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Setler [38]
3 years ago
13

WILL MARK BRAINLIEST PLEASE HELP AND SHOW ALL WORK

Physics
1 answer:
jok3333 [9.3K]3 years ago
6 0

Answer:

Same reading.

Explanation:

Assume that after the string breaks the ball falls through the liquid with constant speed. If the mass of the bucket and the liquid is 1.20 kg, and the mass of the ball is 0.150 kg,

A.) Before the string break, the total weight = weight of the can + weight of the water.

According to Archimedes' Principle which state that: “A body immersed in a liquid loses weight by an amount equal to the weight of the liquid displaced.” Archimedes principle also states that: “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it

B.) After the string break.

The scale will have the same reading as before the string break.

You might be interested in
The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d
r-ruslan [8.4K]

Answer:

0 degrees

Explanation:

Let F_1\ and\ F_2 are two forces. The resultant of two forces acting on the same point is given by :

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}

Where \theta is the angle between two forces

When \theta=0 i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}

F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}

When \theta=90^{\circ} i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}

F_R=\sqrt{F_1^2+F_2^2}

When \theta=180^{\circ} i.e. when two forces are parallel to each other,

F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}

F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

8 0
3 years ago
Please Help! When a metal is heated, the free electrons gain _____________.
lakkis [162]
It’s B potential energy
7 0
3 years ago
Read 2 more answers
What items are inside the nucleus?
Natalija [7]

Answer:

protons and neutrons

Explanation:

8 0
3 years ago
A 26-kg sled is on a snow-covered slope. The coefficients of friction between the sled’s runners and the snow are µs = 0.096 and µ
sweet-ann [11.9K]

Answer:

Explanation:

Given

mass of sled =26 kg

coefficient of static friction \mu _s=0.096

coefficient of kinetic friction \mu _k=0.072

In order to move sled from rest we need to provide a force greater than static friction which is given by

f_s=\mu mg=0.096\times 26\times 9.8=24.46 N

After Moving Sled kinetic friction comes in to play which is less than static friction

f_k=\mu _kmg=0.072\times 26\times 9.8=18.34 N

therefore minimum force to keep moving sledge at constant velocity is 18.34 N

3 0
3 years ago
An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000
Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

     (1.6 x 10⁵)²  =  (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

     (2.56 x 10¹⁰) - (4.624 x 10⁷)  =  2.0884 x 10⁻⁴ f²


Frequency² =   (2.56 x 10¹⁰ - 4.624 x 10⁷)  /  2.0884 x 10⁻⁴

                    =       2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

                    =          1.224 x 10¹⁴ 

                    =          122,400 GHz          <== my calculation

                                      11.1 MHz           <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation. 
  











7 0
3 years ago
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