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Zielflug [23.3K]
3 years ago
9

If two variables have a non-related relationship and one of the variables is changed, how will the other variable change?

Physics
1 answer:
PolarNik [594]3 years ago
6 0

If one of the variables is changed, that tells nothing about what happens to the other one, or IF anything happens, or when, or how long it lasts. Because they are UN-RELATED. You just said so yourself.

None of the choices says this.

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A biker is pedaling at a constant speed of 36 km/h. During the last 10 s of the race, he increases his speed with a constant acc
adell [148]

Answer:

54 km/h

Explanation:

given,

speed of the biker = 36 Km/h

time = 10 s

acceleration = 0.5 m/s²

speed at which it crosses the finish line  = ?

v = 36 x 0.278 = 10 m/s

using equation of motion

v = u + a t

v = 10 + 0.5 x 10

v = 15 m/s

v = 15 x 3.6 = 54 km/hr

speed at which the biker crosses the finish line is equal to 54 km/h

4 0
3 years ago
A cubic sample of a new kind of artificial tissue is subject to an increase in pressure of 160 kPa which results in a reduction
grin007 [14]

Answer:

0.82 MPa

Explanation:

the change in pressure 'σ'=160kPa

K= σ/∈Ф_v => σ/3∈Ф_L

K= 160/(3 x 0.065)

K=820 kPA=0.82 MPa

Thus,the bulk modulus of the tissue 'K' is 0.82 MPa

3 0
2 years ago
Tire or false
spayn [35]
1 and 4 are tire.
2 and 3 are not.
8 0
3 years ago
A washing machine heats 10kg of water in each wash cycle. How much energy is saved by washing at 30'c instead of 50'c if the sta
Volgvan
The equation for this is very simple you add then you subtract then you get the answer then you divide then it all works out for you
6 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
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