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Rzqust [24]
3 years ago
11

The path followed by a projectile is called its _____.

Physics
2 answers:
aniked [119]3 years ago
6 0

Answer:

Trajectory

Explanation:

Trajectory is the path followed or traced by a projectile.

It follows a parabola shape. That is the shape got in javeline.

xxTIMURxx [149]3 years ago
5 0

The path followed by a projectile is called its <em>trajectory. (C)</em>

In the most common school situation<em> </em>... <u>with</u> gravity but <u>without</u> air resistance, the trajectory of a projectile is the shape of an inverted parabola (nose  pointing up).  That's the result of constant horizontal velocity and accelerated vertical velocity.

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List the three main things necessary to build an electromagnet.
Arturiano [62]

Answer:

Electromagnets are special types of magnets that are made by passing current through coils of wire. To make an electromagnet, the minimum requirements are:

1. A nail (usually made of iron, steel or zinc)

2. Dry cell batteries

3. Wire (Usually copper wire)

Other things could be:

1. Electric tape to hold both ends of the wire properly at the battery terminals.

2. Scissors to cut the wire into desired length.

3. Iron fillings for testing purposes.

7 0
3 years ago
A ball rolls from point A to point B. The total energy of the ball at point A isn’t the same as the sum of its potential energy
murzikaleks [220]
B. Some of the ball’s energy is transformed to thermal energy.

Hope this helps you!
3 0
3 years ago
Read 2 more answers
A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity
Viktor [21]

Answer:

-26 m/s.

Explanation:

Hello,

In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s

With which we compute the maximum height:

y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s

Which is clearly negative since it the projectile is moving downwards the starting point.

Regards.

3 0
3 years ago
Which of the following statements is true?
viva [34]
Statement B is correct.
4 0
3 years ago
Read 2 more answers
A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
HACTEHA [7]

Answer:

(a) 2.85 m

(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

Given:

v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

7 0
3 years ago
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