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3 years ago

The following force diagram represents Newton’s Third Law of Motion:

Physics

1 answer:

Julli [10]3 years ago

4 0

Answer:

FALSE.

Explanation:

Newton's third law states that :

Every action has equal and opposite reaction
That is , the magnitude is the same but the directions are opposite
The action reaction forces DONOT operate on the same body.

For example ,

If a block is kept on the ground , the action force is the normal force acting on it due to the ground. BUT , NOTE THAT : the reaction force isn't the gravitational force on the body ! It is the normal force acting on the ground due to the block !

Thus,

we conclude that action and reaction forces donot act on the same body and therefore , this case has the answer : FALSE

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Erica (38 kgkg ) and Danny (46 kgkg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is

salantis [7]

Answer:

The speed with which just after he grabs her is 2.68 m/s.

Explanation:

Given that,

Mass of Erica, m = 38 m/s

Mass of Danny, m' = 46 kg

Erica reaches the high point of her bounce, Danny is moving upward past her at 4.9 m/s. At this moment, the initial speed of Erica will be 0. The momentum will remain conserved. Using the conservation of linear momentum as :

$mu+m'u'=(m+m')V\\\\0+46\times 4.9=(38+46)V\\\\V=\dfrac{225.4}{84}\\\\V=2.68\ m/s$

So, the speed with which just after he grabs her is 2.68 m/s. Hence, this is the required solution.

6 0

3 years ago

Can someone right me a flooding psa ?

Elan Coil [88]

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7 0

3 years ago

Read 2 more answers

Please help! Average speed. Show work!

KiRa [710]

Answer:

3.78 m/s

Explanation:

Recall that the formula for average speed is given by

Speed = Distance ÷ Time taken

Where,

Speed = we are asked to find this

Distance = given as 340m

Time taken = 1.5 min = 1.5 x 60 = 90 seconds

Substituting the values into the equation:

Speed = Distance ÷ Time taken

= 340 meters ÷ 90 seconds

= 3.777777 m/s

= 3.78 m/s (round to nearest hundredth)

3 0

3 years ago

Read 2 more answers

Velocity vector and acceleration vector in a uniform circular motion are related as.

mr_godi [17]

They are related as $\bold{\underline {v}\,.\,\underline a }= \bold{0}$

In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
In mathematics, 2 vectors ( $\underline p$ , $\underline q$ ) that are perpendicular to each other have a quality that their dot product ( $\underline p\,.\, \underline q$ ) equal to zero vector ( $\bold 0$ ) which is written as $\undeline p\,.\, \underline q = \bold 0$ .
This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner $\underline v\,.\, \underline a =\bold 0$ .

#SPJ4

8 0

11 months ago

Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu

Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

$F_y: T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N$