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3 years ago

What’s is a object that moves through the air space acted on only by gravity

Physics

2 answers:

inessss [21]3 years ago

7 0

This describes projectile motion while neglecting any outside forces other than gravity

Arada [10]3 years ago

4 0

I believe the answer is projectile
As the object reach its maximum height after it thrown, its speed will falls into 0. But due to the gravity force, the speed of the object will start to accelerate again after it came down until it hit another medium.
Example of projectiles: a thrown baseball, a catapult shot.

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If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi

Tcecarenko [31]

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

$I\alpha \frac{1}{r^{2} }$

Now according to the question the distance is increased by three times than,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }$

Therefore,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }$

Therefore the intensity will become 1/9 times to the initial intensity.

3 0

3 years ago

Which of the following should you always use to protect yourself from hazardous moving parts of power tools and equipment

Andrew [12]

What are the following answers?

7 0

3 years ago

A car's velocity as a function of time is given by vx(t)=α+βt2, where α=3.00m/s and β=0.100m/s3.

Mice21 [21]

1) Analyze the equation

Vₓ = α + β t²

Vₓ = 3.00 + 0.100 t²

That is a quadratic equation, so the graph is a parabola.

2) Therefore, take into account that the main points to draw a parabola are:

i) vertex

ii) concavity

iii) y-intercepts

iv) x - intercepts

Also, for all graphs you need the domain and the range.

3) Find the y-intercept (t = 0)

t = 0 ⇒ Vₓ = 3.00 + 0.100 (0)² = 3.00

4) Find the x-intercepts (Vₓ = 0)

Vₓ = 0 ⇒ 0 = 3.00 + 0.100 t²

⇒ t² = - 3.00 / 0.100 = -30.0. Since t² cannot be negative, there is not x-intercepts.

5) Concavity

Since, the coefficient of t² is positive, the parabola open upwards.

6) Vertex

It is the local minimum of the equation. You can find it by the first derivative

Vₓ' = 2×0.100 t = 0 ⇒ t = 0

Vₓ = 3.00 + 0.100 (0)² = 3.00 m/s

⇒ vertex = (0,3.00)

7) The domain is given t ∈ [0,5.00]

8) You can also build a table with several points in the domain

t =0; Vₓ = 3.00 + 0.100 (0)² = 0

t = 1; Vₓ = 3.00 + 0.100 (1)² = 3.10

t = 2; Vₓ = 3.00 + 0.100 (2)² = 3.40

t = 3; Vₓ = 3.00 + 0.100 (3)² = 3.90

t = 4; Vₓ = 3.00 + 0.100 (4)² = 4.60

t = 5; Vₓ = 3.00 + 0.100 (5)² = 5.50

9) Range: Vₓ ∈ [ 3.00, 5.50]

10) All that information permits you to put several points in a coordinate sysment and sketch the same graph as the shown in the figure attached.

4 0

3 years ago

Read 2 more answers

A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of

Nesterboy [21]

Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

3 0

2 years ago

Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. what is the magnitud

sergejj [24]

The force on the layer will be equivalent to the weight of water on it. This is given by:
F = mg; m is the mass of water and g is the acceleration due to gravity.

5 0

3 years ago

Read 2 more answers