Question 2 please need help physics

<img src="https://tex.z-dn.net/?f=%5Chuge%5Cbold%7B%5Cpurple%7B%5Cbold%7B%E2%9A%A1Gravitational%20Constant%3F%E2%9A%A1%7D%7D%7D%

baherus [9]

$\huge\underline\mathtt\colorbox{cyan}{G=}$

$6.673 \times {10}^{ - 11}$

And unit is Nm^2/kg^2

8 0

2 years ago

Read 2 more answers

A woman was recently given the opportunity to ride in a porsche race car on their test in Hapeville, Georgia. Below is a graph o

Zinaida [17]

Answer:

During the segments B - C and D - E, the car stopped since the y axis is the distance and the distance stayed the same in between those segments.

For a simpler answer, the flat horizontal lines on the graph are the times when the car was stopped.

6 0

2 years ago

A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con

alukav5142 [94]

Answer:

0.00016 kg

Explanation:

Given:

Power = P = 1.2 × 10⁹ Watts

Power = work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 = 4 × 10⁹ W

Energy = 4 × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c² = (1.44 x 10¹³) / (3 × 10⁸ )²

= 0.00016 kg

6 0

3 years ago

10,500 J of GPE, weight 539 N, how tall is the hill you are sitting on?

maks197457 [2]

Answer:

19.48 m

Explanation:

Gravitational potential energy = mgh

Current weight = 539 N

Weight = mg = 539 N

Mass x Acceleration = 539 N

Mass x 9.81 = 539

Mass = 54.94 kg

Gravitational potential energy = mgh = 10500 J

54.94 x 9.81 x h = 10500

h = 19.48 m

Height of sitting = 19.48 m

6 0

2 years ago

A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,

$Q_A$ = $15\times 10^-6 C$

$Q_B=9\times10^-6C$

$d_A = 1+x cm$

$d_B=x cm$

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

$\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}$

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

#SPJ4

3 0

1 year ago