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hichkok12 [17]
2 years ago
15

Three balls are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density less than the den

sity of water is held below the surface by a cord anchored to the bottom of the container, so that it is fully submerged. Ball 3, of same radius as ball 2, but of greater mass, is suspended from a rope so that it is fully submerged.
Which is true for Ball 1? Choose the best answer.

1) The magnitude of the buoyancy force is exactly one half of that of the ball's weight.
2) The magnitude of the buoyancy force is more than that of the ball's weight.
3) The magnitude of the buoyancy force is less than that of the ball's weight.
4) The magnitude of the buoyancy force is equal to that of the ball's weight.
Physics
2 answers:
goblinko [34]2 years ago
7 0

Answer:

4) The magnitude of the buoyancy force is equal to that of the ball's weight

Explanation:

Since ball 1 floats without sinking then it is in equilibrium. This means that the net downward force (weight) is balanced by the net upward force (buoyancy forces).

Ymorist [56]2 years ago
4 0

Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g  is acting vertically downward

Force of buoyancy FB = ρVdisg is acting vertically upward.

Net force acting on the ball is zero, FB=Fg

Answer

The magnitude of buoyancy force is equal to that of ball's weight.

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What are physical forms in which a substance can exist?
Alecsey [184]
Physical forms are: gas,liquid,and solid
8 0
1 year ago
an input force of 50 Newtons is applied through a distance of 10 meters to machine with mechanical advantage of 3. If the work o
gladu [14]
The output of the machine is

                                      (output work) =  (output force) x (distance)

                                        450 N-m      =  (output force) x (3 meters)

Divide each side
by  3 meters:                Output force = (450 N-m) / (3 m)

                                                           =    150 newtons .

With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.

It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m).  So everything
is nice and hunky-dory.
___________________________________

Well, my goodness !
I didn't even need to go through all of that.

Given:

-- The input force to the machine is 50 newtons.

-- The mechanical advantage of the machine is 3 .

That right there tells us that

-- The output force of the machine is 150 newtons.

We don't need any of the other given information.
5 0
2 years ago
At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
dalvyx [7]

Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}

Part(b):

Also the angular displacement (\Delta \theta) can be written as

&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad

8 0
1 year ago
Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g
lys-0071 [83]

Answer:

a)X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b)Does not affect the long term.

Explanation:

Given that

\dfrac{dx}{dt}=-k(x-A)

A = A0 cos(ωt)

\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))

\dfrac{dx}{dt}+kx=kA_o cos(\omega t)

This is linear equation so integration factor ,I

I=e^{\int kdt}

I=e^{kt}

Now by using linear equation property

e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C

e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C

X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b)

at t= 0

X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C

X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+e^{-kt}\times \left ( X(0)-\dfrac{k^2A_o}{\omega^2+k^2} \right )

So the initial condition does not affect the long term.

5 0
2 years ago
The same physics student jumps off the back of her Laser again, but this time the Laser is
soldi70 [24.7K]

a) The speed of the student after the jump is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

a)

We can solve this problem by applying the law of conservation of momentum: if there are no external forces acting on the system, the total momentum of the student+Laser system must be constant. Therefore, we can write:

p_i = p_f\\0=mv+MV

where

The initial momentum is zero

m = 42 kg is the mass of the Laser

v = 1.5 m/s is the final velocity of the Laser

M = 59 kg is the mass of the student

V is the final velocity of the student

Solving the equation for V, we find the velocity of the student:

V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s

So, the final speed of the student is 1.07 m/s.

b)

In this case, the laser and the student are travelling at 3.1 m/s before the student jumps off: therefore, the total momentum before the jump is not zero.

So, the equation of the conservation of momentum is

(m+M)u=mv+MV

where

m = 42 kg is the mass of the Laser

M = 59 kg is the student's mass

u = 3.1 m/s is the initial velocity of the student and the Laser

V = -2.1 m/s is the velocity of the student after the jump (she jumps backward)

v is the final velocity of the Laser

And solving for v, we find

v=\frac{(m+M)u-MV}{m}=\frac{(42+59)(3.1)-(59)(-2.1)}{42}=10.4 m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

3 0
1 year ago
Read 2 more answers
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