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german
3 years ago
6

What do all the items that light up a bulb have in common? (help)

Physics
1 answer:
kipiarov [429]3 years ago
8 0

Answer:

Light bulbs started off with elements made of carbon, but over the years inventors added new elements such as tungsten, mercury, chlorine and europium to their toolkits

Explanation:

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A star is located at a distance of about 100 million light years from Earth. An astronomer plans to measure the distance of the
nlexa [21]
<span>d. The parallaxes beyond a few thousand light years are
too small to be measured with common instruments.

I'm not sure that parallax can even be used out to a few
thousand light years.

The NEAREST star to Earth has the BIGGEST parallax.
The star is Alpha Centauri.  It's only 4 light years away
from us, and its parallax is  0.000206 of a degree !
I have no idea how astronomers can measure angles
so small ... and that's the BIGGEST parallax angle of
ANY star.</span>
5 0
3 years ago
A baseball is hit from a height of 4 feet above the ground with an initial velocity of 110 feet per second and at an angle of 35
mrs_skeptik [129]

Answer:

Part a)

h = 62.2 ft

Part b)

It will not able to cross the pole

Explanation:

As we know that ball is hit with speed of 110 ft/s at an angle of 35 degree

so here we will say

v_y = 110 sin35

v_x = 110 cos35

now at the maximum height the vertical velocity will become zero

so here we can use kinematics

v_f^2 - v_y^2 = 2 a h

here we have

a = -32 ft/s^2

v_f = 0

v_y = 63.1 ft/s

now we have

0 - 63.1^2 = 2(-32)h

h = 62.2 ft

Part b)

now the height of ball is related to the distance from point of projection is given as

y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}

now we know that

x = 375 ft

y = 375(tan35) - \frac{(32)(375^2)}{2(110^2)(cos35)^2}

y = 262.6 - 277.12 = -14.5 ft

since its coming negative so it will not able to cross the pole

3 0
3 years ago
When a speeding truck hits a stationary car, the car is deformed and heat is generated. What can you say about the kinetic energ
guapka [62]
I can say that the kinetic energy after the collision is less than it was before the collision.  I can say this with confidence because you've said that some energy was used to deform the car, plus there was energy lost from the system in the form of heat.
4 0
3 years ago
You slide a chair across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 156 N dir
Luda [366]

Answer:

626.612 J

Explanation:

Work done by friction on the chair is given as

W-ΔEk = W'..................... Equation 1

Where W' = Work done by friction on the chair, W = Work done on the chair by me, Ek = change in Kinetic energy of the chair as a result of the slide.

From the question,

W = FdcosФ.............. Equation 2

ΔEk = 1/2m(v²-u²)................ Equation 3

Where F = Force applied on the chair, d = distance of slide, Ф = angle between the force and the horizontal, m = mass of the chair, v = final velocity of the chair, u = initial velocity of the chair

Substitute equation 2 and equation 3 into equation 1

W' = FdcosФ-1/2m(v²-u²)........................ Equation 4

Given: F = 156, d = 5 m, Ф = 26°, m = 18.8 kg, v = 3.1 m/s, u = 1.3 m/s

Substitute into equation 4

W' = 156×5×cos26°-1/2×18(3.1²-1.3²)

W' = 701.06-74.448

W' = 626.612 J.

Hence the work done by friction = 626.612 J

8 0
3 years ago
A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with
Ivenika [448]

Complete Question

Complete Question is attached below

Answer:

q=1.558*10^{-9}c

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength E_l=784.75N/m

Right field strength E_r=776.38 N/m

Front field strength E_f=725.5 N/m

Back field strength E_b=749.54 N/m

Top field strength E_t=944.95 N/m

Bottom field strength E_{bo}=1082.58 N/m

Generally, the equation for  Charge flux is mathematically given by

\phi=EAcos\theta

Where

Theta for Right,Left,Front and Back are at an angle 90

cos 90=0

Therefore

\phi =0 with respect to Right,Left,Front and Back

Generally, the equation for  Charge Flux is mathematically also given by

\phi=\frac{q}{e_o}

Where

Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2

Therefore

Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t

Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)

Q_{net}=176N/C m^2

Giving

q=\phi*e_0

q=176N/C m^2*1.558*10^{-12}c

q=1.558*10^{-9}c

4 0
3 years ago
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