Which phase of the moon is represented by position 4?

Directions: Using the T-chart below, compare balanced forces and unbalanced forces.

Salsk061 [2.6K]

Explanation:

unbalanced: a turning vehicle, apple falling on the ground, kicking a ball

balanced: floating on water, fruit hanging from tree, tug of war equally balanced teams

8 0

2 years ago

If a toy car with a coin on top of it is rolling down a hill what force is keeping the coin on top of the car?

forsale [732]

The answer is gravity.

7 0

3 years ago

A. Besides protons, what other particles make up an atom? Write 2 - 3 sentences describing how the electrostatic force acts betw

vodomira [7]

Answer:

For a. Neutrons and electrons also form an atom.

For b. The element is oxygen which is a non-metal and will form a negative ion while forming ionic bond.

Explanation:

For a:

There are 3 subatomic particles which form an atom. These are neutrons, protons and electrons.

Neutrons carry no charge, protons carry positive charge and electrons carry negative charge. Neutrons and protons are present in nucleus and electrons revolve around the nucleus.

The energy which is present between neutrons and protons are nuclear energy and the energy which is present between electrons and protons are electrostatic energy.

For b:

In an element, number of protons is always equal to the number of electrons. The atomic number is equal to the number of protons or electrons. The element which has atomic number 8 is Oxygen.

The electronic configuration of this element is $1s^22s^22p^4$

This element requires only 2 electrons to form a stable electronic configuration. An element which gains electron is considered as a non-metal and forms a negative ion because number of electrons increases.

3 0

3 years ago

Observe: Click Reset. Turn on Show velocity vector and Show velocity components. Set vinitial to 50 m/s and set θ to 45.0 degree

Kobotan [32]

Answer:

v_y = v_{oy} - g t

where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses

Explanation:

In this exercise you are asked to observe the change in velocity in a projectile launch.

If we assume that the friction force is small, the velocity in the x-axis must be constant

vₓ = v₀ₓ

Therefore, the arrow (red) that represents this movement must not change in magnitude.

In the direction of the y axis, the acceleration of gravity is acting, so the magnitude of the velocity in this axis changes

v_y = v_{oy} - g t

where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses

7 0

3 years ago

A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s

Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± $\frac{1}{2} at^2$ ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± $\frac{1}{2} at^2$ ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

From the question;

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - $\frac{1}{2} at^2$

h = 26.54(1.7) - $\frac{1}{2} (10)(1.7)^2$

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m

7 0

3 years ago