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3 years ago

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The two forces in each pair can either both act on the same body or they can act on different bodies. The two forces in each pai

r can either both act on the same body or they can act on different bodies.
a.true
b.false

1 answer:

Marrrta [24]3 years ago

6 0

Answer:

The correct answer in which peer forces each act in a different body

Explanation:

Newton's law of action and reaction states that the two forces have equal magnitude, but in the opposite direction, each acting on a body. Which creates acceleration in these, if they act in the same body it would be canceled so there would be no movement.

The correct answer in which peer forces each act in a different body

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What type of machine is wire cutter pliers?

Sonja [21]

It seems that you have missed the necessary options for us to answer this question, but anyway, here is the answer. The type of machine that a wire cutter pliers is classified is a simple machine. When we say simple machine, this is the type of machine that is considered basic wherein you need to apply force for it to function. Hope this helps.

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2 years ago

Read 2 more answers

As an object moves from point A to point B only two forces act onit: one force is nonconservative and does -30 J of work, the ot

Romashka-Z-Leto [24]

To solve this problem we will apply the principles of energy conservation. On the one hand we have that the work done by the non-conservative force is equivalent to -30J while the work done by the conservative force is 50J.

This leads to the direct conclusion that the resulting energy is 20J.

The conservative force is linked to the movement caused by the sum of the two energies, therefore there is an increase in kinetic energy. The decrease in the mechanical energy of the system is directly due to the loss given by the non-conservative force, therefore there is a decrease in mechanical energy.

Therefore the correct answer is A. Kintetic energy increases and mechanical energy decreases.

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2 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for <em>t</em> to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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3 years ago

Evaluating alternative options and choosing one option among them is called the theory of bounded rationality.

VikaD [51]

It is false, bounded rationality is the idea that rationality is limited when individuals make decisions. ... Limitations include the difficulty of the problem requiring a decision, the cognitive capability of the mind, and the time available to make the decision.

But one thing, NEXT TIME TELL US THE QUESTION FIRST AND DON'T JUST LEAVE BLINDLY ASKING SOMETHING.

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1 year ago

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary

kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0

3 years ago