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2 years ago

Light always travels in a straight line

2 answers:

8 0

Light travels in straight lines. Once a light has been produced, it will keep moving in a straight line until it hits something else. Shadows are evidence of light traveling in straight lines. An object blocks light so that it can’t reach the surface where we see the shadow.

mrs_skeptik [129]2 years ago

5 0

Answer:

No light actually travels in waves

Explanation:

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Which object had more potential energy when it was lifted to a distance of 1000 centimeters? Show your calculation.

RSB [31]

Explanation:

We Know That

POTENTIAL ENERGY= MASS*g*HEIGHT

When the objects are lifted to same height then the object with heavier mass would have the highest potential energy

5 0

2 years ago

Why is it that the two possible faults with inductors are short circuit and open circuit?

lukranit [14]

These are the most common type of faults not just inductors but also with other elements too like resistors,transformers, generators etc.
open circuit fault means the flow of current is disrupted some how in the circuit and the circuit stops operating. and for short circuit fault the current in the system will be pretty high and this short circuit current or fault current will always run back to the fault location, if the inductor got short circuited somehow then the fault current will only run through it because it will then provide a very low impedence path

5 0

3 years ago

A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga

tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₂² v₂

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{2.25^2}{3.75^2} v_2$

$v_1= 0.36 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)$

$32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)$

$v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}$

$v_2=\sqrt{16.084}$

v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)² x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0

3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

2 years ago

What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earths surface?

joja [24]

Answer:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

= ${\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2}$ = 3984378 m / $s^{2}$

Explanation:

The force of gravity on a 700 kg satellite if its 10 km above Earth's surface is given by

= ${\frac{(6.674\times 10^{-11}N. m^2/kg^2)(5.97\times 10^{24}kg) }{(10\times10^3)^2}$ = 3984378 m / $s^{2}$

3 0

2 years ago