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Lorico [155]
3 years ago
5

Suppose an astronaut in outer space wishes to toss a ball against a very massive and perfectly elastic concrete wall and catch i

t as it bounces back. If the ball is as massive as the astronaut, then:_______.
a. the astronaut's time between catches will decrease as the game progresses.
b. the astronaut will never catch the first bounce.
c. the astronaut will catch one bounce only.
d. none of the above.
Physics
1 answer:
madam [21]3 years ago
7 0

Answer:

B. the astronaut will never catch the first bounce.

Explanation:

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A projectile is launched
Flauer [41]

The vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

The given parameters;

  • initial horizontal velocity, vₓ = 16 m/s
  • initial vertical velocity, v_y =0
  • time interval 1 seconds

The components of the velocity can be horizontal or vertical velocity.

The vertical component of the velocity is affected by acceleration due to gravity while the horizontal component of the velocity is not affected by gravity.

The vertical component of the velocity is calculated as;

v_y = v_0_y -gt\\\\v_y = 0 - (1\times 9.8)\\\\v_y = -9.8 \ m/s

The horizontal component of the velocity is constant since it is not affected by gravity.

The horizontal component of the velocity = 16 m/s

Thus, the vertical component of the velocity after the given time is -9.8 m/s while the horizontal component of the velocity is 16 m/s.

Learn more here:brainly.com/question/20349275

4 0
3 years ago
In the simulation above, as the projectile travels downward, how does the vertical velocity change?
NARA [144]

B) is the correct answer
7 0
3 years ago
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What is a tool that can be used to measure the size of a force?
gayaneshka [121]
Newton meter
Torque wrench
Or Just a plain Scale
4 0
2 years ago
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A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest),  = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0
3 years ago
In a track and field event, a hammer thrower accelerates the hammer (mass = 7.30 kg) from rest within four full turns (revolutio
lubasha [3.4K]

Answer:

5.69755 rad/s²

Explanation:

r = Radius = 1.3 m

v = Velocity of the hammer = 22 m/s

n = Number of revolutions = 4

Angular displacement

\theta=n\times 2\pi\\\Rightarrow \theta=4\times 2\pi\\\Rightarrow \theta=8\pi

\omega_i = Initial angular speed = 0

Final angular speed

\omega_f=\frac{v}{r}\\\Rightarrow \omega_f=\frac{22}{1.3}\\\Rightarrow \omega_f=16.92307\ rad/s

Angular acceleration

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{16.92307^2-0^2}{2\times 8\pi}\\\Rightarrow \alpha=5.69755\ rad/s^2

Angular acceleration is given by 5.69755 rad/s²

3 0
3 years ago
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