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3 years ago

Plz, Help!!

Physics

2 answers:

SIZIF [17.4K]3 years ago

7 0

C is the first & the second question is A

andreev551 [17]3 years ago

5 0

The answer to the first question is C the second one is B

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What is An object that changes position?

mel-nik [20]

Explanation:

Motion is when an object changes position over time. The object in motion is usually in front of a reference point-an object that appears to stay in one place. The rate at which an object moves is called speed. Speed depends on both time and distance. The velocity of an object is how fast it is going in one direction

How do you know if an object has changed position?

changes position requires a point of reference. An object changes position if it moves relative to a reference point. To visualize this, picture yourself competing in a 100-m dash. You begin just behind the start line

6 0

3 years ago

Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

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2 years ago

If the volleyball hits the net and comes over to land on the other side it is?

photoshop1234 [79]

Other teams point ......

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4 years ago

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What is the name of the gland that wraps around the front part of the throat at the Adam's apple, and that produces hormones tha

Grace [21]

B. Thyroid Gland is the correct answer

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3 years ago

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A ball is thrown vertically upward and then comes back down. During the ball's flight up and down, its velocity and acceleration

Dovator [93]

During the ball's flight up its velocity and acceleration vectors are in opposite direction and during the ball's flight down its velocity and acceleration vectors are in same direction.

The velocity vector is always in the direction of motion of the object. So, during the ball's flight up its velocity vector is in the upward direction (90°) and during the ball's flight down its velocity vector is in the downward direction (270°).
When there is a positive acceleration in the object the acceleration vector is in the direction of motion of the object. When there is a negative acceleration in the object the acceleration vector is in the opposite direction of motion of the object. So, during the ball's flight up its acceleration vector is in the downward direction (270°) and during the ball's flight down its acceleration vector is in the upward direction (90°).

Velocity vector is the rate of change of position of an object. Acceleration vector is the rate of change of velocity of an object.

Therefore, during the ball's flight up its velocity and acceleration vectors are in opposite direction and during the ball's flight down its velocity and acceleration vectors are in same direction.

To know more about velocity and acceleration vectors

brainly.com/question/13492374

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1 year ago