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3 years ago

2. Describe what would happen to both air temperature and soil temperature if cold weather were to pass through the area.

1 answer:

7 0

The air temperature would drop quickly compared to the soil because the difference between solid and gas. But the ground will stay cooler longer than the air because it contains it better than the air.

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What is a device that does not work with an electric circuit?

storchak [24]

It would be a bike that doesn't use an electric circuit.

8 0

3 years ago

Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above

poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$W_{dis}$ - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

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

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

$f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}$

If $m = 1000\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 40\,m$ , $y_{2} = 25\,m$ and $\Delta s = 400\,m$ , then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0

3 years ago

A missle is fired horizontally with an initial velocity of 45 m/s from the top of a building 75 m high.

NARA [144]

The horizontal range of the missile is b) 176 m

Explanation:

The motion of the missile is a projectile motion, so it consists of two independent motions:

- A uniform motion with constant velocity along the horizontal direction

- A uniformly accelerated motion with constant acceleration (equal to the acceleration of gravity) in the vertical-downward direction

To find the time of flight of the missile, we study the vertical motion. We can use the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where:

s = 75 m is the vertical displacement of the missile (the height of the building)

$u_y=0$ is the initial vertical velocity (the missile is thrown horizontally)

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(75)}{9.8}}=3.91 s$

This means that the missile takes 3.91 s to reach the ground.

Now we study the horizontal motion: the missile moves with a constant horizontal velocity of

$v_x = 45 m/s$

Therefore, the distance covered in a time t is

$d=v_x t$

and by substituting t = 3.91 s, we find the horizontal range of the missile:

$d=(45)(3.91)=176 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago

Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is

marshall27 [118]

Answer:

$x=0.0049\ m= 4.9\ mm$

$d=0.01153\ m=11.53\ mm$

Explanation:

Given:

mass of the object, $m=0.75\ kg$

elastic constant of the connected spring, $k=150\ N.m^{-1}$

coefficient of static friction between the object and the surface, $\mu_s=0.1$

(a)

Let x be the maximum distance of stretch without moving the mass.

The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.

$f=k.x$

$\mu_s.N=k.x$

where:

N = m.g = the normal reaction force acting on the body under steady state.

$0.1\times (9.8\times 0.75)=150\times x$

$x=0.0049\ m= 4.9\ mm$

(b)

Now, according to the question:

Amplitude of oscillation, $A= 0.0098\ m$

coefficient of kinetic friction between the object and the surface, $\mu_k=0.085$

Let d be the total distance the object travels before stopping.

Now, the energy stored in the spring due to vibration of amplitude:

$U=\frac{1}{2} k.A^2$

This energy will be equal to the work done by the kinetic friction to stop it.

$U=F_k.d$

$\frac{1}{2} k.A^2=\mu_k.N.d$

$0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d$

$d=0.01153\ m=11.53\ mm$

is the total distance does it travel before stopping.

7 0

2 years ago

A jetliner flies east 2000 miles from San Francisco to Chicago. It has an average air speed of 600 mph relative to the air. What

Evgesh-ka [11]

The answer is D 600mph east

3 0

3 years ago

Read 2 more answers