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2 years ago

Mtngmrgncseeieeeie what is the unscramble word of this

Physics

2 answers:

Katena32 [7]2 years ago

3 0

Answer:

emergencies

Explanation:

This is what I got when I unscrambeled it

Serhud [2]2 years ago

3 0

i think that would be magnetesium

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A 5.0-n projectile leaves the ground with a kinetic energy of 220 j. at the highest point in its trajectory, its kinetic energy

NikAS [45]

First, we get the difference between the kinetic energies such that,
difference = (220J - 120J)
difference = 100 J
The difference in kinetic energy is the equivalent of the potential energy which is calculated through the equation,
PE = mgh
To calculate for the height, we derive the equation in a form,
h = PE/mg
The product of the mass and acceleration due to gravity is the weight.
h = (100 J) / (5 N)
h = 20 m

<em>Hence, the answer is 20 m. </em>

3 0

2 years ago

14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?

Vladimir79 [104]

The relationship between resistance R and resistivity $\rho$ is
$R= \frac{\rho L}{A}$
where L is the length of the wire and A its cross section.

The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{1.6 mm}{2}=0.8 mm=8\cdot 10^{-4} m$
and the cross section is
$A=\pi r^2 = \pi (8\cdot 10^{-4} m)^2=2.01\cdot 10^{-6} m^2$

From the first equation, we can then find the length of the wire when $R=4.8 \Omega$ (copper resistivity: $\rho = 1.724 \cdot 10^{-8} \Omega m$ )
$L= \frac{AR}{\rho}= \frac{(2.01\cdot 10^{-6} m^2)(1.724 \cdot 10^{-8} \Omega m)}{4.8 \Omega}=7.21 \cdot 10^{-15} m$

4 0

2 years ago

3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P

KIM [24]

Answer:

From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

$E=\frac{F}{Q}$

But the force F

$F= \frac{kQ1Q2}{r^2}$

But the electric field intensity due to a point charge Q at a distance r meters away is given by

$E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }$

<em>From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases</em>

6 0

2 years ago

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0

2 years ago

In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc

Vesna [10]

Answer:

Explanation:

Let T be the tension in the swing

At top point $mg-T=\frac{mv^2}{r}$

where v=velocity needed to complete circular path

r=distance between point of rotation to the ball center=L+\frac{d}{2} (d=diameter of ball)

Th-resold velocity is given by $mg-0=\frac{mv^2}{r}$

To get the velocity at bottom conserve energy at Top and bottom

At top $E_T=mg\times 2L+\frac{mv^2}{2}$

Energy at Bottom $E_b=\frac{mv_0^2}{2}$

Comparing two as energy is conserved

$v_0^2=4gr+gr$

$v_0^2=5gr$

$v_0=\sqrt{5gr}$

$v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}$

6 0

2 years ago