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just olya [345]
3 years ago
6

At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the water on either

the inside of the fishbowl (for points A and B) or the outside of the goldfish (for points C, D, and E).

Physics
1 answer:
GrogVix [38]3 years ago
4 0

Answer:

The direction of the force at A and B is perpendicular to the walls of the container.

The direction of the force at C is down.

The direction of the force in D is up

The direction of the force at E is to the left.

The attached figure shows the forces exerted by the water at points A, B, C, D and E.

Explanation:

The water is in contact with the bowl and with the fish. It exercises at points A, B, C, D and E, but the direction is different from the force.

The fish has a buoyant force on the water and that direction is up. The direction of at point D is up.

The column of water on the fish has a downward force, therefore the direction of the force at point C is down. The water column to the right of the fish has a force to the left, and the direction at point E is to the left.

The water will exert a force on the walls of the container and this force at points A and B is a on the walls of the container.

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A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad
Korolek [52]

Answer:

9.21954 m/s

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Explanation:

r = Radius of arm = 1.5 m

\omega = Angular velocity = 6 rad/s

The horizontal component of speed is given by

v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s

The vertical component of speed is given by

v_v=2\ m/s

The resultant of the two components will give us the velocity of hammer with respect to the ground

v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

3 0
3 years ago
A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a v
ch4aika [34]

Answer

given,

Time period= T = 1.5 s

If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s

v_max=1.00 m/s

we know,

v_ max=A ω  

v = A sin (ωt)

-0.50= -1.00 sin (ωt)

sin (ωt)  =  0.5

\omega t = sin^{-1}(0.5)

\dfrac{2\pi}{T}\times t =0.524

\dfrac{2\pi}{1.5}\times t =0.524

t = 0.125 s

we have time period T=1.5 it is the time to complete one oscillation

means from eq to right,then left,then eq,then left,then from right to eq

time taken for left = t/4 = 0.125/4 = 0.375 s

smallest value of time

=0.375 + 0.125

= 0.50 sec

7 0
3 years ago
A virtual image is formed 17.0 cm from a concave mirror having a radius of curvature of 39.0 cm. (a) Find the position of the ob
Wittaler [7]

Explanation:

Image distance, v = -17 cm (-ve for virtual image)

Radius of curvature of concave mirror, R = 39 cm

Focal length, f = -19.5 cm (-ve for a concave mirror)

(a) Using mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{-19.5}-\dfrac{1}{(-17)}  

u = 132.6 cm    

So, the object is placed 132.6 cm in front of the mirror.

(b) Magnification of the  mirror, m=\dfrac{-v}{u}

m=\dfrac{-17}{132.6}

m = -0.128

Hence, this is the required solution.

7 0
3 years ago
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