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mylen [45]
1 year ago
7

The observed a value for the diameter of a Hydrogen atom is 10.1 nm and the accepted value for this is 10 nm. Was you observatio

n accurate?
A. No but it was precise.
B. Yes, the measurement was very close to the actual value.
C. Not enough information.
D. No not at all it was much smaller.
Physics
1 answer:
professor190 [17]1 year ago
3 0

The value of the diameter of the hydrogen atom as we have it in this experiment is accurate because it is very close to the true value.

<h3>What is accuracy?</h3>

The term accuracy refers to how close the measured value is to the true value. The true value is the same as the accepted value. In this case, the true value of the  diameter of a Hydrogen atom is 10 nm.

The fact that is observed value is obtained to be  10.1 nm implies that the value of the diameter of the hydrogen atom as we have it in this experiment is accurate because it is very close to the true value.

Learn more about accuracy:brainly.com/question/14244630

#SPJ1

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Please help with this, oh and bungee gum contains both the properties of rubber AND gum you see?
Sladkaya [172]
Can I still get 5 points bc u already figured it out
8 0
3 years ago
After the big bang, atoms in gas clouds experienced a greater gravitational pull to each other than atoms in other regions of th
allsm [11]
Answer:
These are the two statements with scientific facts that explain the described phenomenon
<span>
Gravitation between two objects increases when the distance between them decreases.</span>

When the mass of an object increases, its gravitational pull also increases.

Justification:

Those two facts are represented in the Universal Law of Gravity discovered by the scientific Sir Isaac Newton (1642 to 1727) and published in his book <span>Philosophiae naturalis principia mathematica.</span>

That law is represented by the equation:

F = G × m₁ × m₂ / d²

The product of the two masses on the numerator accounts for the fact that the gravitational force is directly proportional to the product of the masses, which is that as the masses increase the attraction also increase.

The term d² (square of the distance that separates the objects) in the denominator accounts for the fact that the gravitational force is inversely proportional to the square of the distance; that is as the separation of the objects increase the gravitational force decrease.


6 0
3 years ago
If you pull up on a bucket with a tension force of 15 N and the bucket has a weight of 15 N, what is the net force acting on the
fredd [130]

Answer:

F = 0 [N]

Explanation:

To solve this problem we must perform a summation of forces in the direction of the vertical axis. Where the positive force is that of the tension of the upward force, while the force exerted by the weight is directed downward with a negative sign.

ΣF = 0

15 - 15 + F = 0

F = 0 [N]

3 0
3 years ago
A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert
ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

F_y=F-mg=8.5-0.46\times 9.8=3.992\ N

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2

Vertical displacement of rocket is same as the height of loop. So, y=20\ m

There is no initial velocity in the vertical direction. So, u_y=0\ m/s

Now, applying equation of motion in vertical direction. we have:

y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0
4 years ago
A particle accelerator fires a proton into a region with a magnetic field that points in the +x-direction (a) If the proton is m
pochemuha

Answer:

a) -z direction, b) +z direction, c) F=0  

Explanation:

The magnetic force is given by the expression

        F = q v x B

the bold indicate vectors, this equation can be separated in its module

        F = a v B sin θ

and where θ is the angles between the speed and the magnetic field.

The direction of the force can be found with the right-hand rule. For a positive charge, the thumb goes in the direction of speed, the fingers extended in the direction of the magnetic field and the palm points in the direction of the force, if the charge is negative the force is in the opposite direction.

a) Let's apply this to our case

the proton is positively charged

moves in the direction of + x

The magnetized field goes in the direction of y

therefore applying the right hand rule the force must be in the direction of the negative part of the z-axis (-z)

The right-hand rule is used to find this address.

b)  in this case it indicates that the proton moves in the recode of -y

again we apply the right hand rule and the force is in the direction of + z

c)   The proton moves in the x direction

In this case the force is zero because the angle between the field and the speed is zero and the sine is zero, therefore the force is zero

4 0
3 years ago
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