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denpristay [2]
2 years ago
14

The length of a piece of tissue is 42 meters. How big is 3/7 of this piece?​

Physics
1 answer:
iris [78.8K]2 years ago
7 0

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

⊱┈────────────────────────┈⊰

\large \bold {ANSWER}

  • \large \boxed { \large \sf \green{18}}

3/7 of the fabric equals 18.

\large \bold {SOLUTION}

To find the solution to your task, we will have to find 3/7 of 42, that is. How much is 3/7 of 42?

So, to find the value we must do the following way, <u>multiply the value of the length of the fabric 42, by the numerator of the fraction 3 and divide by the denominator 7, </u>See:

\begin{gathered}\Large\begin{array}{c}\tt\dfrac{3}{7}\:in\: 42\\\\ \tt 42\cdot 3=126\\\\ \tt 126\div 7=\boxed{\boxed{\tt 18}}\end{array}\end{gathered}

Hence, 3/7 of the fabric equals 18.

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Which objects can an S wave travel through? Check all that apply.
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5 0
3 years ago
Read 2 more answers
.
beks73 [17]

Answer:

2 m/s²

Explanation:

the equations of motion are

S= ut +½at²

v² = u²+ 2as

v = u + at

s = (u+v)/2 × t

From the parameters given

u = 0m/s this is because it starts from rest

Distance (s)  = 9m

Time (t)  = 3s

Based on this the first equation would be used

s = ut + ½at²

Input values

9 = 0×3 + ½ × a x 3²

9 = 0 + 9a/2

9 = 4.5a

Divide both sides by 4.5

a = 9 / 4.5 m/s²

a = 2 m/s²

I hope this was helpful, please mark as brainliest

3 0
4 years ago
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

6 0
3 years ago
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