. A dog walks 14 m to the east and then 20 m back to the west. What is the distance? What is the displacement?

What's the difference between meteoroids, meteorites, and meteors?

SVEN [57.7K]

You're talking about a grain of sand or a stone or a rock that's drifting in space, and then the Earth happens to get in the way, so the stone falls down to Earth, and it makes a bright streak of light while it's falling through the atmosphere and burning up from the friction.

-- While it's drifting in space, it's a meteoroid.

-- While it's falling through the atmosphere burning up and making a bright streak of light, it's a meteor.

-- If it doesn't completely burn up and there's some of it left to fall on the ground, then the leftover piece on the ground is a meteorite.

4 0

3 years ago

A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th

Pavel [41]

Answer:

The block's mass should be $3m$

Explanation:

Given:

Cart with mass $m$

From the conservation of energy before mass is added,

$\frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}$

Where $A =$ amplitude of spring mass system, $k =$ spring constant

$A = v\sqrt{\frac{m}{k} }$

Now new mass $M$ is added to the system,

$\frac{1}{2} (m +M ) v^{2} = \frac{1}{2} k A^{2}$

$A = v \sqrt{\frac{m +M }{k} }$

Here, given in question frequency is reduced to half so we can write,

$f' = \frac{f}{2}$

Where $f =$ frequency of system before mass is added, $f' =$ frequency of system after mass is added.

$\omega ' = \frac{\omega}{2}$

$\sqrt{\frac{k}{m +M} } = \frac{\sqrt{\frac{k}{m} } }{2}$

$\frac{k}{m +M } = \frac{k}{4m}$

$M = 3m$

Therefore, the block's mass should be $3m$

8 0

3 years ago

g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin

DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as $mgh=0.5mv^{2}$ since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then $v=\sqrt {2gh}$ and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that $v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s$

6 0

3 years ago

It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar had a constant accelerati

Soloha48 [4]

Answer: 47.6 m/s

Explanation: Please see attached for the calculation and formula.

5 0

3 years ago

The position of a car at time t is given by the function p(t)=t2 2t−4. What is the velocity when p(t)=11? assume t≥0

AysviL [449]

The velocity when function p(t)=11 is 8 .

According to the question

The position of a car at time t represented by function :

$p(t)=t^{2} +2t-4$

Now,

When function p(t) = 11 , t will be

$p(t)=t^{2} +2t-4$

11 = t²+2t-4

0 = t² + 2t - 15

or

t² +2t-15 = 0

t² +(5-3)t-15 = 0

t² +5t-3t-15 = 0

t(t+5)-3(t+5) = 0

(t-3)(t+5) = 0

t = 3 , -5

as t cannot be -ve as given ( t≥0)

so,

t = 3

Now,

the velocity when p(t)=11

As we know velocity = $\frac{position}{time}$

therefore to get the value of velocity from function p(t)

we have to differentiate the function with respect to time

$\frac{d(p(t))}{dt} =\frac{d}{dt} (t^{2} +2t-4)$

v(t) = 2t + 2

where v(t) = velocity at that time

as t = 3 for p(t)=11

so ,

v(t) = 2t + 2

v(t) = 2*3 + 2

v(t) = 8

Hence, the velocity when function p(t)=11 is 8 .

To know more about function here:

brainly.com/question/12431044

#SPJ4

4 0

2 years ago

. A dog walks 14 m to the east and then 20 m back to the west. What is the distance? What is the displacement?​

. A dog walks 14 m to the east and then 20 m back to the west. What is the distance? What is the displacement?