Physics College

2. An object is dropped from rest. Calculate its velocity after 2.5s if it is dropped:
a.On Earth, where the acceleration due to gravity is 9.8m/s?
b. On Mars, where the acceleration due to gravity is 3.8m/s?

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Answers

Answer 1

Answer:

a=24.5 b=9.5

Explanation:

Related Questions

A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a small pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds

Answers

Answer:

[tex]a=2.5\ m/s^2[/tex]

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]

So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].

ocean currents are always cold true or false

Answers

false, since ocean currents act as conveyer belts of warm and cold water.

Fill in the graph for 50 points

Answers

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Answer:

Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5

Explanation:

a forward horizontal force of 50 N is applied to crate a second horizontal force of 180 N is applied to crate in the opposite direction determine the magnitude and direction of the resultant force acting on the crate
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Answers

Answer:

130n on the 2nd horizontal

Explanation:

The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?

Answers

Answer:

2π/[28 x (10^-3)]

Explanation:

Angular speed : ω=2π/T

T = 28ms = 28 x (10^-3) s

Angular speed = 2π/[28 x (10^-3)]


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Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle

Answers

Answer:

[tex]\tau=0.03\ N-m[/tex]

Explanation:

Given that,

Force acting, F = 6N

The radius of the path, [tex]r=10^{-2}\ m[/tex]

Angle, [tex]\theta=30^{\circ}[/tex]

We need to find the amount of torque acting on the object. The formula for torque is given by :

[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]

So, the required torque is equal to 0.03 N-m.

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