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adoni [48]
3 years ago
13

2) Write two examples in which an object undergoes more than one type of motion at the same time.

Physics
1 answer:
worty [1.4K]3 years ago
7 0

1) when a Car Moves; its Wheels exhibit Both <em><u>Rotational</u></em><em> </em>and <em><u>Rectilinear</u></em> Motion

2) When Earth rotates Around the sun it is executing <em><u>Periodic</u></em> motion through revolving around the sun; at the same time it is executing <em><u>Rotatory</u></em> motion by rotating on its own axis.

Hope it helps...

Regards,

Leukonov/Olegion.

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A car accelerates from rest at 3.00 m/second squared What is the velocity after 5 secs ? What is the displacement after 5 second
andrey2020 [161]

Explanation:

after 5 seconds, the velocity is (5s)(3m/s²) = 15m/s

The displacement after 5s is

x=vo + (1/2)at²

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Juli2301 [7.4K]

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8 0
3 years ago
Read 2 more answers
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at
andrew-mc [135]

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

6 0
3 years ago
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
NikAS [45]

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.

7 0
3 years ago
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