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3 years ago

1.) dependent variable A.)the variable representing the

Physics

1 answer:

My name is Ann [436]3 years ago

4 0

Answer:

Explanation:

the answer is b

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A bicycile traves with an average velocty 15km/h north for 20min what is his displasment

Mila [183]

Answer:

5000 m equivalent to 5 Km

Explanation:

Average velocity = $\frac{Displacement}{Time}$

so, Displacement =Average velocity × time

We should convert Km/h to m/s so Km/h ⇒ $\frac{1000}{60*60}$ m/s, also convert time to second so, 20min ⇒(20* 60)seconds

Displacement = (15 × $\frac{1000}{60*60}$ ) × (20×60) =5000m OR 5Km

6 0

2 years ago

Observing and experimenting are two ways that ecologists can answer scientific questions. What are some differences between thes

Sav [38]

Answer:OBSERVING IS WHATCHING A OBJECT VERY CLOSELY AND EXPIERIMENTING IS WER YOUTEST ON A CERTAIN THING OR CREATURE OR MASS OR ELEMENT

Explanation:IM THE MYSTERY MAN WHOOSH

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3 years ago

A student is pushing a box across the room. To push the box three times farther, the student needs to do how much work?

Travka [436]

Answer:

Removing some of the books reduced the mass of the box, and less force was needed to push it across the floor.

8 0

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

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

$f_n=3.75N$

5 0

2 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago