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UNO [17]
3 years ago
9

What happens to Earth’s axis of rotation as Earth orbits the Sun?

Physics
1 answer:
Dominik [7]3 years ago
7 0
Stays lined up with North Star
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An atom undergoes nuclear decay, but its atomic number is not changed.
andreev551 [17]

Answer:

A. Gamma decay

Explanation:

A form of nuclear decay in which the atomic number is unchanged is a gamma decay.

The atom has undergone a gamma decay.

In a gamma decay, no changes occur to the mass and atomic number of the substance.

  • Gamma rays have zero atomic and mass numbers.
  • When they cause decay, they cause no change to the mass and atomic numbers.
  • They simply produce gamma rays during such reactions and these rays are very energetic.
5 0
3 years ago
Rocket-powered sleds are been used to test the responses of humans to acceleration. Starting from rest, one sled can reach a spe
Greeley [361]

Answer:cho  v₀ =0s  

α=Δv/Δt

Explanation:

\frac{0-495}{2,16-1,78}

=-1302,631579

chuyển động chậm dầnđều

3 0
3 years ago
1. Tonya had a hard time deciding between the Big Burger and the Crispy Chicken sandwiches, her two favorites.
Radda [10]
Bro the md that she lost was the md boneless burger
8 0
3 years ago
26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the
denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

U=mg \Delta h

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

\Delta h = 4 m + 0.02 m = 4.02 m is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J

4 0
3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
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