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2 years ago

A car starts with an initial speed of 12 m/s and accelerates at 3 m/s/s for 5 seconds

Physics

1 answer:

Nostrana [21]2 years ago

3 0

Explanation:

Vi = 12 m/s

a = 3 m/s^2

t = 2 s

Vf = Vi + a × t = 12 + 3 ×2 = 18 m/s

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All contraceptives prevent STIs true or false

NNADVOKAT [17]

False. Not all contraceptives prevent STIs.

Contraceptives are items used or taken to prevent conception. These are birth control pill, patch, ring, IUD, birth control shots, and condoms. Among these contraceptives, condoms are the only type of birth control that prevents STIs or Sexually Transmitted Infections.

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3 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

Semenov [28]

Answer:

$f_n=3.75N$

Explanation:

From the question we are told that:

Frictional force $F=0.150N$

Coefficient of kinetic friction $\mu=0.04$

Generally the equation for Normal for is mathematically given by

$f_n=\frac{F}{\mu}$

Therefore

$f_n=\frac{0.150}{0.04}$

$f_n=3.75N$

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2 years ago

James took two pea plants, placing one in a dark closet and the other on a sunny window sill. Both are located in air-conditione

Fudgin [204]

The constant is the temperature of the air that the plants get.

The independent variable is the thing that YOU control. That's the amount of sunlight each plant gets.

The <em>dependent variable</em> is anything that's caused by changes in the independent variable. That's the growth of the plants.

5 0

3 years ago

Read 2 more answers

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

An engineer wants to set up simple password protection with no usernames for some switches in a lab, for the purpose of keeping

Andre45 [30]

Answer:

A. A login vty mode subcommand

Explanation:

since we are protecting co-workers from connecting to the switches from their desktop PCs, we would need a Telnet line which is used to connect to devices remotely from other network devices on the same network segment as the device we want to connect to. A login local vty subcommand configures a local username for login access but since our design constraint is to configure without usernames, option A is the correct answer.

8 0

3 years ago