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4 years ago

5

HURRY PLEASEEEE a salvage crew is using a giant magnet on a surface of the ocean to raise sunken ship . They have multiple ships

at different depths that need to be raised. Which ships will be attracted to the magnet the fastest?

1 answer:

matrenka [14]4 years ago

7 0

The ship closest to the magnet

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If a transverse wave travels 10 meters in 5 seconds,what is its speed?

djverab [1.8K]

Well if we divide both numbers by 5, you get 2 meters per second. Multply this by 3600seconds per hour and you get 7200 meters per hour. There are 1609.34 meters in a mile, so divide 7200 by 1609 to get 4.5 miles per hour

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3 years ago

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An object is placed at a known distance in front of a mirror whose focal length is also known. You apply the mirror equation and

Licemer1 [7]

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the image is behind the mirror, so the image is virtual and upright.

Explanation:

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4 years ago

System uses 49J of energy to do work in the change of internal energy is 58 j. how much heat was added to the system?

Alika [10]

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3 years ago

The motion of an electron is given by x(t)=pt3+qt2+r, with p = -2.3 m/s^3 ,q = +1.5 m/s^2 , and r = +9.0 m.A) Determine its velo

alexandr402 [8]

Answer:

$v(0)=0\\v(1)=-3.9\ m/s\\v(2)=-21.6\ m/s\\v(3)=-53.1\ m/s$

Explanation:

<u>Instant Velocity </u>

Given the position as a function of time x(t), the instant velocity is the derivative of the function:

$v(t)=x'(t)$

We are given the position as

$x(t)=-2.3t^3+1.5t^2+9$

The derivative of x is

$v(t)=x'(t)=-6.9t^2+3.0t$

A) Let's compute v(0)

$v(0)=-6.9(0)^2+3.0(0)=0$

B)

$v(1)=-6.9(1)^2+3.0(1)$

$v(1)=-3.9\ m/s$

C)

$v(2)=-6.9(2)^2+3.0(2)$

$v(2)=-21.6\ m/s$

D)

$v(3)=-6.9(3)^2+3.0(3)$

$v(3)=-53.1\ m/s$

4 0

3 years ago

A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s

____ [38]

Answer:

Speed of both blocks after collision is 2 m/s

Explanation:

It is given that,

Mass of both blocks, m₁ = m₂ = 1 kg

Velocity of first block, u₁ = 3 m/s

Velocity of other block, u₂ = 1 m/s

Since, both blocks stick after collision. So, it is a case of inelastic collision. The momentum remains conserved while the kinetic energy energy gets reduced after the collision. Let v is the common velocity of both blocks. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{1\ kg\times 3\ m/s+1\ kg\times 1\ m/s}{2\ kg}$

v = 2 m/s

Hence, their speed after collision is 2 m/s.

7 0

4 years ago