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2 years ago

7

Which graph represents a nonlinear relationship? 40 40 40 40 30 30 30 30 20 از مسیر 20 20 20 10 10 10 10 0 0 0 0 2 4 6 0 2 4 6 0

2. 4. 6 0 2 4 6 0

1 answer:

Doss [256]2 years ago

7 0

Explanation:

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maka kina di terima jadi kariawan nya

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karen tidal boleh

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In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

a)

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

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2 years ago

By looking at the relative positions of the elements calcium, Ca, fluorine, F, sulfur, S, and oxygen, O, in the Periodic Table,

SVEN [57.7K]

Answer:

id say the first option.

Explanation:

hope this helps you!

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2 years ago

Use the dot product to find the magnitude of u if u = 6i - 3j

LuckyWell [14K]

Vector u :
u = 6 i - 3 j
The magnitude of vector u :
| u | = $\sqrt{6 ^{2}+(-3) ^{2} } = \sqrt{36+9}= \\ \sqrt{45}= \sqrt{9*5}=3 \sqrt{5}$
Answer:
The magnitude of vector u is 3√5.

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2 years ago

What part of a circuit has the main job of conducting the electricity?

Vesnalui [34]

The main job of conducting electricity is the power source.

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3 years ago

Read 2 more answers

Two identical conducting spheres, having charges of opposite sign, attract each otherwith a force of 0.108 n when separated by 5

lisov135 [29]

Let's start from the final situation. After the two spheres are connected with the conducting wire, the total charge distributes equally between the two spheres (because they are identical). We can call the charge on each sphere $Q/2$ , with Q being the total charge.
The electrostatic force in this situation is 0.0360 N, so we can write
$F=k \frac{( \frac{Q}{2} )^2}{r^2}$
where k is the Coulomb's constant and $r=50.0 cm=0.50 m$ is the separation between the two spheres. Using F=0.0360 N, we can find the value of Q, the total charge shared between the two spheres:
$Q= \sqrt{ \frac{4Fr^2}{k} } = \sqrt{ \frac{4(0.0360 N)(0.50 m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}} }=2.0 \cdot 10^{-6}C$

Now let's go back to the initial situation, before the conducting wire was attached; in this situation, the two spheres have a charge of $q_1$ and $q_2$ , whose sum is Q:
$Q=q_1 + q_2$
The electrostatic force between the two spheres in the initial situation is:
$F=k \frac{q_1 q_2}{r^2}$
And since we know F=0.108 N, we find
$q_1 q_2 = \frac{Fr^2}{k} = \frac{(0.108 N)(0.50 m)^2}{8.99 \cdot 10^9 N m^2 C^{-2}}=3.0 \cdot 10^{-12} C$
But the problem tells us that the two spheres have charges of opposite sign, so we must put a negative sign:
$-3.0 \cdot 10^{-12} C = q_1 q_2$

So now we have basically a system of 2 equations:
$2.0 \cdot 10^{-6} C = Q = q_1 + q_2$
$-3.0 \cdot 10^{-12} C = q_1 q_2$
If we solve it, we find the initial charge on the two spheres:
$q_1 = -1 \cdot 10^{-6}C$
$q_2 = +3 \cdot 10^{-6 } C$

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3 years ago