What is the hardest fruit to open??

Johannes Kepler, an apprentice of Brahe, believed in the heliocentric universe but rejected past astronomers' belief in

Rashid [163]

So the question is in what did johannes Kepler reject since he believed in the heliocentric universe. So heliocentric system is a system where the sun is in the center of the Solar system. Geocentric system is the system where Earth, not Sun is in the center of our system. He was against Geocentric system.

5 0

3 years ago

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A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//) → a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂) → a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).

6 0

2 years ago

Help me pls I WILL GIVE BRAINIEST I DONT CARE PLS HELP

PilotLPTM [1.2K]

Answer:

A) was reusable

Explanation:

Check this website out for more information about the space shuttle: https://www.nasa.gov/audience/forstudents/k-4/stories/nasa-knows/what-is-the-space-shuttle-k4.html

3 0

2 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 positive.

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

Explanation:

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 positive.

Given:

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

2 years ago

The direction of the magnetic field is parallel to the field line at any point in space. A small compass will point in a directi

Slav-nsk [51]

I think the answer is false

6 0

2 years ago