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2 years ago

How can this system be brought to equilibrium?

Physics

1 answer:

dezoksy [38]2 years ago

5 0

Hello! The answer is B and B.

When a system is in equilibrium, this means that the net force/total force = 0N. Hence you need to add 10N to the 10N force acting towards the right so that the 20N acting in opposite directions will ‘cancel’ out to get a net force of 0N.

Average velocity is calculated by distance travelled divided by time. Hence, velocity = 50m/5s = 10ms^1

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At the instant shown in the diagram, the car's centripetal acceleration is directed

serg [7]

Answer:

At the instant shown in the diagram, the car's centripetal acceleration is directed is discussed below in detail.

Explanation:

The direction of the centripetal acceleration is in a circular movement is forever towards the middle of the roundabout pathway. In the picture displayed, the East direction is approaching the center. So, the course of the car's centripetal acceleration is (H) toward the east.

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2 years ago

6. A suitcase weighing 120 N sits on a counter 1.8 m high

Liono4ka [1.6K]

Answer:

216 J

Explanation:

h = 1.8

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2 years ago

The Sun being the largest body has the highest gravitational pull on all planets however planets do not collapse into the sun th

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Answer:

A:gravitational of the earth

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2 years ago

Two hicers are 13 miles apart and walking toward each other. They meet in 2 hours. Find the rate of each hiker if one hiker walk

xz_007 [3.2K]

Answer:

Explanation:

Let r be the rate of the slower walker in mph

[r + (r + 1.7)](2) = 13

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4r + 3.4 = 13

4r = 9.6

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2 years ago

MIT’s robot cheetah can jump over obstacles 46. cm high and has speed of 12.0 km/h. a) If the robot launches itself at an angle

labwork [276]

Answer:

(a) $y_{max}=0.423m$

(b) $\alpha =64.3^{o}$

Explanation:

Given data

$v_{i}=12km/h=3.33m/s\\\alpha =60^{o}\\g=9.8m/s^{2}\\Required\\(a)y_{max}\\(b)Angle$

Solution

For Part (a)

As the velocity component in direction of y is given by:

$v_{yi}=v_{i}Sin\alpha \\v_{yi}=3.33Sin60\\v_{yi}=2.88m/s$

The maximum displacement is given by:

$v_{yf}^{2}=v_{yi}^{2}-2gy_{max}\\ y_{max}=\frac{(2.88)^{2}}{2(9.8)}\\ y_{max}=0.423m$

For Part (b)

To reach y=46cm =0.46m apply:

$0=v_{yi}^{2}-2(9.8)(0.46)\\v_{yi}=3m/s\\As\\Sin\alpha =\frac{v_{yi}}{v_{i}}\\\alpha =Sin^{-1}(\frac{v_{yi}}{v_{i}})\\\alpha =Sin^{-1}(\frac{3}{3.33} )\\\alpha =64.3^{o}$

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2 years ago

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