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Irina-Kira [14]
3 years ago
14

Galileo was the first scientist to do which of the following? A. estimate the speed of light B. propose the heliocentric theory

C. discover radioactivity D. define and measure speed E. invent the reflecting telescope
Physics
2 answers:
Delvig [45]3 years ago
7 0

Galileo Galilei was the first scientist to perform experiments in order to test his ideas. He was also the first astronomer to systematically observe the skies with a telescope.

:)

Papessa [141]3 years ago
6 0
<h2>Answer:</h2>

<u>Galileo was the first scientist to invent the reflecting telescope</u>

<h2>Explanation:</h2>

Galileo was very concerned with astronomy and he made his own telescope to observe stars and heavenly objects. He used his newly invented telescope to discover four of the moons circling Jupiter, to study Saturn, to observe the phases of Venus, and to study sunspots on the Sun. His achievements include improvements to the telescope and consequent astronomical observations, and support for Copernicanism.

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(a) Assume an electron in the ground state of the hydrogen atom moves at an average speed of 5.00 × 106 m/s. If the speed is kno
Sonbull [250]

Answer:

The minimum uncertainty in its position is 1.1587 nm

Explanation:

Given;

average speed of electron, v = 5.00 × 10⁶ m/s

percentage of speed uncertainty = 1%

Δv = 0.01( 5.00 × 10⁶ m/s) = 5.00 × 10⁴ m/s

Applying Heisenberg's uncertainty principle, to determine the uncertainty in its position.

ΔxΔP ≥ h/4π

Δx(mΔv)  ≥ h/4π

Δx = h/4πmΔv

where;

Δx is uncertainty in its position

h is Planck's constant

m is mass of electron

Δx ≥ \frac{6.626 *10^{-34}}{4\pi *9.1*10^{-31}*5*10^4} = 1.1587*10^{-9} \ m

Δx ≥ 1.1587 nm

Therefore, the minimum uncertainty in its position is 1.1587 nm

4 0
3 years ago
A 1200 kg car accelerats from reat to 10.0 m/s in a time of 4.50 seconds. Calculate the force that thr car's tires exerted on th
Aleksandr [31]

Answer:

2667 N

Explanation:

<h2>Method 1: Impulse </h2>

We can solve this problem by using the impulse formula.

  • FΔt = mΔv  
  • Δt = time interval, m = mass of the car (kg), Δv = change in velocity

We have three known variables, so we can solve for the fourth: F.

Divide Δt from both sides to isolate F.

  • F = (mΔv)/Δt  

Substitute known values into the equation.

  • F = [(1200 kg)(10 m/s -  0 m/s)] / 4.5 s
  • F = [(1200)(10)]/4.5
  • F = 12000/4.5
  • F = 2666.666667 N

The force that the car's tires exert on the road is 2667 N.

<h2>Method 2: Newton's Second Law</h2>

The force that the car's tires exert on the road is equivalent to the force that the road exerts on the car due to Newton's Third Law of Motion.

We can calculate the force that the car's tires exert on the road by using the formula F = ma, which was derived from Newton's Second Law of Motion.

  • F = ma
  • F = force exerted on the car, m = mass of the car (kg), a = acceleration of the car (m/s²)

We are given the mass of the car, velocity of the car, and the time in which it accelerated.

We can use this equation for acceleration:

  • a = Δv/Δt
  • Δv = final velocity - initial velocity (change in velocity), Δt = time interval

The car started from rest, meaning it had an initial velocity of 0 m/s. Its final velocity was 10 m/s. The time that it took for the car to go from 0 m/s to 10 m/s was 4.5 seconds.

  • a = (10 m/s - 0 m/s) / 4.50 s
  • a = 10/4.5
  • a = 2.222... m/s²

Now we have two known variables, mass and acceleration. We can solve for the force exerted on the car (and thus, the force the car exerts on the road) using the formula F = ma.

  • F = ma
  • F = (1200 kg)(2.222... m/s²)
  • F = 1200 · 2.222...
  • F = 2666.666667 N

The force that the car's tires exert on the road is 2667 N.

7 0
2 years ago
The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2615
mart [117]

Answer:

0.68 kg-m²

Explanation:

F = Force applied by the muscle = 2615 N

r = effective perpendicular lever arm = 2.85 cm = 0.0285 m

α = Angular acceleration of the forearm = 110.0 rad/s²

I  = moment of inertia of the boxer's forearm = ?

Torque is given as

τ = I α                                   eq-1

Torque is also given as

τ = r F                                   eq-2

using eq-1 and eq-2

r F =  I α

(0.0285)(2615) = (110.0) I

I = 0.68 kg-m²

7 0
2 years ago
What are three basic conditions for a hurricane to form
tester [92]
High ocean temperatures of above 26.5 degrees Celsius and coriolis effect( deflection of winds create spiraling motion)
3 0
3 years ago
According to quantum physics, measuring velocity of a tiny particle with an electromagnet
lidiya [134]

Answer:

Option A.

Explanation:

In quantum physics <u>there is a law to relate the position and the momentum of the particle</u>, it says that if we know with precision where is a quantum particle, we can not know the momentum of this particle, in other words, the velocity of the particle. So, when we measure the velocity of the particle we find the correct value of the particle, but we can not determine with accuracy where is the particle. This law is known as the Heisenberg's uncertainty principle and, its expressed as follows:    

\Delta x \Delta p \geq \frac{h}{4 \pi}

<em>where Δx: is the position's uncertainty, Δp: is the momentum's uncertainty and h: is the Planck constant.</em>  

Therefore, the correct answer is A: measuring the velocity of a tiny particle with an electromagnet has no effect on the velocity of the particle. It only affects the determination of the particle's position.      

I hope it helps you!

4 0
2 years ago
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