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2 years ago

The buoyant force on an object fully submerged in a liquid depends on (select all that apply)

1 answer:

8 0

The buoyant force on an object fully submerged in a liquid depends on
the density of the liquid, and the density of the object. But the density
of the object depends on the object's volume and the object's mass.

So the only item on this list that it DOESN't depend on is the mass of
the liquid.

I guess that means that the buoyant force on a fully submerged object is
the same whether it's submerged in a cup of water or the Pacific Ocean.

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A rock is rolling down a hill. At position 1, it’s velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, it’s velo

mr Goodwill [35]

Answer

Hi,

correct answer is {D} 3.5 m/s²

Explanation

Acceleration is the rate of change of velocity with time. Acceleration can occur when a moving body is speeding up, slowing down or changing direction.

Acceleration is calculated by the equation =change in velocity/change in time

a= {velocity final-velocity initial}/(change in time)

a=v-u/Δt

The units for acceleration is meters per second square m/s²

In this example, initial velocity =2.0m/s⇒u

Final velocity=44.0m/s⇒v

Time taken for change in velocity=12 s⇒Δt

a= (44-2)/12 = 42/12

3.5 m/s²

Best Wishes!

5 0

1 year ago

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is referred to as

Dafna11 [192]

When carrying extra weight, the space formed between the top of your head and the two axles of the motorcycle is called "load triangle". Because of a motorcycle's size and weight<span> and the fact that it has only two wheels, how to carry extra load is very important. One has to make sure that they are keeping the weight low and close to the middle of the motorcycle and keep the load evenly from side to side. Heavier items should be in the "load triangle".</span><span> </span>

3 0

1 year ago

Disturbed by speeding cars outside his workplace, Nobel laureate Arthur Holly Compton designed a speed bump (called the "Holly h

Bezzdna [24]

:<span> </span><span>30.50 km/h = 30.50^3 m / 3600s = 8.47 m/s

At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg)

So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..

Fn = mg - mv²/R
Fn = m(g - v²/R) .. .. 1800kg (9.80 - 8.47²/20.20) .. .. .. ►Fn = 11 247 N (upwards)
(b)
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero.

ie .. mg = mv²/R .. .. v² = Rg .. .. 20.20m x 9.80 = 198.0(m/s)²

►v = √198 = 14.0 m/s</span>

3 0

2 years ago

Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass

zvonat [6]

Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value into the formula

$8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v$

$v=\dfrac{-19.375}{14.25}$

$v=-1.35\ m/s$

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2$

$K.E_{i}=239.96\ J$

We need to calculate the final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2$

$K.E_{f}=12.98\ J$

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

$\Delta K.E=K.E_{f}-K.E_{i}$

Put the value into the formula

$\Delta K.E=12.98-239.96$

$\Delta K.E=-226.98\ J$

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

8 0

1 year ago

Read 2 more answers

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago