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3 years ago

Which events may occur when ocean salinity increases? Check all that apply.

Physics

2 answers:

sergey [27]3 years ago

7 0

Answer:

1)The mass of water increases.

2) Freezing point of water decreases.

4) bouyancy of objects in the water increases.

Salinity in the oceanic water is attributed by the water coming from other sources like rivers carrying rocks and soil containing salt. Salinity decreases the temperature of water, making it cold and more denser. The cold water settles at the bottom of the ocean. It flows across the oceanic basins as slow currents. The objects being less in density flows over the surface of water. Hence,bouyancy of objects in water increases.

Pavlova-9 [17]3 years ago

3 0

The answer is number two, number four, and number one

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PLEASEEEE HEEEEELP!!!!!

8090 [49]

Answer:

Before:

$p_{truck}=16400\ kg.m/s$

$p_{car}=10000\ kg.m/s$

After:

$p_{truck}=8000\ kg.m/s$

$p_{car}=8400\ kg.m/s$

$v_{fcar}=8.4\ m/s$

$F=9333.33 \ Nw$

Explanation:

Conservation of Momentum

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

$p_1=m_1v_{1o}+m_2v_{2o}$

After the collision, they have speeds of v1f and v2f and the total momentum is

$p_2=m_1v_{1f}+m_2v_{2f}$

Impulse J is defined as

$J=F.t$

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

$J=\Delta p$

As the total momentum is conserved:

$p_1=p_2$

$m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}$

We can compute the speed of the second object by solving the above equation for v2f

$\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{ m_2 }$

The given data is

$m_1=2000\ kg$

$m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s$

a) The impulse will be computed at the very end of the answer

b) Before the collision

$p_{truck}=2000\cdot 8.2=16400\ kg.m/s$

$p_{car}=1000\cdot 0=0\ kg.m/s$

c) After collision

$p_{truck}=2000\cdot 4=8000\ kg.m/s$

Compute the car's speed:

$\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }$

$v_{2f}=8.4\ m/s$

And the car's momentum is

$p_{car}=1000\cdot 8.4=8400\ kg.m/s$

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

$J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s$

The force can be computed as

$\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw$

The force on the car has the same magnitude and opposite sign

7 0

4 years ago

A frictionless toy car is placed on a ramp, which is inclined at an unknown angle with respect to the horizontal. Starting from

NikAS [45]

The final speed of the toy car at the end of the given time period is 3.58 m/s.

The given parameters;

distance traveled by the car, s = 1.2 m
time of motion of the car, t = 0.67 s
initial velocity of the car, u = 0

The acceleration of the car is calculated as;

$s = ut + \frac{1}{2} at^2\\\\1.2 = 0 + 0.5\times a\times (0.67)^2\\\\1.2 = 0.225a\\\\a = \frac{1.2}{0.225} \\\\a = 5.33 \ m/s^2$

The final velocity of the toy car is calculated as;

$v_f^2 = u^2 + 2as\\\\v_f^2 = 0 + 2\times 5.33 \times 1.2\\\\v_f^2 = 12.792\\\\v_f = \sqrt{12.792} \\\\v_f = 3.58 \ m/s$

Thus, the final speed of the toy car at the end of the given time period is 3.58 m/s.

Learn more here: brainly.com/question/20352766

6 0

2 years ago

Which statement is a hypothesis?

olga_2 [115]

B, do earthworms prefer bright light or darkness!

3 0

2 years ago

Read 2 more answers

B c _ - _ a b ( a=3 b=-5 c=6)

Pachacha [2.7K]

(-5)/3 - 6/(-5)
You can solve it now :)

7 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

3 years ago

Read 2 more answers