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2 years ago

PLEASE ANSWER ASAP!! YOU WILL GET BRAINLIEST IF RIGHT!

Physics

2 answers:

Elena-2011 [213]2 years ago

6 0

D just east of station b

yarga [219]2 years ago

3 0

D will be your answer because if I am right point a is the pressure and tge line is heading to the east side of point b. I hope this helps

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Two obiect accumulated a charge of

tamaranim1 [39]

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

$q_1=4.5\mu C=4.5\cdot 10^{-6}C$ is the magnitude of the 1st charge

$q_2=2.8\mu C=2.8\cdot 10^{-6}C$ is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

$F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N$

So, the closest answer is

A) 181.24 N

3 0

3 years ago

Which of Newton's motion laws BEST explains WHY a rock falls when it is dropped from a bridge?

erma4kov [3.2K]

It’s A because it stays in motion whenever you drop it

4 0

2 years ago

Read 2 more answers

Please Help!!

kotegsom [21]

Answer:

Potential energy of spring = 24 Joules.

Explanation:

Given the following data;

Spring constant = 85N/m

Extension, e = 0.75m

Mass = 25kg

To find the potential energy of a spring

Potential energy of a spring is given by the formula;

P.E = ½ke²

Substituting into the equation, we have

P.E = ½*85*0.75²

P.E = 42.5 * 0.5625

P.E = 23.91 ≈ 24 Joules

P.E = 24 Joules

8 0

3 years ago

One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched

mash [69]

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

$c=\frac {F}{y}$

Since F is gm but taking the given mass to be F

$c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930$

By substitution now considering F to be 3.3 kg

$y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm$

8 0

3 years ago

A runner traveling with an initial velocity of 1.1 m/s accelerates at a constant rate of 0.8 m/s2fora time of 2.0 s.(a).What is

pychu [463]

Answer:

The final velocity of the runner at the end of the given time is 2.7 m/s.

Explanation:

Given;

initial velocity of the runner, u = 1.1 m/s

constant acceleration, a = 0.8 m/s²

time of motion, t = 2.0 s

The velocity of the runner at the end of the given time is calculate as;

$v = u + at$

where;

v is the final velocity of the runner at the end of the given time;

v = 1.1 + (0.8)(2)

v = 2.7 m/s

Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.

7 0

2 years ago