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2 years ago

Which of the following statements best describes the movement of particles in a liquid?

Physics

2 answers:

Vesna [10]2 years ago

5 0

Answer:

The particles in a liquid are close together but move past each other

Explanation:

Rufina [12.5K]2 years ago

5 0

Answer:

The partials are far apart but moving fast

Explanation:

g o o g l e

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A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of

OLEGan [10]

Answer:

c) $F_{net} = 0.640 kN$

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

$F_{net} = ma$

here we know

m = 80 kg

for circular motion acceleration is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{40^2}{200} = 8 m/s^2$

now we have

$F_{net} = 80 \times 8$

$F_{net} = 640 N$

$F_{net} = 0.640 kN$

7 0

3 years ago

A baseball is thrown at a 28° angle and an initial velocity of 70 m/s. Assume no air resistance. What is the vertical component

icang [17]

Answer:

answer is 61.8 m/s; 32.9

l am not sure

3 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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3 years ago

Which of these energy transformations is not part of how a nuclear power plant

frutty [35]

I think the second one

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3 years ago

What shape is the unit cell of ruby?

sleet_krkn [62]

Ruby, a variety of the mineral corundum is in the trigonal crystal system, with hexagonal scalenohedra crystals

7 0

4 years ago