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Alborosie
3 years ago
8

1,021 kg x 9.8m/s/s =

Physics
2 answers:
drek231 [11]3 years ago
3 0
The anwser is 10005.8 newtons
Sergeu [11.5K]3 years ago
3 0
<span>10 005.8 newtons is your answer, hope this helps!</span>
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Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4
arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

  • initial velocity of the car, u = 76 m/s
  • acceleration of the car, a = - 9 m/s²
  • time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

  • height above the ground, h = 500 m
  • velocity of spaceship, u = 150 m/s
  • time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) -  (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m

Learn more here: brainly.com/question/24527971

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2 years ago
What types of sound waves diffract more than others
cricket20 [7]

Answer:

Infrasound

but the longer the wavelength of a soundwave, the more it diffracts

Explanation:

7 0
3 years ago
A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back
saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

K.E =\dfrac{1}{2}mv^2

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

K.E=\dfrac{1}{2}\times0.5\times(4)^2

K.E = 4\ J

The average force exerted by the floor on the ball = change in kinetic energy over collision time

F = \dfrac{4}{0.05}

F=80\ N

Hence, The force exerted by the floor is 80 N.

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I got this answer from the internet

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Slav-nsk [51]
The first step is the fusion of hydrogen into Deuterium.
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