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SVETLANKA909090 [29]

3 years ago

3

Whether or not the information on an online source is credible and factual, is up to the user to decide.

2 answers:

Neporo4naja [7]3 years ago

5 0

Answer:

Informational references provide only a limited amount of specific statistical information. false. Information should be________to ensure more than one source provides the same factual information.

Explanation:

sergiy2304 [10]3 years ago

4 0

Answer:

The statement presented above is TRUE. Yes, it's up to us whether to accept the statement whether as reliable or not. At the end of the day, it is our responsibility to know the facts and not let other people dictate what you have to think about. It has to be taken into account that you can decide for yourself.

Explanation:

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For your answer to this problem, just type in the numerical magnitude of the momentum - no units.

stepan [7]

Answer:

120 kg•m/s.

Explanation:

From the question given above, the following data were obtained:

Case 1

Mass of object = M

Velocity of object = V

Momentum = 15 kg•m/s

Case 2

Mass of object = 2M

Velocity of object = 4V

Momentum = ?

Momentum is defined as follow:

Momentum = mass × velocity

The momentum of object in case 2 can be obtained as follow:

From case 1

Momentum = mass × velocity

15 = M × V

15 = MV ....... (1)

From case 2:

Momentum = mass × velocity

Momentum = 2M × 4V

Momentum = 8MV ....... (2)

Finally , substitute the value of MV in equation 1 into equation 2.

Momentum = 8MV

MV = 15

Momentum = 8 × 15

Momentum = 120 kg•m/s

Therefore, an object with a mass of 2M and 4V would have a momentum of 120 kg•m/s

3 0

3 years ago

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position

Illusion [34]

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Place of particles

$X(t)=\int V_{x}(t)dt$

$=\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C$

$\to t=0\\\\ \to X(0)=2.3 \ m$

$\to X(0)=0+C\\\\ \to C=2.3\ m$

$\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3 \\\\$

$= \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\ = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m$

In point b:

when $t=2.2 \ s$

the Particle velocity $(V)=2 \times 2.22 =9.68\ \frac{m}{s}$

In point c:

Calculating the Particle acceleration:

$\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2 =8.8 \ \frac{m}{s^2}$

8 0

3 years ago

Please help with both will give brainlist worth 20

REY [17]

Answer:

Second bubble

Explanation:

Its going up

7 0

3 years ago

a coin is dropped from the top of a tall building. determine the coin's (a) velocity and (b) displacement after 1.5 sec.

Lilit [14]

1) v = gt = 10*1.5 = 15 m/s
2) r = gt^2 /2 = 10*(1.5)^2 / 2 = 11.25 meters

8 0

3 years ago

Anyone please.??????

Ira Lisetskai [31]

Answer:

468.42572 Wavelength In Metres

Explanation:

8 0

3 years ago

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