(01.06 MC)

The height of a transverse wave from the midpoint to the crest or trough is the .

Sedaia [141]

Answer:Amplitude

Explanation:

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3 years ago

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Newton’s cradle is a contraption where metal balls hang from a frame. When one ball is pulled and released, the collision causes

Marysya12 [62]

Answer: A

Explanation: the principle of friction with swinging and colliding balls.

6 0

2 years ago

a hot piece of copper is placed in an insulated cup. what is the final temperature of the water and copper?

slavikrds [6]

Answer:

Option C. 30°C.

Explanation:

The following data were obtained from the question:

Mass of water (Mw) = 0.5 Kg

Specific heat capacity of water (Cw) = 4.18 KJ/Kg°C

Initial temperature of water (Tw1) =

22°C

Change in temperature (ΔT) = T2 – Tw1 = T2 – 22°C

Mass of copper (Mc) = 0.5 Kg

Specific heat capacity of copper (Cc) = 0.386 KJ/kg°C

Initial temperature of copper (Tc1) = 115°C

Change in temperature (ΔT) = T2 – Tc1 = T2 – 115°C

Final temperature (T2) =..?

Note: Both the water and the piece copper will have the same final temperature and the heat will be zero since the water will cool the piece of copper.

Thus, we can determine the final temperature of the water and copper as follow:

Q = MwCwΔT + McCcΔT

0 = 0.5 x 4.18 x (T2 – 22) + 0.5 x 0.386 x (T2 – 115)

0 = 2.09 (T2 – 22) + 0.193 (T2 – 115)

0 = 2.09T2 – 45.98 + 0.193T2 – 22.195

Collect like terms

2.09T2 + 0.193T2 = 45.98 + 22.195

2.283T2 = 68.175

Divide both side by the coefficient of T2 i.e 2.283

T2 = 68.175/2.283

T2 = 29.8 ≈ 30°C

Therefore, the final temperature of water and copper is 30°C.

8 0

3 years ago

Make a Book Quilt Project about the book Bud, Not Buddy

Leya [2.2K]

Answer:

Explanation:

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A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N

Nonamiya [84]

Answer:

A) 667 J

B) 381.4 J

C) 0 J

D) 245.4 J

E) 40.2J

F) 2 m/s

Explanation:

Let g = 9.81 m/s2

A) The work done on the suitcase is the product of the force applied and the distance travelled:

w = Fs = 145 * 4.6 = 667 J

B) The work done by gravitational force the dot product between the gravity vector and the distance vector

$W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J$

C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0

D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient

$W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J$

E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work

$W = w – W_g – W_f = 667 – 381.4 – 245.4 = 40.2 J$

F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:

$E_k = W = 40.2 j$

$mv^2/2 = 40.2$

$20v^2/2 = 40.2$

$10v^2 = 40.2$

$v^2 = 4.02$

$v = \sqrt{4.02} = 2 m/s$

3 0

3 years ago