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3 years ago

5

A minimum temperature thermometer in a standard weather instrument shelter measures an overnight low temperature of 36 degrees F

ahrenheit. Explain why frost was observed on the ground beneath the shelter, even though the recorded minimum temperature was above the freezing point.

1 answer:

Rudiy273 years ago

5 0

Answer:

Explanation:

Even though the minimum temperature is more than the freezing point, Frost was observed on the ground because the ground will cool rapidly as cool air tends to move towards the ground, the temperature of the ground is lower than the atmosphere a few feet above it.

As the thermometer is kept some feet above the ground so ground temperature may be lower than the minimum recorded temperature.

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In t-ball, young players use a bat to hit a stationary ball off a stand. The 140 g ball has about the same mass as a baseball, b

fomenos

Question is missing. Found on google:

<em>"Part A What is the acceleration of the ball? Express your answer to two significant figures and include the appropriate units. </em>

<em>Part B </em>

<em>What is the net force on the ball during the hit? </em>

<em>Express your answer to two significant figures and include the appropriate units."</em>

Solution:

A) $6000 m/s^2$

The acceleration of the ball is given by

$a=\frac{v-u}{t}$

where

v = 12 m/s is the final velocity

u = 0 is the initial velocity (the ball is stationary)

t = 2.0 ms = 0.002 s is the time of contact

Substituting,

$a=\frac{12-0}{0.002}=6.0 \cdot 10^3 m/s^2$

B) $8.4\cdot 10^2 N$

The force on the ball can be found by using Newton's second law:

$F=ma$

where

m = 140 g = 0.14 kg is the mass of the ball

$a=6.0\cdot 10^3 m/s^2$ is the acceleration

Substituting,

$F=(0.14)(6.0\cdot 10^3)=8.4\cdot 10^2 N$

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3 years ago

(Look at the minerals in the linked picture)

Crank

Both diamond and coal are formed by changes in pressure and temperature below the Earth's surface. The step in the formation of the minerals is <span>atoms break up in extreme heat. The answer is letter A.</span>

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3 years ago

g Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the bu

dlinn [17]

Answer:

the observed frequency will reduce but the wavelength will increase

Explanation:

As we know

fo = fs (v/(v-vs))

fo = observed frequency

vs = velocity of source

As per this equation,

When an observer moves away from the stationary source, the observed frequency reduces. Since the observer in the balloon is moving away from the source which itself is moving in opposite direction, the observed frequency will reduce.

Since wavelength = V/fs . The source frequency is unchanged but the velocity is increasing as it is moving in downward direction. Hence, the wavelength will increase

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3 years ago

A mover loads a 100 kg box into the back of a moving truck by

NeX [460]

Answer:

2.7

Explanation:

The following data were obtained from the question:

Mass (m) of box = 100 Kg

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:

Mechanical Advantage = Lenght / height

MA= L/H

With the above formula, we can obtain the mechanical advantage of the ramp as follow:

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

MA = 4/1.5

MA = 2.7

Therefore, the mechanical advantage of the ramp is 2.7

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3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

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3 years ago