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NeTakaya
3 years ago
10

A 6.00 kg object is released from rest while fully submerged in a liquid. The liquid displaced by the submerged object has a mas

s of 3.45 kg. How far and in what direction does the object move in 0.200 s, assuming that it moves freely and that the drag force on it from the liquid is negligible
Physics
1 answer:
slava [35]3 years ago
7 0

Answer:

The object moves 0.0833 m downwards.

The object moves 8.33 cm downwards.

Explanation:

Weight of object = W = 6 × 9.8 = 58.8 N

Mass of liquid displaced = 3.45 kg

Meaning that buoyant force = Fb = 3.45 × 9.8 = 33.81 N

drag force on the object from the liquid is negligible.

Net force acting on the object = ma

Force balance on the object gives

Net force = W - Fb + D

W = 58.8 N

Fb = 33.81 N

D = 0

Net force = 58.8 - 33.81 = 24.99 N

ma = 24.99

6a = 24.99

a = (24.99/6) = 4.165 m/s²

Now, using one of the equations of motion

u = 0 m/s

a = 4.165 m/s²

y = ?

t = 0.200 s

y = ut + ½at²

y = 0 + (0.5 × 4.165 × 0.2²)

y = 0.0833 m

Since the Weight of the body is more than the buoyant force exerted by the liquid on the object, the object moves downwards.

Hope this Helps!!!

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A 25 kg bear slides, from rest, 12 m down a lodgepole pine tree, moving with a speed of 5.6 m/s just before hitting the ground.
Anuta_ua [19.1K]

Answer:

(A) -2940 J

(B) 392 J

(C) 212.33 N

Explanation:

mass of bear (m) = 25 kg

height of the pole (h) = 12 m

speed (v) = 5.6 m/s

acceleration due to gravity (g) = 9.8 m/s

(A) change in gravitational potential energy (ΔU) = mg(height at the bottom- height at the top)

height at the bottom = 0

         = 25 x 9.8 x (0-12) = -2940 J

(B) kinetic energy of the Bear (KE) = 0.5mv^{2}

           = 0.5 x 25 x 5.6^{2}  = 392 J

(C) average frictional force = \frac{change in thermal energy}{height} = \frac{-(ΔKE+ΔU)}{h}

  • change in KE (ΔKE) = initial KE - final KE
  • ΔKE = 0.5mv^{2} - 0.5mvf^{2}            
  • when the Bear reaches the bottom of the pole, the final velocity (Vf) is 0, therefore the change in kinetic energy becomes  ΔKE = 0.5x25x5.6^{2} - 0 = 392 J

 \frac{-(ΔKE+ΔU)}{h}[/tex] = \frac{-(392 + (-2940))}{12}

=  \frac{(-392 + 2940)}{12} = 212.33 N

5 0
4 years ago
1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t
tiny-mole [99]

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

6 0
3 years ago
A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.50 107 m/s and experiences an acceleration
KIM [24]

Answer:

A) B = 0.009185 T

B) Drection is negative y-direction

Explanation:

A) We are given;

Speed(v) = 2.5 x 10^(7) m/s

Acceleration (a) = 2.2 x 10^(13) m/s²

We also know that charge of proton(q) = 1.6 x 10^(-19)

Mass of proton(m) = 1.67 x 10^(-27)

Now, Since the proton is moving by circular motion, this force is equal to the centripetal force which is given as;

F = qvBsinθ = ma

Since perpendicular, θ = 90°

And so, sinθ = sin 90 = 1

Thus, qvB = ma

Making B the subject gives;

B = ma/qv

B = (1.67 X 10^(-27) X 2.2 X 10^13)) / (1.6 X 10^(-19) X 2.5 X 10^(7))

= 0.009185 T

B) By use of Flemings right hand rule, we can see that the middle finger points toward negative y-direction, so the magnetic field is in the negative y-direction

4 0
4 years ago
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7 0
3 years ago
A lawnmower engine running for 20 minutes does 4560000 j of work. What is the power output of the engine
Alenkasestr [34]

<em>Answer: </em>

tim e (t) = 20 min.

            = 20 × 60 = 1200 s ,

Work ( W) = 4560000 J

                 = 4560 KJ ,

Determine:

                 Power output (P) = Work ÷ time

                                              = 4560 ÷  1200

                                         <em>   P  = 3.8 KW</em>


3 0
4 years ago
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