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OleMash [197]
1 year ago
5

If you drop a ball off a cliff, it starts out at 0 m/s. after 1 s, it will be traveling at about 10 m/s. if air resistance is re

moved, what will happen in 2 s?
Physics
1 answer:
anastassius [24]1 year ago
5 0

The ball dropped of a cliff will have a final velocity of 19.6 m/s at 2s

The free fall formula and the procedure we will use is:

fv = g* t

Where:

  • fv = final velocity
  • g = acceleration due to gravity
  • t = time taken
  • h = height traveled

Information of the problem:

  • g = 9.8 m/s².
  • t = 2 s
  • vf = ?

Applying final velocity formula we get:

fv = g* t

fv = 9.8 m/s² * 2 s

fv = 19.6 m/s

<h3>What is free fall?</h3>

It is when the object or mobile falls from a height (h) with a positive acceleration equal to the gravity, describing a vertical rectilinear travel.

Learn more about free fall at brainly.com/question/11257529

#SPJ4

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What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use
marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force F_{1}=18\ N

Force F_{2}=26\ N

Force F_{3}=14\ N

We need to calculate the torque due to force F₁

Using formula of torque

\tau_{1}=-F_{1}d_{1}

\tau_{1}=-F_{1}\times\dfrac{a}{2}

Put the value into the formula

\tau_{1}=-18\times\dfrac{0.2}{2}

\tau_{1}=-1.8\ N-m

We need to calculate the torque due to force F₂

Using formula of torque

\tau_{2}=F_{2}d_{2}

\tau_{2}=F_{2}\times\dfrac{a}{2}

Put the value into the formula

\tau_{2}=26\times\dfrac{0.2}{2}

\tau_{2}=2.6\ N-m

We need to calculate the torque due to force F₃

Using formula of torque

\tau_{3}=F_{3}d_{3}

\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}

Put the value into the formula

\tau_{3}=0.1(14\sin45+14\cos45)

\tau_{3}=1.92\ N-m

We need to calculate the net torque on the square plate

\tau=\tau_{1}+\tau_{2}+\tau_{3}

\tau=-1.8+2.6+1.92

\tau=2.72\ N-m

Hence, The net torque on the square plate is 2.72 N-m.

3 0
3 years ago
How many electrons are in the outer energy level of an atom of carbon
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Answer:

4 electrons

Explanation:

Carbon, is the group 14 element, with four electrons in its outer shell. Carbon always shares electrons to reach a complete valence shell, making bonds with other atoms.

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Does it take more force to slow something down than to speed it up.
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To bring something to a stop the same force that was applied to speed it up can be used to stop it. If a greater force is used it will stop quicker.
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An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases
egoroff_w [7]

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

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