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pochemuha
1 year ago
5

What data will the simulation provide about your design? You will also need a control (something you don’t change) that you can

use as a comparison to see how effective your design is. Explain how you will set up a control.
Physics
1 answer:
Ksenya-84 [330]1 year ago
6 0

Answer:

yes you will need a control

Explanation: all i know is that you need a control i don't know how to set it up lol sorry

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You are sitting on a merry-go-round at a distance of 2m from its center. It spins 15 times in 3 min. What distance do you move a
soldier1979 [14.2K]

Answer:

A) 12.57 m

B) 5 RPM

C) 3.142 m/s

Explanation:

A) Distance covered in 1 Revolution:

The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:

s = rθ

where,

s = distance covered = ?

r = radius of circle = 2 m

θ = Angle = 2π radians  (For 1 complete Revolution)

Therefore,

s = (2 m)(2π radians)

<u>s = 12.57 m</u>

B) Angular Speed:

The formula for angular speed is given as:

ω = θ/t

where,

ω = angular speed = ?

θ = angular distance covered = 15 revolutions

t = time taken = 3 min

Therefore,

ω = 15 rev/3 min

<u>ω = 5 RPM</u>

C) Linear Speed:

The formula that gives the the linear speed of an object moving in a circular path is given as:

v = rω

where,

v = linear speed = ?

r = radius = 2 m

ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s

Therefore,

v = (2 m)(1.571 rad/s)

<u>v = 3.142 m/s</u>

8 0
3 years ago
A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an
AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0
3 years ago
Questions are in the picture ^ (please answer separately)
Klio2033 [76]
A: The battery is a store of internal energy (shown as chemical energy). The energy is transferred through the wires to the lamp, which then transfers the energy to the surroundings as light. These are the useful energy transfers - we use electric lamps to light up our rooms.

B: In the case of the light bulb the 95J of energy transferred as heat is wasted energy as it is not useful because the purpose of the device is to produce light.

SORRY I ONLY HAVE ANSWERS FOR A AND B
5 0
2 years ago
Read 2 more answers
The heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water
Naily [24]

Answer:

A. 4148 J/K/Kg

B. 4148 J/K/L

Explanation:

A. Heat capacity per unit mass is known as the specific heat capacity, c.

C = Heat capacity/mass(kg)

C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg

B. Volume of water = mass/density

Density of water = 1 Kg/L

Volume of water = 0.125 Kg/ 1Kg/L

Volume of water = 0.125 L

Heat capacity per unit volume = (523 J/K) / 0.125 L

Heat capacity per unit volume = 4148 J/K/L

5 0
3 years ago
1. f=
ahrayia [7]
1. F = 8 Hz
2. V = 8 m/s
3. D = 8 m
4. T = 8 sec
5 0
2 years ago
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