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sattari [20]
3 years ago
13

An object falling under the influence of gravity and no other force is said to be

Physics
1 answer:
miv72 [106K]3 years ago
7 0

Answer:

true

Explanation:

free fall is said to be the downward movement of an object under the force of gravity only

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A 20 kg shopping cart moving at a velocity of 0.5 m/s collides with a store wall and
MAXImum [283]

Answer:

<h2>10 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 20 × 0.5

We have the final answer as

<h3>10 kg.m/s</h3>

Hope this helps you

7 0
2 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t
inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

h(t)=35t-0.83t^2

We know that the rate of change of displacement is equal to the velocity of an object.

v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t

Velocity of the arrow after 3 seconds will be :

v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0
3 years ago
Explain how VSEPR theory predicts the shapes of molecules. Match the words in the left column to the appropriate blanks in the s
vlabodo [156]

Answer:

The VSEPR theory and how it predicts the shapes of molecules:

Explanation:

The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used in chemistry to predict the shapes of individual molecules by the number of electron pairs that they have in the center of the atom. This theory is also based on the notion that the electrons around the atom repel one another. The Valence electrons on the outermost layer of the molecule are the most important in defining the geometry as they are the first to interact with other atoms and will be involved in bonding.

8 0
3 years ago
An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reach
matrenka [14]

Answer:

H = 34.43 m

Explanation:

Given that,

Initial speed of the object, u = 30 m/s

The angle of projection, \theta=60^{\circ}

We need to find the maximum height reached by the object. Let it is H. Using the formula for maximum height reached by the projectile.

H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m

So, the maximum height reached by the object is 34.43 m.

6 0
3 years ago
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