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Studentka2010 [4]
4 years ago
15

Speed (mph) 58 72 55 65 70 81 66 What car goes the fastest? Put the car speed in order, from fastest to slowest miles per hour.

A) 55, 58, 65, 66, 70, 72, 81 B) 81, 72, 70, 66, 65, 58, 55 C) 81, 70, 66, 65, 58, 55 D) 81, 72, 70, 66, 58, 55
Physics
1 answer:
RoseWind [281]4 years ago
7 0

Answer:

The answer to your question is letter B

Explanation:

Data

Order from fastest to slowest speed

           58 72 55 65 70 81 66

The fastest is  81, then, 72, 70, then, 66, 65 finally 58 and 55

The correct order

       81 mi/h, 72 mi/h, 70 mi/h, 66 mi/h, 65 mi/h, 58 mi/h, 55 mi/h

The other options are from slowest to fastest or a different order.

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A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of
Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

  1.29 kg/m³   =  <u>      mass    </u>

                             1800 m³

mass =  1.29 kg/m³  ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0
3 years ago
Read 2 more answers
The pages of a book are numbered 1 to 200 and each
never [62]

Answer:

10.4mm

Explanation:

2 pages = 1 leaf

200 pages = 100 leaves

100 × 0.10 = 10 mm thickness

Total thickness = 2(0.20) +10 = 0.4+10 = 10.4mm

6 0
3 years ago
Hello can someone please help me with this.
user100 [1]

Answer:

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Explanation:

8 0
3 years ago
Two astronauts are playing catch in a zero gravitational field. Astronaut 1 of mass m1 is initially moving to the right with spe
Ede4ka [16]

The final velocity (v_1_f) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (v_2_f) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts <em>after throwing the ball</em>.

The given parameters;

  • Mass of the first astronaut, = m₁
  • Mass of the second astronaut, = m₂
  • Initial velocity of the first astronaut, = v₁
  • Initial velocity of the second astronaut, = v₂ > v₁
  • Mass of the ball, = m
  • Speed of the ball, = u
  • Final velocity of the first astronaut, = v_f_1
  • Final velocity of the second astronaut, = v_f_2

The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.

m_1v_1 + m_2v_2 = m_2v_2_f + m_1v_1_f

if v₂ > v₁, then v_1_f > v_2_f, to conserve the linear momentum.

Thus, the final velocity (v_1_f) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (v_2_f) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

Learn more here: brainly.com/question/24424291

5 0
3 years ago
A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
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