Which describes the difference between ionic and covalent bonds?

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that

Natasha2012 [34]

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

$T = \frac{t}{n}$

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

$T = \frac{135\ s}{98}$

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

$T = 2\pi \sqrt{\frac{l}{g}}$

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

$(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}$

g = 11.2 m/s²

3 0

2 years ago

A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward p

Alexxx [7]

Answer:1) the total distance is the sum of the two distances

60 km + 45 km = 105 km

2) The displacement is the net movement, or the difference between the initial position and the final position

Call x the initial position, then the final position is x + [60km - 45km]

And the displacement is x + (60km - 45km) - x =60km -45 km = 15 km

Explanation:

6 0

3 years ago

Since the Wagon is being pulled down hill is it increasing (C)

Ket [755]

Since the wagon is being pulled down hill with a constant velocity, all the forces of the wagon would be (C) increasing.
You are correct! **

4 0

3 years ago

I just need number 2

il63 [147K]

We will apply the conservation of linear momentum to answer this question.

Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Note that v₁ and v₁' is the velocity of m₁ before and after the collision.

Let's choose m₁ and v₁ to represent the bullet's mass and velocity.

m₂ and v₂ represents the wood block's mass and velocity.

The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'

m₁v₁ + m₂v₂ = m₁v' + m₂v'

m₁v₁ + m₂v₂ = (m₁+m₂)v'

Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.

m₁v₁ = (m₁+m₂)v'

Here are the given values:

m₁ = 0.005kg

v₁ = 500m/s

m₂ = 5kg

Plug in the values and solve for v'

0.005×500 = (0.005+5)v'

v' = 0.4995m/s

v' ≅ 0.5m/s

4 0

3 years ago

Which of the following is an example of proper movement for a biceps curl?

DanielleElmas [232]

i think the answer is : Bend the arm at the elbow with the back straight

5 0

3 years ago