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3 years ago

Why can making observations be a very good way to come up with a scientific question?

Physics

1 answer:

astra-53 [7]3 years ago

5 0

Because observation is a guess that you make and study on that certain topic.

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A solid conducting sphere with radius R = 0.390 m carries a net charge of +0.650 nC.

Tom [10]

Answer:

38.5 N/C

Explanation:

The electric field generated by a charged sphere at a point outside the sphere is equivalent to the electric field generated by a single point charge, and it is given by

$E=k\frac{Q}{r^2}$

where

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

Q is the net charge

r is the distance from the centre of the sphere

In this problem, we have

$Q=+0.650 nC=0.65\cdot 10^{-9}C$

$r=0.390 m$

Substituting into the equation, we find

$E=(9\cdot 10^9)\frac{(0.65\cdot 10^{-9}C)}{(0.390m)^2}=38.5 N/C$

6 0

3 years ago

What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?

klasskru [66]

P.E=0.0675 J

Explanation:

Elastic potential Energy=Force × distance of displacement

The formula to apply is;

P.E=1/2 ks²

where k is the spring constant given as; 1.5 *10^-2 N/m and s is the displacement

In this case,

s=3

P.E= 1/2 * 1.5 × 10^-2 ×3²

P.E=1.5×10^-2*4.5

P.E=0.0675 J

Learn More

Elastic potential energy:brainly.com/question/1352053

Keywords: approximate,value,elastic potential energy,spring, elongated

#LearnwithBrainly

7 0

3 years ago

Give some example acostic using "ELEMENTS"

Jet001 [13]

E=ERBIUM
L=LITHIUM
E=EINSTEINIUM
M=MAGNESIUM
E=EUROPIUM
N=NITROGEN
T=TITANIUM
S=SULPHURDIOXIDE

5 0

3 years ago

How is an electric field similar to a gravitational field?

kap26 [50]

They both-
1) are vector fields
2) they both transmit forces between objects

I hope this helps :))

7 0

3 years ago

Read 2 more answers

A point on a wheel of radius 40 cm that is rotating at a constant 5.0 revolutions per second is located 0.20 m from the axis of

Margarita [4]

Answer:

$197.2 m/s^2$

Explanation:

The centripetal acceleration of a point moving by circular motion is given by:

$a=\omega^2 r$

where

$\omega$ is the angular velocity

r is the distance from the axis of rotation

The point on the wheel makes 5.0 revolutions per second, so the frequency is

$f=\frac{5}{1}=5 Hz$

and the angular velocity is

$\omega=2\pi f = 2\pi (5)=31.4 rad/s$

While the distance of the point from the axis of rotation is

$r=0.20 m$

Substituting, we find the acceleration:

$a=(31.4)^2(0.20)=197.2 m/s^2$

5 0

3 years ago

Read 2 more answers