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Ira Lisetskai [31]
3 years ago
15

3. The pressure at the bottom of the ocean is great enough to crush submarines with steel walls that are 10 centimeters thick. S

uppose a submarine is at a depth of 1,000 meters. The weight of water above each square meter of the submarine is 9,800,000 newtons.
Physics
2 answers:
Arada [10]3 years ago
8 0

Answer:

answer one

Explanation:

Elden [556K]3 years ago
3 0

Answer:

just awnsered this one your awnser is the the second option

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An incident ray of light strikes water at an angle of 30°. The index of refraction of air is 1.0003, and the index of refraction
defon

Answer: It’s A

Explanation:

22 degrees

6 0
3 years ago
Read 2 more answers
Un paracaidista desciende desde 6 000 m de altura. Si la masa, con su equipo, es de 65 kg, ¿cuánto valdrá su energía potencial e
iris [78.8K]

Explanation:

PRIMERO ENCUENTRAS EL PESO DEL PARACAIDISTA

F_{peso} = 65 kg(9.80 m/s^{2}) = 637 N

CON LA FÓRMULA DE LA ENERGÍA POTENCIAL

U = 637 N(6000 m - 3500 m) = 1592500 J

3 0
3 years ago
A coin is placed on a vinyl stereo record that is making 33 1/3 revolutions per minute on a turntable. (a) In what direction is
mamaluj [8]

Answer: a) The acceletarion is directed to the center on the turntable. b) 5 cm; ac= 0.59 m/s^2; 10 cm, ac=1.20 m/s^2; 14 cm, ac=1.66 m/s^2

Explanation: In order to explain this problem we have to consider teh expression of the centripetal accelartion for a circular movement, which is given by:

ac=ω^2*r where ω and r are the angular speed and teh radios of the circular movement.

w=2*π*f

We know that the turntable is set to 33  1/3 rev/m so

the frequency 33.33/60=0.55 Hz

then w=2*π*0.55=3.45 rad/s

Finally the centripetal acceleration at differents radii results equal:

r= 0.05 m  ac=3.45^2*0.05=0.50 m/s^2

r=0.1  ac=3.45^2*0.1=1.20 m/s^2

r=0.14 ac=3.45^2*0.14=1.66 m/s^2

4 0
3 years ago
The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?
Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

                                               solve for L = .457 m/s^2  DOWNWARD

                                                        so L = - .457 m/s^2

4 0
2 years ago
-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a
bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1

The velocity of the carts after the event is 1 m/s

3 0
3 years ago
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