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Naddik [55]
3 years ago
6

I need to know which is represented as a element

Physics
1 answer:
Tema [17]3 years ago
6 0

Answer:

the yellow one

Explanation:

2 of the same elements resolute as the same element

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I believe the answer is CONVECTION.




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Understanding that if we say something unkind to someone else his or her
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Answer: Moral

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Which phenomenon occurs when the moon and Earth are aligned with the sun? A. Retrograde motion B. Solstice C. Equinox D. Eclipse
muminat

Answer: Eclipse

Explanation: A lunar eclipse occurs when the full moon moves through the shadow of the Earth. This can only happen when the Earth is between the Moon and the Sun and all three are lined up in the same plane, called the ecliptic. The ecliptic is the plane of Earth's orbit around the Sun.

7 0
3 years ago
A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa
kotykmax [81]
Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is v_i=180~m/s, the final velocity is v_i=0~m/s, and the total time of the motion is \Delta t=0.02~s, so the acceleration is given by
a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2 
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
S=v_i t + \frac{1}{2}at^2 = 180~m/s \cdot 0.02~s + \frac{1}{2}(-9000~m/s^2)(0.02~s)^2=1.8~m
So, the bullet penetrates the sandbag 1.8 meters.
5 0
2 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
2 years ago
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