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beks73 [17]
3 years ago
13

What force causes a moving object to move in a curved or circular path?

Physics
1 answer:
jasenka [17]3 years ago
5 0
THE UNBALANCED FORCE THAT CAUSES OBJECTS TO MOVE IN A CIRCULAR PATH IS CALLED A CENTRIPETAL FORCE. GRAVITY PROVIDES THE CENTRIPETAL FORCE THAT KEEPS OBJECTS IN ORBIT. THE WORD CENTRIPETAL MEANS "TOWARD THE CENTER
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Complete the electron-pushing mechanism for the e1 reaction when 2-methylbutan-2-ol is treated with 20% sulfuric acid.
Sav [38]
E1 reaction works in the mechanism that the removal of an HX substituent results in the formation of a double bond. The E1 reaction for 2-methylbutan-2-ol is shown in the figure. This reaction is called acid-catalyzed dehydration of a tertiary alcohol.

The mechanism works in three major steps:
1. The OH group of the main reactant is hydrated by H2SO4 so it becomes H2O. 
2. The H2O leaves taking electrons with it. This results to a carbocation intermediate on the carbon atom where it was attached.
3. Another H2O protonates the beta carbon. This is the carbon atom next to the carbocation. It will donate its electrons to the neighboring C-C bond, as indicated by the arrow. The carbons are rehybridized from sp3 to sp2, which is a pi bond. As a result, a double bond forms.

The product is 2-methyl-2-butene.


6 0
4 years ago
Read 2 more answers
What is the relationship between the speed of impact and the dropping height?
SVETLANKA909090 [29]
The difference is that speed means how fast u go and height is how tall you are..
8 0
3 years ago
An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the —z direction. The electric field has an amplitude of
joja [24]

Answer:

a) 6.9*10^14 Hz

b) 9*10^-12 T

Explanation:

Given that

The wavelength of the wave, λ = 435 nm

Amplitude of the electric field, E(max) = 2.7*10^-3 V/m

a)

The frequency of the wave can be found by using the formula

c = fλ, where c = speed of light

f = c/λ

f = 3*10^8 / 435*10^-9

f = 6.90*10^14 Hz

b)

E(max) = B(max) * c, magnetic field amplitude, B(max) =

B(max) = E(max)/c

B(max) = 2.7*10^-3 / 3*10^8

B(max) = 9*10^-12 T

c)

1T = 1 (V.s/m^2)

8 0
3 years ago
A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, r
V125BC [204]

Answer:

(a) whether slipping occurs between the belt and the cylinder i think is the answer dont hate if you get it wrong please and thank you.

Explanation:

i am just guessing otay.

7 0
3 years ago
A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.
kipiarov [429]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g =  acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t²        (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t)      (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s     ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

3 0
3 years ago
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