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Andrei [34K]
3 years ago
8

Question 1 of 10

Physics
1 answer:
Annette [7]3 years ago
7 0

Answer:

A. 2.0

Explanation:

You might be interested in
A man is 40 kg . calculate his weight.(g=10m/s^2)
Dafna1 [17]

{ \large{ \bf{Weight = mg}}}

{ \large{ \bf{Mass = 40 \:  kg \:  \:  \:  \: (given)}}}

{ \large{ \bf{Weight = 40 \times 10}}}

{ \large{ \therefore{ \bf{ \red{ Weight = 400N}}}}}

5 0
3 years ago
Imagina que compras una placa rectangular de metal de 2mm de alto, 10mm de ancho x 50mm de largo, y una masa de 0.02kg. El vende
shusha [124]

Answer:

Densidad de la placa = 20 g/cm³.

La placa no es de oro.

Explanation:

Para encontrar la densidad de la placa rectangular primero debemos hallar su volumen:

V = 2 mm*10 mm*50 mm = 1000 mm^{3}*\frac{1 cm^{3}}{(10 mm)^{3}} = 1 cm^{3}      

Ahora, encontremos al densidad de la placa:

d = \frac{m}{V} = \frac{20 g}{1 cm^{3}} = 20 g/cm^{3}

Dado que la densidad del oro es 19.32 g/cm³ y que la densidad de la placa rectangular calculada es 20 g/cm³, podemos decir que dicha placa no es de oro.                  

Espero que te sea de utilidad!

6 0
3 years ago
What type of fingerprints is invisible to the naked eye?
luda_lava [24]
Well latent fingerprints are made of oil and sweat and generally materials that you can't see very easily, so it should be that.

Hope this helps :D

3 0
3 years ago
What is the mass of an object that requires a force of 30 N to accelerate at a rate of 5 m/sec2 ?
Anastaziya [24]
M=F/A
Which means 30 divided by 5 m/s is 6kg(mass)
8 0
3 years ago
Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea
Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

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

Where,

L = Vibrating length string

T = Tension in the string

\mu = Linear mass density

At the same time we have the expression for the number of beats described as

n = |f_1-f_2|

Where

f_1 = First frequency

f_2 = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

f \propto \sqrt{T}

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

n = f_2-f_1

f_2 = n+f_1

Replacing 3/sfor n and 202Hz for f_1,

f_2 = 3/s + 202Hz

f_2 = 3/s(\frac{1Hz}{1/s})+202Hz

f_2 = 206Hz

The frequency of the tightened is 205Hz

7 0
3 years ago
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