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4 years ago

14

A(n) ____ is a place or object used for comparison to determine of something is in motion

2 answers:

Andre45 [30]4 years ago

4 0

<span>reference point is the answer.</span>

tresset_1 [31]4 years ago

3 0

A reference point is a place or object used for comparison to determine of something is in motion

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Komok [63]

Answer:

A

Explanation:

6 0

3 years ago

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What are two examples of goods and two examples of services

julsineya [31]

The goods and the services make up the basis of every economy. The goods can simply be defined as merchandise or possessions. The services can be defined as the actions through which help is provided, or work is done for someone else. Example of goods are the food and furniture, with the food being crucial for the survival of the people, while the furniture is an essential part of every home and its practicality and decor. Examples of services are teaching and car repairing. The teaching is crucial for the development of the societies, as through it the people get education, while the repairing of cars is very important as lot of people have them, can not afford to buy new ones all the time, and they need for their daily movement over longer distances.

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4 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

Which change would result in a stronger electromagnet?

Sonbull [250]

I'd say B.) Increasing the voltage of the battery.

6 0

3 years ago

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Calculating the Mass of a Granite Monument

Volgvan

Answer:

m = 684,865,8 g

Step-by-step explanation

V = 25,365.4 cm^3 Is volume

r = 27g/cm^3 Is density

To calculate mass you use formula:

m= V*r

m = 25,365.4 x 27

m = 684,865,8 g

8 0

3 years ago