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3 years ago

11

Which will result in positive buoyancy and cause the object to float?

2 answers:

Wewaii [24]3 years ago

4 0

the answer should be:

When the buoyant force is equal to the force of gravity

natali 33 [55]3 years ago

4 0

The answer is when the buoyant force is greater than gravity

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A student weighing 120 lbs climbs a 12 ft flight of stairs in 9 seconds. how much power did the student create?

Alex Ar [27]

Power can be calculate through the equation,
Power = Force x velocity

It should be noted that velocity is calculated by dividing displacement by time. Thus, from the given in this item we can calculate for the power.
Power = (120 lb) x (12 ft/9 s)
<em> </em><span><em>Power = 160 lb.ft/s</em></span>

7 0

3 years ago

Red shift data shows that galaxies are O expanding O shrinking O moving away O moving closer

Anon25 [30]

Answer:

Should be moving away

Explanation:

Red is a longer wavelength therefore further away. Wavelength is stretched out more and on the red end. I hope this is right. I decided to research and answer since you didn’t have other answers. Are you taking this on edg? I hope I helped!

7 0

2 years ago

An acorn with a mass of 0.300 kilograms falls from a tree. What is its kinetic energy when it reaches a velocity of 5.oo m/s?

vladimir1956 [14]

Answer: 3.75 joules

Explanation:

Given that:

Mass of acorn = 0.300 kilograms

velocity = 5.oo m/s

Kinetic energy = ?

Since, kinetic energy is the energy possessed by a moving object, its value depends on the mass M and velocity V of the acorn.

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 0.300kg x (5.00m/s)^2

= 0.5 x 0.3kg x (5.00m/s)^2

= 0.15 x (5.00m/s)^2

= 3.75 joules

Thus, the kinetic energy of the falling acorn is 3.75 joules

5 0

2 years ago

A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth

svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

$\omega = \sqrt{\dfrac{k}{m}}$

now from equation (3)

$a_x = \dfrac{k}{m}x$

$k = \dfrac{m a_x}{x}$

$k = \dfrac{0.45 \times (-14)}{0.24}$

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

8 0

3 years ago

Read 2 more answers

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago