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2 years ago

If you're going 80 mph how long does it take to 80 miles

Physics

1 answer:

natka813 [3]2 years ago

6 0

It would take you an hour, but only if you keep your speed.

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Which pair of terms correctly fills the blanks in the following sentence? ________ is a property of an object while ________ is

Murljashka [212]

Answer:

D) resistance, resistivity

Explanation:

Resistance is a physical quantity that indicates the opposition of an object to conduct electricity, this quantity depends on different factors such as temperature, material, object length, among other things. The resistance of two objects of the same material may be different, because it depends on the specifications of the object.

On the other hand, resistivity is a more general quantity, since it is assigned to materials and depends only on the nature of the material and its temperature.

So the resistivity is related to the meterial rather than the object.

The answer is: Resistance is a property of an object while resistivity is a property of a material.

8 0

3 years ago

We know that the motion of the Moon around the Earth is due

nadezda [96]

Answer:

YES

Explanation:

Gravity acts as the centripetal force and the velocity earth has keeps it from falling on the sun.

3 0

3 years ago

A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =

Ratling [72]

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

$v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2$

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

$v = \omega R\\a = \alpha R$

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

$v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft$

One revolution is equal to the circumference of the drum. So, total number of revolutions is

$x / (2\pi R) = 6/(\pi R)$

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

$a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2$

b) $a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2$

5 0

3 years ago

HELP ASAP!! 15 POINTS!!

Allisa [31]

Answer:

The bottom one has the highest amplitude

3 0

2 years ago

Read 2 more answers

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

2 years ago