PLZ HELP THE CORRECT ANSWER GETS BRAINLEAST!

1 answer:

4 0

The answer is B because the tire is circular. it will have more momentum and of course it can’t be on snow since it’s not smooth enough for the ice. Plus the sled will go slower on a grass hill since sled needs smooth flooring for no friction.

You might be interested in

A car drives toward the right over the top of a hill, as shown below. An illustration of car at the top of a hill pointing right

Sav [38]

Answer: X

Explanation:

This situation can be illustrated as a car in circular motion (image attached).

In circular motion the acceleration vector $\vec{a}$ is always directed toward the center of the circumference (that's why it's called centripetal acceleration).

So, in this case the arrow labeled X is the only that points toward the center, hence it represents the car's centripetal acceleration

6 0

2 years ago

Read 2 more answers

A biker travels at an average speed of 18 km/hr along a 0.30 km straight segment of a bike path. How much time (in hours) does t

lawyer [7]

Answer: 0.016 h

Explanation:

$\text{Average speed} = \frac{\text {Total Distance}}{\text {total time taken}}$

It is given that, biker has an average speed = 18 km/h

Total distance traveled = 0.30 km

Therefore, time taken by biker to travel this distance:

$\Rightarrow \text{total time taken} = \frac{0.30 km}{18 km/h}=0.016 h$

Thus, the biker takes 0.016 hours to travel the segment of 0.30 km at an average speed of 18 km/h.

7 0

2 years ago

he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold

bekas [8.4K]

Answer:

Relative population is 2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-E_{1} }{KT} } }{g_{2}e^{\frac{-E_{2} }{KT} }}$