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Scorpion4ik [409]

4 years ago

9

4748.36242611007 round to the nearest tenth

2 answers:

IceJOKER [234]4 years ago

3 0

Answer:4748.36

Explanation:

Harman [31]4 years ago

3 0

Answer:

4748.36242611007 rounded to the nearest tenth is 4,750

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If a car is traveling forward at 15 m/s, how fast will it be going in 1.2 seconds if the acceleration is

Law Incorporation [45]

Answer:

3

Explanation:

v = v⁰ (its original speed) + a (negative acceleration) X t² (time)

v = 15 - 10 x 1.2 = 15 - 12 = 3 (it's slowing down)

3 0

2 years ago

You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l

Ugo [173]

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

$E_p = E_{kv} + E_{k\omega}$

$mgh = mv^2/2 + I\omega^2/2$

where m is the mass, $I = mr^2/2$ is the moments of inertia of the solid cylinder $\omega = v / r$ is the angular speed of the cylinder

$mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}$

$mgh = mv^2/2 + mv^2/4 = 3mv^2/4$

$h = 3v^2/(4g)$

So if you plot a liner chart of h vs $v^2$ and get a slope of 6.42 then that means 3/(4g) = 6.42 so $g = 6.42*4/3 = 8.56 m/s^2$

The gravitational acceleration on this planet is 8.56 m/s2

3 0

4 years ago

19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,

sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

$t=15.2 s$ is the time it takes to the child to reach the bottom of the slope

$V_{o}=0$ is the initial velocity (the child started from rest)

$\theta=15\°$ is the angle of the slope

$d$ is the length of the slope

Now, the Force exerted on the sled along the ramp is:

$F=ma$ (1)

Where $m$ is the mass of the sled and $a$ its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

$F=mg sin \theta$ (2)

Where $g=9.8 m/s^{2}$ is the acceleration due gravity

Then:

$ma=mg sin \theta$ (3)

Finding $a$ :

$a=g sin \theta$ (4)

$a=9.8 m/s^{2} sin(15\°)$ (5)

$a=2.5 m/s^{2}$ (6)

Now, we will use the following kinematic equations to find $d$ :

$V=V_{o}+at$ (7)

$V^{2}=V_{o}^{2}+2ad$ (8)

Where $V$ is the final velocity

Finding $V$ from (7):

$V=at=(2.5 m/s^{2})(15.2 s)$ (9)

$V=38 m/s$ (10)

Substituting (10) in (8):

$(38 m/s)^{2}=2(2.5 m/s^{2})d$ (11)

Finding $d$ :

$d=288.8 m$

6 0

3 years ago

A motorist drives north for 35.0 minutes at 85.0 kph. He then stops for 15.0 minutes. The motorist then drives 130.0 km in 2.0 h

o-na [289]

Answer:

A. 216.36 $\frac{km}{h}$

B. 96.56 $\frac{km}{h}$

Explanation:

Let $s_{1}$ be the distance in first part.

$s_{1}$ = velocity × time

$s_{1}$ = 85 × $\frac{35}{60}$

$s_{1}$ = 49.58 km

Let $s_{2}$ be the distance in first part.

$s_{2}$ = 130 km

Average velocity = $\frac{Total displacement}{Total time}$

When second leg of the trip is

A. Toward north

Average velocity = $\frac{[tex] s_{1} + s_{2}$ }{Total time} [/tex]

Average velocity = $\frac{130+49.58}{0.25+0.58}$

Average velocity =216.36 $\frac{km}{h}$

B. Toward south

Average velocity = $\frac{[tex] s_{1} - s_{2}$ }{Total time} [/tex]

Average velocity = $\frac{130-49.58}{0.25+0.58}$

Average velocity =96.56 $\frac{km}{h}$

3 0

3 years ago

A(n)<br> reverses the flow of current through an electric motor.

Andrews [41]

Answer:

Commutator reverses the flow of current through an electric motor

Explanation:

4 0

4 years ago

Read 2 more answers