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VladimirAG [237]

3 years ago

14

a moving object has the clockwise function: s = 5 + 2t (no S.I). determine its position when t = 7 s B) in which moment does the

moving object surpass the position s = 25m? Need an answer pls :)

1 answer:

Montano1993 [528]3 years ago

4 0

25m hope this help it is c

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What are some examples of how we use science in or daily lives?

mamaluj [8]

Answer: Science is involved in cooking, eating, breathing, driving, playing, etc. The fabric we wear, the brush and paste we use, the shampoo, the talcum powder, the oil we apply, everything is the consequence of advancement of science. Life is unimaginable without all this, as it has become a necessity.

Explanation:

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4 years ago

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A car speeds up as it rolls down a hill. which is this an example of? positive acceleration negative acceleration relative veloc

alex41 [277]

Positive acceleration

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4 years ago

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A ball of mass 0.250 kg and a velocity of + 5.00 m/s collides head-on with a ball of mass 0.800 kg that is initially at rest. No

Mademuasel [1]

Answer:

$v_1=-2.616\ m/s$

Explanation:

Given that,

Mass of ball 1, $m_1=0.25\ kg$

Initial speed of ball 1, $u_1=5\ m/s$

Mass of ball 2, $m_2=0.8\ kg$

Initial speed of ball 2, $u_2=0$ (at rest)

After the collision,

Final speed of ball 2, $v_2=2.38\ m/s$

Let $v_2$ is the final speed of ball 1.

Initial momentum of the system is :

$p_i=m_1u_1+m_2u_2$

$p_i=0.25\times 5+0$

$p_i=1.25\ m/s$

Final momentum of the system is :

$p_f=m_1v_1+m_2v_2$

$p_f=0.25\times v_1+0.8\times 2.38$

$p_f=0.25 v_1+1.904$

According the law of conservation of linear momentum :

initial momentum = final momentum

$1.25=0.25 v_1+1.904$

$v_1=-2.616\ m/s$

So, the final velocity of ball 1 is (-2.616)m/s.

8 0

3 years ago

During a certain comet’s orbit around the Sun, its closest distance to the Sun is 0.6 AU, and its farthest distance from the Sun

Ray Of Light [21]

Answer:

At the closest point

Explanation:

We can simply answer this question by applying Kepler's 2nd law of planetary motion.

It states that:

"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"

In this problem, we have a comet orbiting around the Sun:

- Its closest distance from the Sun is 0.6 AU

- Its farthest distance from the Sun is 35 AU

In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: $A\propto vr$ , therefore if r is larger, then v (velocity) must be lower).

On the other hand, when the the comet is closer to the Sun the line must move faster ( $A\propto vr$ , if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.

7 0

3 years ago

Three resistors of value 4 ω, 10 ω and 7 ω are placed in parallel. they are connected to a 6 ω resistor in series. what is the e

asambeis [7]

resistance of 4 ohm, 10 ohm and 7 ohm is connected in parallel.

so net resistance in parallel is given by

$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$

$\frac{1}{R} = \frac{1}{4} + \frac{1}{10} + \frac{1}{7}$

$R = \frac{140}{69} ohm$

now it is connected to 6 ohm resistance in series

so net resistance is given by

$R_{net} = R_1 + R_2$

$R_{net} = 6 ohm + \frac{140}{69} ohm$

$R_{net} = 8.03 ohm$

4 0

3 years ago