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3 years ago

if an object starts at rest and moves 60 meters north along a straight line in 2 seconds, what is the average velocity?

Physics

2 answers:

VashaNatasha [74]3 years ago

8 0

30 m/s because you would divide 60 meters by 2 seconds. 60 divided by two equals 30

Thepotemich [5.8K]3 years ago

5 0

Velocity =
(distance between start-point and end-point) / (time to run the course)
in the direction of
(direction from the start-point to the end-point).

Distance from start-point to end-point = 60 meters
Time to run the course = 2 seconds
Direction = north

Velocity = 30 m/s north

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Answer:

Explanation:

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3 years ago

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What is the net force exerted on a 28.8 kg shopping cart that accelerates at a rate of 2.88 m/s^2?

Mama L [17]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 28.8 × 2.88 = 82.944

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3 years ago

Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist

Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

$i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$ , where $i$ max= $\frac{E}{R}$

Given that, at i(2)= $\frac{imax}{2} =\frac{E}{2R}$

⇒ $\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})$

⇒ $\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}$

⇒ $\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}$

⇒ $log(2)=\frac{2}{\frac{L}{R}}$

⇒ $\frac{L}{R}=\frac{2}{log2}$

Now substitute $i(t)=\frac{3}{4}imax=\frac{3E}{4R}$

⇒ $\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})$

⇒ $\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}$

⇒ $\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}$

Applying logarithm on both sides,

⇒ $log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}$

⇒ $log(4)=\frac{t}{\frac{L}{R}}$

⇒ $t=log4\frac{L}{R}$

now subs. $\frac{L}{R}=\frac{2}{log2}$

⇒ $t=log4\frac{2}{log2}$

also $log4=log2^{2}=2log2$

⇒ $t=2log2\frac{2}{log2}$

⇒ $t=4$

5 0

3 years ago

A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t

vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

$\mu =7.20 g/m = 0.0072 kg/m$ is the mass linear density

Solving the equation,

$f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz$

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

$f=\frac{v}{2L}$

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

$L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m$

4 0

3 years ago

Find the radioactivity of a 1 g sample of 226Ra given that <img src="https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D1620" id="TexFormul

dangina [55]

Answer:

Explanation:

No of atoms of Ra in 1 g of sample = 6.023 x 10²³ / 226

N = 2.66 x 10²¹

disintegration constant λ = .693 / half life

half life = 1620 x 365 x 60 x 60 x 24 = 5.1 x 10¹⁰ s

disintegration constant λ = .693 / 5.1 x 10¹⁰

radioactivity dn / dt = λN

= (.693 / 5.1 x 10¹⁰ ) x 2.66 x 10²¹

= .3614 x 10¹¹ per sec

= 3.614 x 10¹⁰ / s

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2 years ago