If arts gravitational pool just got three times stronger what would happen to our weight

Hatshy [7]

Answer:

No we cannot

Explanation:

But what causes a ring to appear around the moon? This phenomenon is called a "moon halo." According to the National Weather Service, this ring of light, which is actually an optical illusion, forms around the moon when moonlight refracts off ice crystals in cirrus clouds, high up in the Earth's atmosphere.

8 0

2 years ago

Read 2 more answers

A force F is exerted at a distance D from the axis of rotation of a wrench and is exerted at an angle θ to the wrench (θ is the

Katyanochek1 [597]

Answer:

Torque = R X F = R * F sin theta

-f * 2R will exert an equal opposite torque

-f * 2 = F sin theta

f = -F sin theta / 2

4 0

3 years ago

A 9 newtwon charge is Felt on a (1x10^-4) C charge from another charge that is 3 meters away. What is the magnitude of that othe

Marina86 [1]

The magnitude of that other charge will be 9×10⁻⁵ C. The force on the charge is inverse of the distance.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The magnitude of that other charge is found as;

$\rm F = \frac{Kq_1q_2}{r^2} \\\\ \rm 9 = \frac{9 \times 10^9 \times 1\times 10^{-4}\times q_2}{(3)^2} \\\\ q_2=9\times 10^{-5} \ C$

Hence, the magnitude of that other charge will be 9×10⁻⁵ C.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

6 0

2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.0 m/s at ground level.

BARSIC [14]

Answer:

There is an interval of 24.28s in which the rocket is above the ground.

Explanation:

$y_{i}=0m$

$v_i=80\frac{m}{s}$

$a=4\frac{m}{s^2}$

$y_{e}=1000m$

$g=9.8\frac{m}{s^2}$

From Kinematics, the position $y$ as a function of time when the engine still works will be:

$y(t)=v_it+\frac{1}{2}at^2$

At what time the altitud will be $y_{e}=1000m$ ?

$v_it+\frac{1}{2}at^2=y_{e}$ ⇒ $\frac{1}{2}at^2+v_it-y_{e}=0$

Using the quadratic formula: $t_1=10s$ .

How much time does it take for the rocket to touch the ground? No the function of position is:

$y(t)=y_{e}+v_et-\frac{1}{2}gt^2$

Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).