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2 years ago

Two examples of a stopping motion

2 answers:

7 0

Add the ball going at a fast speed then slowing down and coming to a complete stop.

Then you will have a car going down a slope the the speed of the car will slow down to the minimum speed of 0.

So these two will be used as examples of stopping motion

Anton [14]2 years ago

5 0

Pause?Freeze?
Stop?
Halt?

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A car starts moving from the position of rest with uniform acceleration of 8m/s^calculate the distance travelled by it during 10

ad-work [718]

We calculate the coordinates at t₁ = 9 min and t₂ = 10 min, since the 10th minute is between t₁ and t₂.

As it leaves from rest, it means that the initial speed is zero

t₁=9 min=540 s

t₂=10 min=600 s

x₁=at₁²/2=8*540²/2=4*291600=1166400 m

x₂=at₂²/2=8*600²/2=4*360000=1440000 m

Δx=x₂-x₁=1440000-1166400=273600 m represents the distance traveled by the car in the 10th minute of travel

8 0

3 years ago

The decomposition of water into hydrogen gas H2 and oxygen gas O2 can be modeled by the balanced chemical equation

Volgvan

2H2O->2H2+O2

This balanced chemical equation represents the decomposition of water into hydrogen gas and oxygen gas

3 0

3 years ago

Read 2 more answers

A parallel-plate capacitor consists of plates of area 1.5 x 10^-4 m^2 separated by 2.0 mm The capacitor is connected to a 12-V b

Katen [24]

Answer:

4.78 x 10^-11 J

Explanation:

A = 1.5 x 10^-4 m^2

d = 2 mm = 2 x 10^-3 m

V = 12 V

Let C be the capacitance of the capacitor

C = ε0 A / d

C = (8.854 x 10^-12 x 1.5 x 10^-4) / (2 x 10^-3)

C = 6.64 x 10^-13 F

Energy stored, U = 1/2 CV^2

U = 0.5 x 6.64 x 10^-13 x 12 x 12

U = 4.78 x 10^-11 J

4 0

3 years ago

An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75 in +ve x- direction

$F_1 = \dfrac{GM_1 m}{R_1^2}$

$F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}$

$F_1 = 2.019\times 10^{8}\ m$

Force applied due to mass 14 Kg in -ve x-direction

$F_2 = \dfrac{GM_2 m}{R_2^2}$

$F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}$

$F_2 = 0.857\times 10^{8}\ m$

net force

F = F₁ + F₂

$F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m$

$F = 1.162\times 10^{8}\ m$

Using newton second law

$a = \dfrac{F}{m}$

$a = \dfrac{ 1.162\times 10^{8}\ m}{m}$

$a =1.162\times 10^{8} \ m/s^2$

b) As the acceleration of mass comes out to be +ve hence, the direction will be toward the mass of 8.75 Kg

6 0

3 years ago

The automobile beltway around many cities is approximately circular. Suppose that you start driving at point A on the east side

lord [1]

speed is the magnitude of velocity which is given as 90 km/hr and it does not change. only the direction change , the direction at any time is given by the tangent to the circle at that time and location.

from the diagram , at north side, the velocity is directed in west direction

5 0

3 years ago

Read 2 more answers