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Olenka [21]
3 years ago
9

Physics Help ASAP Please!

Physics
2 answers:
stellarik [79]3 years ago
6 0

Answer:

9.3N

Explanation:

Fgrav = m • g = 14.7 N (g is 9.8 m/s/s)

a = v2 / R = (4 • Fgrav)/1 = 16 m/s/s

Fnet = m • a = 1.5 kg •16 m/s/s = 24 N, down

Fnet = Fgrav + Ftens, so

Ftens = Fnet - Fgrav

Ftens = 24 N - 14.7 N = <u>9.3 N</u>

balu736 [363]3 years ago
5 0

Answer:

23N

Explanation:

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A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.
stira [4]

Answer:

-8.04 m/s2

Explanation:

To find the answer to this, you have to use the 4th kinematic equation:

v^{2} = v^{2}_{0}  + 2ax

You plug into the equation to get:

0 = 15^{2} + 2a(14)

solve for a to get

-8.04 m/s2

3 0
3 years ago
If for every action force an equal and opposite reaction force exists, how can anything ever be accelerated?
Anon25 [30]

The action and reaction forces act on different objects, therefore the net force is not always zero

Explanation:

Newton's third law of motion states that:

<em>"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"</em>

From the statement above, it is clear that the action and the reaction forces act on different objects. In fact, the action force acts on object B, while the reaction force acts on object A. This means that the two forces never appear in the same free-body diagram of the same object, and therefore when computing the net force acting on an object, we take into account only one of the two forces, never both.

For instance, imagine you are pushing a box along the floor with a force F (action). The box is exerting back on you a reaction force of equal magnitude in the opposite direction, - F. However, if we want to compute the net force on the box, we just take into account the action (F), not the reaction, because the reaction is acting on you, not on the box.

Therefore, the correct answer is

The action and reaction forces act on different objects, therefore the net force is not always zero

Learn more about Newton's third law:

brainly.com/question/11411375

#LearnwithBrainly

3 0
3 years ago
While a block slides forward 1.35 m, a force pulls back at a 135 direction, doing -17.8 J of work. what is the magnitude of the
Mademuasel [1]
<h2>18.6467 N</h2>

Explanation:

       The Work done by any force is defined as the force applied times the displacement of point of application of force in the direction of force.

       This is better represented as a scalar product of Force vector and displacement vector.

       Work\textrm{ = }\vec{F}.\vec{s}

Here, the angle between force and displacement is 145^{o}.

       Work=|F||s|\cos\theta

       -17.8\textrm{ J = }|F|\times 1.35\textrm{ }m\times cos(135^{o})

       |F|=18.6467N

∴ Magnitude of force = 18.6467N

3 0
3 years ago
1. Explain the following terms i. Design ii. Possible solution iii. Specification.
Furkat [3]

Answer:

phpgodigpfogpglxjccxnckxjccll

3 0
3 years ago
An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec
Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0
3 years ago
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