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2 years ago

In order to be considered a vector, a quantity must include what two components? A. Magnitude

Physics

2 answers:

kaheart [24]2 years ago

5 0

Answer:

Magnitude and direction.

Explanation:

There are two types of physical quantity first one is vector and other is scalar.

Vectors are those physical quantities that have both magnitude as well as direction. Scalars are that type of physical quantities having only magnitude.

Some of the examples of vector quantities are velocity, acceleration, force etc.

So, in order to be considered a vector, a quantity must include its magnitude as direction.

shepuryov [24]2 years ago

3 0

In order to be considered a vector, a quantity must include Magnitude (A) and Direction (D).

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kykrilka [37]

Answer:

C would be the answer

Explanation:

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2 years ago

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Which of the following is not an example of a physical change?

Mazyrski [523]

Answer:

Explanation:

Cutting a string in half because

b is irreversible

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2 years ago

Janet jumps off a high diving platform with a horizontal velocity of 2.06 m/s and lands in the water 1.8 s later. How high is th

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Initial height of platform is 15.87m or 16m.

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2 years ago

Beings on spherical asteroid have observed that a large rock is approaching their asteroid in a collision course. At 7514 km fro

luda_lava [24]

Answer:

c. $4.582\times10^{21} kg$

Explanation:

$r_{i}$ = Initial distance between asteroid and rock = 7514 km = 7514000 m

$r_{f}$ = Final distance between asteroid and rock = 2823 km = 2823000 m

$v_{i}$ = Initial speed of rock = 136 ms⁻¹

$v_{f}$ = Final speed of rock = 392 ms⁻¹

$m$ = mass of the rock

$M$ = mass of the asteroid

Using conservation of energy

Initial Kinetic energy of rock + Initial gravitational potential energy = Final Kinetic energy of rock + Final gravitational potential energy

$(0.5) m v_{i}^{2} - \frac{GMm}{r_{i}} = (0.5) m v_{f}^{2} - \frac{GMm}{r_{f}} \\(0.5) v_{i}^{2} - \frac{GM}{r_{i}} = (0.5) v_{f}^{2} - \frac{GM}{r_{f}} \\(0.5) (136)^{2} - \frac{(6.67\times10^{-11}) M}{(7514000)} = (0.5) (392)^{2} - \frac{(6.67\times10^{-11}) M}{(2823000)} \\M = 4.582\times10^{21} kg$

8 0

3 years ago

The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl

Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, $m_1=5\ kg$

Initial speed of block A, $u_1=5\ m/s$

Mass of block B, $m_2=8\ kg$

Initial speed of block B, $u_2=0$

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s$

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

8 0

3 years ago