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3 years ago

What is a chunk of space rock

Physics

2 answers:

Igoryamba3 years ago

7 0

The answer to your question is,

A meteoroid.

-Mabel <3

vitfil [10]3 years ago

7 0

It is typically a piece of Asteroid or Comet!!

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An oceanic ridge might be compared to what continental landform? A) delta B) great plains C) mountain range Eliminate D) mountai

Pavlova-9 [17]

A mountain range because an ocean ridge is an underwater mountain hope this helps you

6 0

3 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav

Vikki [24]

Answer:

The force constant is $k =1.316 *10^{7} \ N/m$

The energy stored in the spring is $E = 1.68 *10^{7} \ J$

Explanation:

From the question we are told that

The mass of the object is $M = 4.8*10^{5} \ kg$

The period is $T = 1.2 \ s$

The period of the spring oscillation is mathematically represented as

$T =2 \pi \sqrt{ \frac{M}{k}}$

where k is the force constant

So making k the subject

$k = \frac{4 \pi ^2 M }{T^2}$

substituting values

$k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}$

$k =1.316 *10^{7} \ N/m$

The energy stored in the spring is mathematically represented as

$E = \frac{1}{2} k x^2$

Where x is the spring displacement which is given as

$x = 1.6 \ m$

substituting values

$E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2$

$E = 1.68 *10^{7} \ J$

7 0

3 years ago

An empty 2,500 kg train car is headed northbound at a velocity of 5 m/s. Ahead of the first car, an empty 1,500 kg car is headed

Tema [17]

Let the mass of 2500 kg car be $m_1$ and it's velocity be $v_1$ and the mass of 1500 kg car be $m_2$ and it's velocity be $v_2$ .

After the bumping the mass be M and it's velocity be V.

By law of conservation of momentum we have

$m_1v_1+m_2v_2 = MV$

2500 * 5 + 1500 * 1=4000 * V

V = 14000/4000 = 7/2 = 3.5 m/s

So the velocity of the two-car train = 3.5 m/s

9 0

3 years ago

Which material is a conductor? A. chalk B. lead C. leather D. paper E. rubber

Elan Coil [88]

B. Lead is a conductor....

8 0

3 years ago

A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn

Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J

Explanation:

We have given the player turntable initially rotating at speed of $33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec$

Now speed is reduced by 75 %

So final speed $\frac{3.49\times 75}{100}=2.6175rad/sec$

Time t = 5.5 sec

From first equation of motion we know that '

$\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2$

(a) Now final velocity $\omega =0rad/sec$

So time t to come in rest $t=\frac{0-3.49}{-0.158}=22sec$

(b) The work done in coming rest is given by

$\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J$

4 0

3 years ago