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igor_vitrenko [27]
2 years ago
8

Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter

val (on Earth) the Moon clock:
A) (9.8/1.6)h
B) 1 h
C) the square root of 9.8/1.6 h
D) (1.6/9.8)h
E) the square root of 1.6/9.8 h
Physics
1 answer:
satela [25.4K]2 years ago
7 0

The time-period of a simple pendulum is

<em>Time =  2 π √(length/grav-accel)</em>

After unraveling the question, then completing it, and working out what I <em>believe</em> it's trying to ask, the choice that correctly answers the question that I have invented is <em>choice-E</em> .

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Init. A
Gekata [30.6K]

Answer:

v = 1/3 m / s = 0.333 m / s

in the direction of the truck

Explanation:

The average speed is defined by the variation of the position between the time spent

           v = Δx / Δt

since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is

           Δx = 20 - 15 +20

           Δx = 25 m

therefore we calculate

         v = 25/75

         v = 1/3 m / s = 0.333 m / s

in the direction of the truck

7 0
2 years ago
The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress
PolarNik [594]

Answer:

T = 0.03 Nm.

Explanation:

d = 1.5 in = 0.04 m

r = d/2 = 0.02 m

P = 56 kips = 56 x 6.89 = 386.11 MPa

σ = 42-ksi = 42 x 6.89 = 289.58 MPa

Torque = T =?

<u>Solution:</u>

σ = (P x r) / T

T = (P x r) / σ

T = (386.11 x 0.02) / 289.58

T = 0.03 Nm.

7 0
2 years ago
Imagine that a small boy and his much larger father are enjoying a winter morning, sliding along the ice on a nearby playground.
kramer
You can write an hypothesis such as this:
The weight of an object has effects on the operating frictional force, the greater the weight, the higher the operating frictional force.
The father is the one with the higher weight while the son has the lower weight. The operating frictional force is the friction that their weights exert.
8 0
2 years ago
Read 2 more answers
An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with
Anna71 [15]

(D) The gravitational force between the astronaut and the asteroid.

Reason :

All the other forces given in the options, except (D), doesn't account for the motion of the astronaut. They are the forces that act between nucleons or atoms and neither of them accounts for an objects motion.

6 0
2 years ago
Consider a ball rolling around in a circular path on the inner surface of a cone. the weight of the ball is shown by the vector
zimovet [89]

Answer:

option 2 is righttttt.................

5 0
1 year ago
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