All categories

3 years ago

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

You might be interested in

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