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horrorfan [7]
3 years ago
5

Explain why it is important to identify a reference point for any description of motion.

Physics
2 answers:
CaHeK987 [17]3 years ago
6 0
N the particular case of motion, a reference point is a stationary point in space. You can then measure the displacement of objects in space from this point. A reference point is important in determining motion because in order to say that something is moving, you need to have something stationary to compare it to.
Leya [2.2K]3 years ago
3 0

A reference is critical because it determines how much motion has occurred. You can't determine how much an objected has moved if you don't know where it came from.

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Definition of distance travelled?​
Mekhanik [1.2K]

Answer:

distance traveled is a total length of the path traveled between two positions.

6 0
3 years ago
Read 2 more answers
A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator
Ivenika [448]
Here you go mate. Hope it helps u. Pls follow me in reddit lol username: RobloxNoob2006

5 0
3 years ago
An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in ser
s2008m [1.1K]

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

V_L=V_oe^{-\frac{Rt}{L}}

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s

hence, after 1.6s the inductor will have a potential difference of 24V

3 0
3 years ago
Two particles move along an x axis. The position of particle 1 is given by x ! 6.00t 2 # 3.00t # 2.00 (in meters and seconds); t
s344n2d4d5 [400]

Answer:

15.6m/s

Explanation:

V1=\frac{dx}{dt}=\frac{d}{dt}(6t^{2}+3t+2) because the derivate of the position is the velocity

V1=12t+3

V2=20+\int\limits^_ {} \,-8t because the integral of the acceleration is the velocity

V2=20-4t^{2}

V1=V2 to see when the velocities of particles match

12t+3=20-4t^2

4t^2+12t-17=0 we resolve this with \frac{-b+-(\sqrt{b^{2} -4ac} )}{2a}

and we take the positif root

t=1.05 sec

if we evaluate the velocity (V1 or V2) the result is 15.6m/s

7 0
3 years ago
4. What is the electric field strength 1.4 nm from a charge of 4.7 cC?
pentagon [3]

The electric field strength is 2.16\cdot 10^{26} N/C

Explanation:

The strength of the electric field produced by a single point charge is given by:

E=k\frac{q}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge at which the field strength is calculated

For the charge in the problem, we have:

q=4.7 cC = 0.047 C is the charge

r=1.4 nm = 1.4\cdot 10^{-9} m

Therefore, the electric field strength is

E=(8.99\cdot 10^9)\frac{0.047}{(1.4\cdot 10^{-9})^2}=2.16\cdot 10^{26} N/C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
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