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4 years ago

Please help!! (Picture attached)

Physics

1 answer:

KonstantinChe [14]4 years ago

3 0

Well it is definitely answer B because when light is on for a long time it heats up a lot.the thermometer obviously went up which means the light bulb had more energy and was hotter than the start of it

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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you.Your reaction time

shtirl [24]

Consider the motion of the car before brakes are applied:

v₀ = maximum initial velocity of the car before the brakes are applied

t = reaction time = 0.50 s

x₀ = distance traveled by the car before brakes are applied

since car moves at constant speed before brakes are applied

Using the equation

x₀ = v₀ t

x₀ = v₀ (0.50)

Consider the motion after brakes are applied :

v₀ = initial velocity of the car before the brakes are applied

a = acceleration = - 10 m/s²

v = final velocity of the car after it comes to stop = 0 m/s

x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)

Using the equation

v² = v²₀ + 2 a x

inserting the values

0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))

v²₀ = 20 (38 - v₀ (0.50))

v₀ = 23 m/s

3 0

3 years ago

What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?

Tcecarenko [31]

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy, $U = 13.0 J$

The volume, $V = 6.00 mm^3 = 6.00*10^{-9}m^3$

The potential energy per unit volume is defined as the energy density.

$u = \frac{U}{V}$

$u= \frac{(13.0 J)}{(6.00*10^{-9} m^3)}$

$u= 2.167109 J/m^3$

The energy density related with electric field is given by

$u = \frac{1}{2} \epsilon_0 E^2$

Here, the permitivity of the free space is

$\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2$

Therefore, rerranging to find the electric field strength we have,

$E = \sqrt{\frac{2u}{\epsilon_0}}$

$E = \sqrt{\frac{2(2.167109)}{8.85*10^{-12}}}$

$E = 2.211010 V/m$

Therefore the electric field is 2.21V/m

8 0

3 years ago

Please Help!!!!

Len [333]

Answer:

v = 0.92 c

Explanation:

Here, we will use the time dilation formula from Einstein's theory of relativity to find the speed of traveling of the friend:

$t =\frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}} \\\\\\\sqrt{1-\frac{v^2}{c^2}}=\frac{t_o}{t}\\\\$

where,

v = speed of traveling = ?

c = speed of light

t = time of return = 10 years

t₀ = time passed on earth = 4 years

Therefore,

$\sqrt{1-\frac{v^2}{c^2}} = \frac{4\ years}{10\ years}\\\\ 1-\frac{v^2}{c^2}=(\frac{2}{5})^2\\\\\frac{v^2}{c^2} = 1-\frac{4}{25}\\\\\frac{v^2}{c^2} = \frac{21}{25}\\\\v^2 = 0.84c^2\\\\$

8 0

3 years ago

If earth had no atmosphere, would a falling object ever reach terminal velocity?

stepan [7]

No, because terminal velocity is when the acceleration of the Earth’s gravity is balanced by the air resistance of the atmosphere.

5 0

3 years ago

Represent 7468 N with SI units having an appropriate prefix. Express your answer to four significant figures and include the app

Musya8 [376]

Answer:

7.468 kN

Explanation:

Here the force is given in Newton

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

Tera = 10¹²

The number is 7468.0

Here, the only solution where the number of significant figures is kilo. If any other prefix is chosen then the significant figures will increase.

1 kilonewton = 1000 Newton

$1\ Newton=\frac{1}{1000}\ kilonewton$

$\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton$

So, 7468 N = 7.468 kN

7 0

3 years ago