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svlad2 [7]
2 years ago
12

What is the average speed of a kangaroo that hops 60 m in 5 s ?

Physics
2 answers:
jok3333 [9.3K]2 years ago
6 0
The formula for average speed is
v_{avg}= \frac{total distance}{total time}
So we can just substitute our data.
v_{avg}= \frac{60}{5}=12 \frac{m}{s} - its the result
ololo11 [35]2 years ago
6 0

<u></u>
12 m/s that's really easy man, just divide 60 by 5.


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When an object threw to the free space to make an angle of 25 degree at an initial speed of 15 m/sec, the ball takes time to rea
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Answer:

The horizontal distance traveled by the projectile is 15.23 m.

Explanation:

Given;

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initial velocity of the projectile, u = 15 m/s

time of flight, t = 1.12 s

The the travelling path of the object is calculated as the range of the projectile

R = u_x t\\\\R = (15\ Cos \ 25^0) \times 1.12\\\\R = 13.595 \times 1.12\\\\R = 15.23 \ m

Therefore, the horizontal distance traveled by the projectile is 15.23 m.

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2 years ago
A car is moving at a velocity of 25 km/h and increases velocity to 1200 km/h in 2 min. what is the acceleration?
OlgaM077 [116]

Answer:

a = 2.72 [m/s2]

Explanation:

To solve this problem we must use the following kinematics equation:

v_{f} =v_{o} + a*t

where:

Vf = final velocity = 1200 [km/h]

Vo = initial velocity = 25 [km/h]

t = time = 2 [min] = 2/60 = 0.0333 [h]

1200 = 25 + (a*0.0333)

a = 35250.35 [km/h2]

if we convert these units to units of meters per second squared

35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]

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3 years ago
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