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4 years ago

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The speed of sound in liquids, gases, and solids is as follows, from fastest to slowest a Gas, liquid, solid b- Liquid, solid, g

as c- Solid, liquid, gas

1 answer:

Hunter-Best [27]4 years ago

7 0

Answer:

c- Solid, liquid, gas

Explanation:

The inter molecular gap in solid is the lowest so sound travels fastest through solids

The inter-molecular gap in liquids is higher than solids so sound travels slower when compared though solids.

The inter-molecular gap in gasses is the highest among the three states of matter which makes sound travel through it the slowest.

So, the order from fastest to slowest will be Solid, liquid, gas

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A hot-air balloon with a mass of 400 kilograms moves across the sky with 3,200 joules of kinetic energy. The velocity of the bal

kotegsom [21]

Answer:

4 m/s

Explanation:

KE =

Velocity of balloon will be 4 m/s.

!! Hope It Helps !!

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3 years ago

For each of the following velocity vectors, give the vector component, the scalar component, and the magnitude. (a) v = 48.0j m/

Oduvanchick [21]

Obtaining the component part of the letter 'i' warns that it is the velocity is oriented on the X axis. At the same time the numeric part, in this particular case will represent the scalar part and therefore the magnitude of the vector.

The velocity is

$\vec{v} = 48.0\hat{i} \text{ m/s }$

The vector component of the velocity is

$v_x = 48.0m/s$

The scalar component of the velocity is

$v = 48.0m/s$

The magnitude of the velocity is

$v = 48.0m/s$

3 0

4 years ago

A 2000 kg car moves along a horizontal road at speed vo

cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

$\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a$ (Eq. 1)

$\Sigma F_{y} = N-m\cdot g = 0$ (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

$a = -\mu_{k}\cdot g$ (Eq. 3)

Where:

$a$ - Deceleration of the car, measured in meters per square second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $\mu_{k} = 0.0735$ and $g = 9.807\,\frac{m}{s^{2}}$ , then deceleration of the car is:

$a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )$

$a = -0.721\,\frac{m}{s^{2}}$

The stopping distance of the car ( $\Delta s$ ), measured in meters, is determined from the following kinematic expression:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (Eq. 4)

Where:

$v_{o}$ - Initial speed of the car, measured in meters per second.

$v$ - Final speed of the car, measured in meters per second.

If we know that $v_{o} = 15.9\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $a = -0.721\,\frac{m}{s^{2}}$ , stopping distance of the car is:

$\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}$

$\Delta s = 175.319\,m$

The shortest possible stopping distance of the car is 175.319 meters.

8 0

4 years ago

A force of 10N is applied by a boy while lifting a 20,000g mass. How much does it accelerate by?

Tanya [424]

Answer:

0.5 $m/s^{2}$

Explanation:

20000g to kg = $\frac{20000}{1000}$ = 20 kg

f = ma

a = $\frac{f}{m}$

a = $\frac{10}{20}$

a = $\frac{1}{2}$

a = 0.5 $m/s^{2}$

8 0

3 years ago

a nonposionous beetle raises its backside in the air to make a predator thinks its posionous this is an example of what

zalisa [80]

Its an example of survival

7 0

3 years ago