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madam [21]
3 years ago
15

What decimal power does the abbreviation p represent?

Physics
1 answer:
Sindrei [870]3 years ago
7 0
Pico (:
i think so i dont know for sure

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Graphing data is most used to help you to observe_____ *
kotykmax [81]

Answer: Graphing data is used to display data because it is easier to see trends in the data when it is displayed visually compared to when it is displayed numerically in a table.

5 0
3 years ago
a doghouse has a mass of 98 kg, and the coefficient of friction between it and the ground is 0.95. what is the maximum force of
MaRussiya [10]
The maximum force of friction:

F = μ* N, where μ is the friction coefficient and N id the force of the doghouse normal to the surface.

N for a flat surface would be: N = m*g, where m is the mass and g is the gravitational acceleration = 9,81 m/s².

Combined:

F = μ * g * m
3 0
4 years ago
Read 2 more answers
4. A force of 5 N gives a mass m,, an acceleration of 10 m's, and a mass
Art [367]

Answer:

For mass m  1  newton 2nd law

F=m  1  a  1

​5=m  1  ×10

m  1  =  2 1 kg

For mass m  2

​F=m  2  a 2

​5=m  2 ×20

m  2 =  4 1  kg

if tied together  

Total mass =m  1  +m  2  =  1/2 +1/4=3/4kg  

Now

F=M T  Q T

 a  T    =  5/m T =  5×4/3  =  3 /20m/s^2

Explanation:

8 0
3 years ago
Equations to use: v= λ ∙ f v=d/t
Margarita [4]

b. 460.8 m/s

Explanation:

The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

f=\frac{v}{2L}

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

v=2Lf=2(0.90 m)(256 Hz)=460.8 m/s

c. 18,000 m

Explanation:

The relationship between speed of the wave, distance travelled and time taken is

v=\frac{d}{t}

where

v = 6,000 m/s is the speed of the wave

d = ? is the distance travelled

t = 3 s is the time taken

Re-arranging the formula and substituting the numbers into it, we find:

d=vt=(6,000 m/s)(3 s)=18,000 m

3 0
3 years ago
The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the
ycow [4]

F = 5.93×10^{13}\:\text{N}

Explanation:

Given:

m_1= 2×10^{16}\:\text{kg}

m_2= 4×10^{22}\:\text{kg}

r = 30000\:\text{km} = 3×10^7\:\text{m}

Using Newton's universal law of gravitation, we can write

F = G\dfrac{m_1m_2}{r^2}

\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}

\:\:\:\:= 5.93×10^{13}\:\text{N}

3 0
3 years ago
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