All categories

3 years ago

Is wind electrically charged

Physics

1 answer:

nataly862011 [7]3 years ago

5 0

No that's not true I believe only lightning is electrically charged because it can shock you.

You might be interested in

When dogs see police dogs do they think "oh no its the cops"?

ira [324]

No they say "Watch out it's the fuzz"

4 0

2 years ago

Read 2 more answers

A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force

Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0

3 years ago

Read 2 more answers

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago

Holding onto a tow rope moving parallel to a frictionless ski slope, a 70.1 kg skier is pulled up the slope, which is at an angl

lara [203]

Answer:

given,

mass of the skier = 70.1 Kg

angle with horizontal, θ = 8.6°

magnitude of the force,F = ?

a) Applying newton's second law

$m g sin\theta - F_{rope} = ma$

velocity is constant, a = 0

$m g sin\theta - F_{rope} =0$

$F_{rope} = m g sin\theta$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0$

$F_{rope}= 102.73\ N$

b) now, when acceleration, a = 0.135 m/s²

$F_{rope}-m g sin\theta = ma$

velocity is constant, a = 0.135 m/s₂

$F_{rope} = m g sin\theta+ma$

$F_{rope} = 70.1\times 9.8\times sin 8.6^0+70.1\times 0.135$

$F_{rope}= 112.19\ N$

7 0

3 years ago

Nora walks down a street and sees a ball dropped from a building

Grace [21]

It is gravity¿ what is the question?

5 0

3 years ago