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4 years ago

I need help for a review sheet

Physics

1 answer:

siniylev [52]4 years ago

3 0

Tin is Sn, atomic number 47 is Silver, Mass of sodium is 22.9 u

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A 35-gram stainless steel ball on a track is moving at a velocity of 9 m/s. On the same track, a 75-gram stainless steel ball is

elena-s [515]

Answer:

4.2 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v

315 g m/s − 525 g m/s = -525 g m/s + (75 g) v

315 g m/s = (75 g) v

v = 4.2 m/s

5 0

4 years ago

Read 2 more answers

A waitperson carrying a tray with a platter on it tips the tray at an angle of 12 degrees below the horizontal. If the gravitati

Tasya [4]

Answer:1.04 N

Explanation:

Given

Gravitational Force on the Platter is $5 N$

Tray makes an angle of $\theta =12^{\circ}$

This gravitational Force has components along and Perpendicular to Platter

Perpendicular Force $W_p=W\cos \theta$

$W_p=5\times \cos 12=4.89 N$

Along the Tray

$W_{along}=W\sin \theta$

$W_{along}=5\times \sin 12=1.04 N$

Thus 1.04 N is the magnitude of force that will cause Platter to slide down

7 0

3 years ago

Which of the following atmospheric conditions are needed to create the winds?

Talja [164]

Answer:

is there suppose to be a pic?

Explanation:

5 0

3 years ago

Read 2 more answers

A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 83.1 N

Kay [80]

Answer:

The work done by the 83.1 N force is 4687.5 J.

The magnitude of the work done by the force of friction is 1187.5 J.

Explanation:

Given:

Mass of the block, $m=15.7$ kg

Force acting on it, $F=83.1$ N

Angle of application of force, $\theta = 25.7$ °

Displacement of the block, $d=62.6$ m

Coefficient of friction, $\mu =0.161$

Acceleration due to gravity, $g=9.8$ m/s²

Work done by a force is given as:

$Work,W=F\times d\times \cos\theta$

So, work done by the constant force is given as:

$W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}$

Now, in order to find work done by friction, we need to evaluate friction.

For the vertical direction, the net force is zero as there is no vertical motion.

Therefore,

$N + F\sin\theta = mg\\ N = mg-F\sin\theta$

Frictional force is given as:

$f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}$

Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.

Therefore, work done by friction is:

$W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}$

Negative sign indicates that frictional force is acting opposite to motion.

So, the magnitude of the work done by the force of friction is 1187.5 J.

8 0

4 years ago

Hi i need answers for this. Thank you!! this is also really important and is due tomorrow at 9am!!

Allushta [10]

Answer:

You're four sentences should include about how the roller coaster has the most potential energy at the top of the track, and the opposing energy, "kinetic" has the most kinetic energy when going down the hill.

Explanation:

Kinetic - In-Motion.

Potential - Gathering Energy to go into Motion.

( I'll try to answer questions to clear up confusion. )

7 0

3 years ago