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3 years ago

List 6 Physical properties List 6 Chemical properties

1 answer:

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Physical-
Appearance, taste, texture, temperature, state of matter and odor

Chemical-
Reactivity, concentration, Chemical composition, charge, <span>Heat of combustion, and toxicity</span>

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5. A family leaves for an evening walk around the neighborhood. They leave at 5 pm and return at 6 pm. They traveled a total of

vlabodo [156]

This question involves the concepts of average speed, distance, and time.

The average speed of the family is "2.5 mi/hr".

<h3>AVERAGE SPEED</h3>

The average speed is defined as the ratio of the total distance covered with the total time taken to cover it.

$v=\frac{s}{t}$

where,

s = distance covered = 2.5 miles
t = total time taken = 1 hour
v = average speed = ?

Therefore,

$v=\frac{2.5\ mi}{1\ hr}$

v = 2.5 mi/hr

Learn more about average speed here:

brainly.com/question/12322912

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2 years ago

Which change indicates that the universe is expanding?

Aleonysh [2.5K]

Red shift of distant galaxies

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3 years ago

Read 2 more answers

T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?

Debora [2.8K]

Answer:

$P_J=195N$

Explanation:

From the question we are told that

$Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\$

Generally the equation for momentum is mathematically given by

$P=mv$

Therefore

T-Joe momentum $P_J$

$P_J=65*3$

$P_J=195N$

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2 years ago

Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp

erma4kov [3.2K]

Answer:

c. 981 watts

$P=981\ W$

Explanation:

Given:

horizontal speed of treadmill, $v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}$

weight carried, $w=588.6\ N$

grade of the treadmill, $G\%=10\%$

<u>Now the power can be given by:</u>

$P=v.w$

$P=588.6\times\frac{5}{3}$ (where grade is the rise of the front edge per 100 m of the horizontal length)

$P=981\ W$

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3 years ago

A soccer player pumps air into a soccer ball until no more air can be pushed inside. Describe the air inside the soccer ball com

inessss [21]

Answer:

the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

Explanation:

This exercise asks to describe the inflation situation of a spherical fultball.

Initially the balloon is deflated, therefore the internal pressure is equal to the pressure of the air outside, atmospheric pressure, when it begins to inflate the balloon with a pump this creates a pressure in the inlet valve and as it is greater than the pressure inside, the air enters it, this is repeated in each filling cycle, manual pump.

When the ball is full we have two forces, the one created by the external walls and the one aired by the pressure of the pump, these forces are directed towards the inside, but the air molecules exert a pressure towards the outside, which translates into a force. When these two forces are equal, the pump is no longer able to continue introducing air into the balloon.

Consequently the filling stops when the pressure of the pump equals the pressure of the interior air plus the pressure of the walls.

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3 years ago