How far does a runner run if he runs for 60 seconds at 5 m/s

A box of mass m is held at rest on a frictionless surface with force F up the ramp. The ramp has an angle

frutty [35]

Answer:

Gv

Explanation:

8 0

3 years ago

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A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago

1. What is evaporation and how does it affect weather?

alina1380 [7]

Answer:

Explanation:don’t know

8 0

3 years ago

A circular test track for cars has a circumference of 3.5 kmkm . A car travels around the track from the southernmost point to t

elixir [45]

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

Explanation:

We have given circumference of the circular track = 3.5 km

Circumference is given by $2\pi r=3.5$

r = 0.557 km

(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to $=\frac{3.5}{2}=1.75km$

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

8 0

3 years ago

A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors

ale4655 [162]

Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b) When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, $F_1$ = 140 N

let the second force = $F_2$

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

$F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N$

$F_2 = 280 \ East$

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

$F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N$

$F_2 = 560 \ West$

8 0

3 years ago