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professor190 [17]
3 years ago
10

What is the weight of a 4.5 kg mass on Earth?

Physics
1 answer:
swat323 years ago
4 0

Answer:

7.535×10^25 earth mass

Explanation:

for an approximate result,divide the mass value by 9.223e+18

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A steel piano string of mass per unit lenght 4×10-⁴kgm-¹ was struck, if the speed of the sound on the string is 300ms-¹. Calcula
attashe74 [19]

Answer:

36 N

Explanation:

Velocity of a standing wave in a stretched string is:

v = √(T/ρ),

where T is the tension and ρ is the mass per unit length.

300 m/s = √(T / 4×10⁻⁴ kg/m)

T = 36 N

3 0
3 years ago
What factor about the planets caused you to weigh more or less?
ad-work [718]

usually gravity is what causes us to make us weigh more or less depending on which planet we are on

7 0
3 years ago
All of the following would be questions that could be scientifically investigated except:
dimulka [17.4K]

Answer:

"What is the best advice a parent can give a child?"

Explanation:

The other answers have one clear response. Ex: Orange popsicles melt faster than grape popsicles. That would be a fact.

But parental advice can vary, depending on your opinion. I may say that all parents must teach their children not to talk to strangers, while someone else may say that parents should advice their kids to treat everyone fairly. Nothing can be proven as the only appropriate response.

Hope this helps!

5 0
3 years ago
A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia
slega [8]

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

8 0
3 years ago
We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

R = 2(1.5 -1)\ 30

R = 30 cm

hence, the radii of curvature is 30 cm.

3 0
3 years ago
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