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3 years ago

Define Plateau period.

1 answer:

5 0

A plateau period refers to a period in life where you are kind of stuck in the middle. There is little growth or decline so you are kind of at a standstill.

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Jason walks 20 m East, turns around and 20 m West, Finally, he walks 10 rn North. This takes 20 s. what is Jason's velocity

serious [3.7K]

Answer:

0.5 m/s north

Explanation:

Take east to be +x, west to be -x, north to be +y, and south to be -y.

His displacement in the x direction is:

x = 20 m − 20 m = 0 m

His displacement in the y direction is:

y = 10 m

His total displacement is therefore 10 m north.

His velocity is equal to displacement divided by time.

v = 10 m north / 20 s

v = 0.5 m/s north

3 0

3 years ago

What's formular for calculating a mass <br>

katrin [286]

force times gravity (FG) =mass times gravity (mg)

6 0

2 years ago

What happens whenever energy is transformed from one form to another?

andriy [413]

Answer:

Energy transformation is when energy changes from one form to another – like in a hydroelectric dam that transforms the kinetic energy of water into electrical energy. While energy can be transferred or transformed, the total amount of energy does not change – this is called energy conservation.

Explanation:

Hope This Helps!!

God Bless!!

~DuffyDuck~

3 0

2 years ago

A 6 m/s vector pointing North is added to a 2 m/s vector pointing East. What are the magnitude and direction of the resultant?

irina [24]

Answer:

A + B = C Ax = 2 Ay = 0 Bx = 0 By = 6

Ax + Bx = Cx = 2

Ay + By = Cy = 6

C = (2^2 + 6^2)^1/2 = 6.32

Tan Cy / Cx = 6 / 2 = 3

Cy at 71.6 deg

6 0

3 years ago

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)