The big bang theory of the information and expansion of the universe is supported by the observed

Ok I have no clue for this one I’m not sure what to make out of this one please help

MatroZZZ [7]

Helium, Neon, and Xenon are all part of the same column on the Periodic Table. Such a column is referred to as a Group, because they have the same number of valence electrons in their outermost shell. Hope this helps!

5 0

3 years ago

A bird sits on top of a 639 m tall tower. If it's gravitational potential energy up there is 2033 J, what is its mass?

iris [78.8K]

The mass of the bird is 0.32 kg.

<u>Explanation:</u>

Gravitational potential energy, the energy exhibited by an object at rest due to the influence of gravitational force. So the increase in distance of object from the surface of earth leads to increase in the gravitational potential energy. Thus,

$\text {Gravitational potential energy}=m \times \text { Acceleration } \times \text { Distance of bird from bottom }$

So, as the gravitational potential energy is given as 2033 J and the position of bird placed on the tall tower is 639 m away from the bottom, then the mass (m) of the bird can be found as below.

$m o f \text { bird }=\frac{\text {Gravitational potential energy}}{a \times \text {Distance}}=\frac{2033}{9.8 \times 639}=\frac{2033}{6262.2}$

So, finally we get the bird's mass as,

m of bird = 0.32 kg

7 0

3 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h

Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

$v_{1}=\dfrac{3.0\times10^{3}}{3600}$

$v_{1}=0.833\ m/s$

The force F₁is constant acceleration is also a constant.

$F_{1}=ma_{1}$

We need to calculate the acceleration

Using formula of acceleration

$a_{1}=\dfrac{v}{t}$

$a_{1}=\dfrac{0.833}{10}$

$a_{1}=0.083\ m/s^2$

Similarly,

$F_{2}=ma_{2}$

For total force,

$F_{3}=F_{2}+F_{1}$

$ma_{3}=ma_{2}+ma_{1}$

The speed of second tugboat is

$v=\dfrac{11\times10^{3}}{3600}$

$v=3.05\ m/s$

We need to calculate total acceleration

$a_{3}=\dfrac{v}{t}$

$a_{3}=\dfrac{3.05}{10}$

$a_{3}=0.305\ m/s^2$

We need to calculate the acceleration a₂

$0.305=a_{2}+0.083$

$a_{2}=0.305-0.083$

$a_{2}=0.222\ m/s^2$

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

$\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}$

$\dfrac{F_{1}}{F_{2}}=3.7$

$F_{1}=3.7F_{2}$

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0

3 years ago

Explain why free fall acceleration near earths surface is constant

Naya [18.7K]

Because gravity is constant
<span>the only force acting in free-fall is gravity which points downward at 9.8 m/s</span>

4 0

3 years ago

How do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate?

Wittaler [7]

Sky diving involves free fall under gravity along with the drag due to air molecules pushing against the body slowing the rate of fall of a body. This is actually a significant amount of force. The drag force depends on the contact surface area and weight of the body. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position. This is because of the less contact surface area of the body with the air molecules while in the former case. Since no two persons have identical body shape and weight, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.

3 0

3 years ago