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makkiz [27]
3 years ago
5

A ball is thrown downward with an initial speed of 6m/s. the ball's velocity after 4 seconds is m/s. (g=-9.8m/s^2)

Physics
1 answer:
7nadin3 [17]3 years ago
6 0

Answer:

8

Explanation:

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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0
3 years ago
In Example 2.7, we investigated a jet landing on an aircraft carrier. In a later maneuver, the jet comes in for a landing on sol
igor_vitrenko [27]

Answer:

a) 20 seconds

b) No.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s (given in the question)

v = Final velocity = 0 (because the jet will stop at the end)

s = Displacement of the jet (Distance between the moment the jet touches the ground to the point the point it stops)

a = Acceleration = -5.00 m/s² (slowing down, so it is negative)

a) Equation of motion

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

s=ut+\frac{1}{2}at^2\\\Rightarrow s=100\times 20+\frac{1}{2}\times -5\times 20^2\\\Rightarrow s=1000\ m

The distance it requires for the jet to stop is 1000 m so in a small tropical island airport where the runway is 0.800 km long the plane would not be able to land. The runway needs to be atleast 1000 m long here the runway on the island is 1000-800 = 200 m short.

5 0
3 years ago
Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e
NeTakaya

We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

Q=\frac{E}{Nr^2}

       

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the  <em>magnitude </em>of charge moved, Q   is mathematically given as

Q=\frac{E}{Nr^2}

Therefore

An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

 Q=\frac{E}{Nr^2}

 

For more information on this visit

brainly.com/question/16517842

3 0
2 years ago
Which of the following has to happen before the eardrum begins to vibrate with the same frequency as the source of the sound wav
777dan777 [17]
I just took the test it is D




I am 100% sure
8 0
3 years ago
Read 2 more answers
What do you think that astronomers mean when they use the term observable universe?
iren2701 [21]
The observable universe consists of galaxies and other matter that can, principally, be seen from Earth because the light signals have had time to reach us. Not everything in the sky is the way it is when we see it, because of the distance the light travels to reach us. 

Hope this helps :)
6 0
3 years ago
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