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GalinKa
4 days ago
8

A dust particle floats in front of a silent loudspeaker as shown in the figure. The loudspeaker is turned on and plays a constan

t low frequency note.
Physics
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During a snowball fight two balls with masses of 0.4 and 0.6 kg, respectively, are thrown in such a manner that they meet head-o
Yuliya22 [3333]

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

Speed of 1 Kg mass post-collision =?

Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

8 0
3 months ago
On a hot summer day, you decide to make some iced tea. First, you brew 1.50 LL of hot tea and leave it to steep until it has rea
Yuliya22 [3333]

Explanation:

Please refer to the attachment for the solution.

6 0
2 months ago
two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire
Sav [3153]
Thanks for asking your question here. I hope this response provides clarity. Feel free to ask additional questions. The moment resulting from the two forces about point O is 376 lb-ft counterclockwise.
6 0
2 months ago
A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,
Sav [3153]
Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.
6 0
2 months ago
A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of
Sav [3153]

Answer:

A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.

Explanation:

τchild=τrock  

We will utilize the formula for torque:

(F)child(d)child)=(F)rock(d)rock)

The gravitational force acts equally on both objects.

(m)childg(d)child)=(m)rockg(d)rock)

We can eliminate gravity from both sides of the equation for simplification.

 (m)child(d)child)=(m)rock(d)rock)  

Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.

(25kg)(1m)=(50kg)drock

Solve for the distance where the rock should be positioned in relation to the seesaw's center.

drock=25kg⋅m50kg

drock=0.5m

6 0
3 months ago
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