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3 years ago

Help with 3 and check the other ones please :(

Physics

2 answers:

gladu [14]3 years ago

5 0

The other person is correct I just checked their work to see if they were doing it right mark them as brainlest

Kobotan [32]3 years ago

4 0

Answer:

number 3 The wind because the wind has strong pressure which causes to change peoples directions in where they are going

Explanation:

hope this helped by the way i think you answered b and 4 i don't know if this is a test and i sent this at the wrong time :(

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| The electric field 5.0 cm from a very long charged wire is (2000 N/C, toward the wire). What is the charge (in nC) on a 1.0-cm

Serggg [28]

Answer:

The charge is 0.056 nC.

Explanation:

Given that,

Electric field = 2000 N/C

Distance = 5.0 cm

We need to calculate the charge density

Using formula of charge density

$E=\dfrac{\lambda}{2\pi\times\epsilon_{0}r}$

$\lambda=2\pi\times\epsilon_{0}\times r\times E$

Put the value into the formula

$\lambda=2\pi\times8.85\times10^{-12}\times5.0\times10^{-2}\times2000$

$\lambda=5.56\times10^{-9}\ C/m$

We need to calculate the charge in 1.0 cm

Using formula of charge

$Charge = \lambda\times\text{length of segment}$

$Charge =5.56\times10^{-9}\times1.0\times10^{-2}$

$Charge=0.056\times10^{-9}\ C$

$Charge=0.056\ nC$

Hence, The charge is 0.056 nC.

3 0

3 years ago

Which is a characteritic of a physical change?

blagie [28]

Some thing that is done physically like with a persons hands and such

4 0

3 years ago

3B with worked out equations please

bekas [8.4K]

Answer:

b) 1500N

c) to the right

Explanation:

Resultant force= 2000 -500= 1500N

Resultant force is a combination of 2 or more forces acting on an object.

4 0

3 years ago

A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=

Mrac [35]

Answer:

a) 0.67 rad/sec in the clockwise direction.

b) 98.8% of the kinetic energy is lost.

Explanation:

Let us take clockwise angular speed as +ve

For first cylinder

rotational inertia $I$ = 2.0 kg-m^2

angular speed ω = +5.0 rad/s

For second cylinder

rotational inertia $I$ = 1.0 kg-m^2

angular speed = -8.0 rad/s

The rotational momentum of a rotating body is given as = $I$ ω

where $I$ is the rotational inertia

ω is the angular speed

The rotational momenta of the cylinders are:

for first cylinder = $I$ ω = 2.0 x 5.0 = 10 kg-m^2 rad/s

for second cylinder = $I$ ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s

The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s

When they are coupled coupled together, their total rotational inertia $I_{t}$ = 1.0 + 2.0 = 3 kg-m^2

Their final angular rotational momentum after coupling = $I_{t}$ $w_{f}$

where $I_{t}$ is their total rotational inertia

$w_{f}$ = their final angular speed together

Final angular momentum = 3 x $w_{f}$ = 3 $w_{f}$

According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum

this means that

2 = 3 $w_{f}$

$w_{f}$ = final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s

From the first statement, the direction is clockwise

b) Rotational kinetic energy = $\frac{1}{2} Iw^{2}$

where $I$ is the rotational inertia

$w$ is the angular speed

The kinetic energy of the cylinders are:

for first cylinder = $\frac{1}{2} Iw^{2}$ = $\frac{1}{2}*2*5^{2}$ = 25 J

for second cylinder = $\frac{1}{2}*1*8^{2}$ = 32 J

Total initial energy of the system = 25 + 32 = 57 J

The final kinetic energy of the cylinders after coupling = $\frac{1}{2}I_{t}w^{2} _{f}$

where

where $I_{t}$ is the total rotational inertia of the cylinders

$w_{f}$ is final total angular speed of the coupled cylinders

Final kinetic energy = $\frac{1}{2}*3*0.67^{2}$ = 0.67 J

kinetic energy lost = 57 - 0.67 = 56.33 J

percentage = 56.33/57 x 100% = 98.8%

6 0

3 years ago

A helicopter descends from a height of 600 m with uniform negative acceleration, reaching the ground at rest in 5.00 minutes. de

uranmaximum [27]

Given:
h = 600 m, the height of descent
t = 5 min = 5*60 = 300 s, the time of descent.

Let a = the acceleration of descent., m/s².
Let u = initial velocity of descent, m/s.
Let t = time of descent, s.
The final velocity is v = 0 m/s because the helicopter comes to rest on the ground.
Note that u, v, and a are measured as positive upward.

Then
u + at = v
(u m/s) + (a m/s²)*(t s) = 0
u = - at
u = - 300a (1)

Also,
u*t + (1/2)at² = -h
(um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600
ut + (1/2)at² = 600 (2)
From (1), obtain
-300a +(1/2)(a)(90000) = -600
44700a = -600
a = - 1.3423 x 10⁻² m/s²

From (1), obtain
u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s

Answer:
The acceleration is 0.0134 m/s² downward.
The initial velocity is 4.0 m/s upward.

3 0

3 years ago