Identifying Sources Used to Study Earth's Interie

In the above diagram, the block is at static equilibrium on the ground. What is the value of the force exerted by the ground on

Liula [17]

Answer: C. -12 newtons

Explanation: Since, the block is in static equilibrium,

the net force on the block is zero i.e. balance forces are acting on the block.

In the horizontal direction,

force applied+force due to friction = 0

20 N + (-20N) = 0

In the vertical direction as well, net force is zero:

Force due to gravity + force exerted by the ground = 0

12 N + force exerted by the ground = 0

⇒force exerted by the ground = 0 -12 N = -12 N

Thus, the correct option is C. -12 N where negative sign means that the direction of the force due to ground is opposite to the force of gravity.

8 0

3 years ago

1.- Escribir los siguientes números aplicando la notación científica:

Kaylis [27]

Answer is 5,000,000

4 0

3 years ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$

$\Rightarrow q_1=5\times 10^{-8}-q_2$

The equation (ii) become:

$(5\times 10^{-8}-q_2)q_2=6\times10^{-16}$

$\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0$

$\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}$

From equation (iii)

$q_1=2\times10^{-8}, 3\times10^{-8}$

So, the magnitude of initial charges on both the sphere are $3\times10^{-8}$ Coulombs $=0.03 \mu C$ and $2\times10^{-8}$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

4 0

3 years ago

What is the net force acting on an object that is not accelerating in the horizontal axis?

yuradex [85]

Answer:

.......................

Explanation:

A net force = unbalanced force. If however, the forces are balanced (in equilibrium) and there is no net force, the object will not accelerate and the velocity will remain constant.

8 0

3 years ago

An endangered species is a species that has died out and no individuals are left. True or false

valentinak56 [21]

Answer:

the answer is false... ....

8 0

3 years ago

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