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2 years ago

Express 2759 mockingbirds in kilomockingbirds. answer in units of kmockingbird

Physics

1 answer:

Nadya [2.5K]2 years ago

6 0

I would think 2.759kmockinbirds if you consider mockingbirds equal to a metre and kmockinbird to kilometres

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A negative charge, q1, of 6 µC is 0. 002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the e

prohojiy [21]

Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

<h3>What is electrical force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is

q₁ is the negative charge = 6 µC=6×10⁻⁶ C

q₂ is the positive charge = 3 µC=3×10⁻⁶ C

r is the distance between the charges=0.002 m

$F_E$ is the electric force =?

The value of electric force will be;

$\rm F_E= \frac{Kq_1q_2}{r^2} \\\\ F_E= \frac{9\times 10^9\times 6\times 10^{-6}\times3\times10^{-6}}{(0.002)^2}\\\\ \rm F_E=4.05\times10^4\;N$

Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

To learn more about the electrical force refer to the link;

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7 0

2 years ago

A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict

lakkis [162]

The frictional force is 39.4 N

Explanation:

We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

Here we know that the box is moving with constant velocity, so its acceleration is zero:

$a=0$

This means that the net force is also zero:

$\sum F=0$

The net force on the block is given by the applied force, forward, and the frictional force, backward:

$\sum F = F_a-F_f=0$

where

$F_a=39.4 N$ is the applied force

$F_f$ is the frictional force

Therefore, solving for $F_f$ ,

$F_f=F_a=39.4 N$

Learn more about friction:

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#LearnwithBrainly

8 0

2 years ago

Read each scenario and then answer the question. Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in

Alexeev081 [22]

The power in every scenario is the same. Power is the measurement of the energy transfer over time.

<h3>What is power?</h3>

Power is the measurement of the energy transfer over time. It can be calculated by the formula,

$P = \dfrac Wt$

Where,

$P$ - power

$W$ - work

$t$ - time

In scenario A,

$P = \dfrac {120 }6 = 20 \rm \ W$

In scenario B,

$P = \dfrac {160 }8 = 20 \rm \ W$

In scenario B,

$P = \dfrac {200 }{10} = 20 \rm \ W$

Therefore, the power in every scenario is the same.

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4 0

2 years ago

Read 2 more answers

You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f

Alborosie

Answer:

3.25 m

Explanation:

t = Time taken = 0.166 seconds

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s²

s = 1 because meter stick is 1 meter in length

$s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s$

Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m$

The distance of the acorn from the ground is 1.87+1.38 = 3.25 m

4 0

2 years ago

A wire of length 6cm makes an angle of 20° with a 3 mT

Crazy boy [7]

Answer:

Approximately $7.3 \times 10^{-3}\; \rm A$ (approximately $7.3\; \rm mA$ ) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let $\theta$ denote the angle between the wire and the magnetic field.

Let $B$ denote the magnitude of the magnetic field.

Let $l$ denote the length of the wire.

Let $I$ denote the current in this wire.

The magnetic force on the wire would be:

$F = B \cdot l \cdot I \cdot \sin(\theta)$ .

Because of the $\sin(\theta)$ term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is $90^\circ$ .)

In this question:

$\theta = 20^\circ$ (or, equivalently, $(\pi / 9)$ radians, if the calculator is in radian mode.)
$B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T$ .
$l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m$ .
$F = 1.5\times 10^{-4}\; \rm N$ .

Rearrange the equation $F = l \cdot I \cdot \sin(\theta)$ to find an expression for $I$ , the current in this wire.

$\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}$ .

5 0

2 years ago