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11Alexandr11 [23.1K]
3 years ago
14

Describe an example of Newton’s 1st Law of Motion (the motion of an object remains the same unless it is acted upon by a force)

Physics
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:

if your in a car and you suddenly stop since you were in motion you fling forward while the car is stopped because an object in motion stays in motion unless it's stopped by an outside force and that's why we have seatbelts

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What real world examples show no work being done? Can you think of examples other than resisting the forces of gravity?
Lesechka [4]
I need more info man then i can help you. 
7 0
3 years ago
Read 2 more answers
The earth’s radius is 6.37 × 10 6 m ; it rotates once every 24 hours. What is the earth’s angular speed? Viewed from a point abo
Advocard [28]

Answer:

a) the angular speed of the Earth's rotation is <em>7.272 × 10⁻⁵ rad/s.</em>

<em></em>

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the <em>angular velocity is positive.</em>

<em></em>

c) Earth's speed at a point on the equator is <em>463.23 m/s.</em>

<em></em>

d)  the speed of a point on the earth’s surface halfway between the equator and the pole is <em>231.61 m/s.</em>

Explanation:

a) The angular speed of the Earth's rotation is:

ω = 2π / T

where

T is the period

ω = 2π / (24 hr × (3600 s / 1 hr))

<em>ω = 7.272 × 10⁻⁵ rad/s</em>

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the <em>angular velocity is positive.</em>

c) Earth's speed at a point on the equator is:

v = r × ω

v = (6.37 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

<em>v = 463.23 m/s</em>

d) The radius of the circle in which the point moves is half of Earth's radius. Therefore,

r = 1/2(6.37 × 10⁶ m)

r = 3.19 × 10⁶ m

Therefore, the speed of a point on the earth’s surface halfway between the equator and the pole is:

v = r × ω

v = (3.19 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

<em>v = 231.61 m/s</em>

6 0
3 years ago
Suatu sistem gas berada didalam ruang yang fleksibel. Pada awalnya gas berada pada kondisi P1 = 1,5 × 10 pangkat 5 N/m2, T1 = 27
enot [183]

Answer:

16.00L

Explanation:

First you calculate the number of moles in the system:

PV=nRT\\\\n=\frac{PV}{RT}\\\\n=\frac{(1.5*10^5N/m^2)(12L)}{(0.082L.atm/mol.K)(300.15K)}=73134.16\ mol

To find the new volume of the system you use the following formula for an isobaric procedure:

T_2-T_1=\frac{P}{nR}(V_2-V_1)\\\\V_2=\frac{nR}{P}(T_2-T_1)+V_1\\\\V_2=\frac{(73134.16\ mol)(0.082L.atm/mol.K)}{1.5*10^5N/m^2}(400.15-300.15)K+12L\\\\V_2=16.00L

hence, the new volume is 16.00L

6 0
4 years ago
Formula for work done​
Alenkinab [10]

Answer:

Work = Force × Distance

-----------------------------------

hope this helps :)

6 0
3 years ago
A handel or knob is fixed at the free end of door. Why?
andriy [413]
Maximum torque.
T=F.r
r is measured from point of turning, which is the hinge. r is maximum when the handle is at the free edge.
5 0
3 years ago
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