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3 years ago

8. A baseball is hit upward and travels along a parabolic arc before it strikes the se

Physics

1 answer:

Usimov [2.4K]3 years ago

8 0

d. The x-component of the velocity of the ball is the same througout the ball's flight

Explanation:

The motion of the baseball is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction.

From this description, we can already see that the correct statement is D: in fact, there is no force acting in the horizontal direction on the ball (if we neglect air resistance), therefore the acceleration in the x-direction is zero, therefore the x-component of the velocity is constant.

The other statements are wrong because:

a. The acceleration of the ball decreases as the ball moves upward. --> false because the acceleration remains the same ( $g=9.8 m/s^2$ downward)

b. The velocity of the ball is zero m/s when the ball is at the highest point in the arc. --> false because only the y-component of the ball's velocity is zero at the highest point, but the x-component is not zero

c. The acceleration of the ball is zero m/s when the ball is at the highest point in the arc. --> false because the acceleration remains the same ( $g=9.8 m/s^2$ downward)

Therefore the correct answer is

d. The x-component of the velocity of the ball is the same througout the ball's flight

Learn more about projectile motion:

brainly.com/question/8751410

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A 100-g toy car is propelled by a compressed spring that starts it moving. the car follows a curved track. what is the final spe

wel

Answer:

0.687 m/s

Explanation:

Initial energy = final energy

1/2 mu² = mgh + 1/2 mv²

1/2 u² = gh + 1/2 v²

Given u = 2.00 m/s, g = 9.8 m/s², and h = 0.180 m:

1/2 (2.00 m/s)² = (9.8 m/s²) (0.180 m) + 1/2 v²

v = 0.687 m/s

5 0

3 years ago

Read 2 more answers

I would love to stretch a wire from our house to the Shop so I can 'call' my husband in for meals. The wire could be tightened t

dezoksy [38]

Note: I'm not sure what do you mean by "weight 0.05 kg/L". I assume it means the mass per unit of length, so it should be "0.05 kg/m".

Solution:
The fundamental frequency in a standing wave is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }$
where L is the length of the string, T the tension and m its mass. If we plug the data of the problem into the equation, we find
$f= \frac{1}{2 \cdot 24 m} \sqrt{ \frac{240 N}{0.05 kg/m} }=1.44 Hz$

The wavelength of the standing wave is instead twice the length of the string:
$\lambda=2 L= 2 \cdot 24 m=48 m$

So the speed of the wave is
$v=\lambda f = (48 m)(1.44 Hz)=69.1 m/s$

And the time the pulse takes to reach the shop is the distance covered divided by the speed:
$t= \frac{L}{v}= \frac{24 m}{69.1 m/s}=0.35 s$

7 0

3 years ago

A light beam travels at 1.94×108 in quartz. The wavelength of the light in quartz is 355 .Part AWhat is the index of refraction

Alja [10]

A) 1.55

The speed of light in a medium is given by:

$v=\frac{c}{n}$

where

$c=3\cdot 10^8 m/s$ is the speed of light in a vacuum

n is the refractive index of the material

In this problem, the speed of light in quartz is

$v=1.94\cdot 10^8 m/s$

So we can re-arrange the previous formula to find n, the index of refraction of quartz:

$n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55$

B) 550.3 nm

The relationship between the wavelength of the light in air and in quartz is

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda$ is the wavelenght in quartz

$\lambda_0$ is the wavelength in air

n is the refractive index

For the light in this problem, we have

$\lambda=355 nm\\n=1.55$

Therefore, we can re-arrange the equation to find $\lambda_0$ , the wavelength in air:

$\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm$

4 0

3 years ago

Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2

wolverine [178]

The particles can undergo small oscillations around x₂.

The given parameters;

<em>initial energy of the particles = E₁</em>
<em>final energy of the particles, E₂ = 0.33E₁</em>

The movement of the particles depends on the kinetic energy of the particles.

When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.

However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.

The maximum position the particle can oscillate is x₅

The half position the particles can oscillate is x₃

Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.

Thus, we can conclude that the particles can undergo small oscillations around x₂.

Learn more here:brainly.com/question/23910777

3 0

2 years ago

A .140kg baseball traveling 35.0m/s strikes the catchers mit bringing the ball to rest, recoils backward 11.0cm what’s the avera

Juli2301 [7.4K]

Average force applied by the call on the glove = 780 N

Explanation:

mass of baseball=0.140 kg

initial velocity = Vi= 35 m/s

Final velocity=Vf= 0

distance traveled=11 cm= 0.11 m

using the kinematic equation Vf²= Vi²+ 2 a d

where a = acceleration

0²= 35²+ 2 a (0.11)

a=-5568.2 m/s²

Now force is given by F= ma

F= 0.140 (5568.2)

F=-780 N

The negative sign signifies that the force acts in the opposite direction.

so the average force= 780 N

6 0

2 years ago