown of the resistor.

Explanation:

This indicated that there will be an abundance of free electrons which will be available freely. This abundance of the electrons in a conductor, will result in the reduction of the overall conductivity. whereas in an insulator or semiconductor this will enhance the overall conductivity.

As the point lies in the thermal breakdown, the Ohm's law is no longer valid and thus no analytical solution can be presented in this regard.

3 0

4 years ago