All categories

3 years ago

I need actual help, not just the answers. I need explanations and formulas please.

1 answer:

3 0

Using newton's law of gravity on a 1kg mass near the star. G is newton's grav constant. g is acceleration of grav at starGMstar/r^2 = gstar

You might be interested in

Help can’t find the answer no where

katen-ka-za [31]

i'm stuck on that question also

5 0

3 years ago

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

q₁ = + 1.25 nC

3 0

3 years ago

What is runoff water?

vitfil [10]

<u>Answer:</u> runoff water is water from rain, snow, or other sources, that flows through the land, and is a major component of the water cycle.

6 0

4 years ago

Read 2 more answers

A bicycle rider traveling east at 10 km/hr sees a blue car pass her appearing to travel west at 50 km/hr. What is the blue car's

BigorU [14]

Answer:

$v_{bR} = 25 km/h$ towards west

Explanation:

As we know that the speed of the blue car as appear to the bicycle rider is given as

$v_{bc} = 50 km/h$ towards west

also it is given that bicycle is moving at speed of 10 km/h towards East

so here we have

$v_{bc} = v_b - v_c$

so we have

$v_b = v_{bc} + v_c$

$v_b = -50 + 10 = 40 km/h$ towards west

now speed of the red car is given as 15 km/h towards west

so here the relative speed of blue car with respect to red car is given as

$v_{bR} = v_b - v_R$

$v_{bR} = 40 - 15 = 25 km/h$ towards west

8 0

4 years ago

What formula should I use?

Elodia [21]

These nuts on your chin

7 0

3 years ago