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kvv77 [185]
3 years ago
9

Help me please

Physics
1 answer:
insens350 [35]3 years ago
6 0

Answer: a or b can u pls give me brainlest

Explanation:A straight line is a curve with constant slope. Since slope is acceleration on a velocity-time graph, each of the objects represented on this graph is moving with a constant acceleration.

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Two trains travel toward each other on the same track, beginning 100 miles apart. One train travels at 40 miles per hour; the ot
Paladinen [302]

D = distance between th two trains at the start of the motion = 100 miles

V = speed of the faster train towards slower train = 60 mph

v = speed of the slower train towards faster train = 40 mph

t = time taken by the two trains to collide = ?

time taken by the two trains to collide is given as

t = D/(V + v)

t = 100/(60 + 40) = 1 h

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7 0
3 years ago
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Which of the following describes the efficiency of real machines?
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Choice-B is the true one.
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How was the formation of the outer planets affected by their distance from the sun?
djverab [1.8K]

The formation of the outer planets are affected by their distance from the sun with having them to maintain the lighter elements that they are composed of such as the hydrogen and helium, having them far away will also make their planet more cooler as the sun is distant from them.

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3 years ago
3. A cat pushes a 0.25-kg toy with a net force of 8 N. According to Newton's second
jek_recluse [69]
  • Mass=0.25kg
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\\ \sf{:}\!\implies F=ma

\\ \sf{:}\!\implies Acceleration=\dfrac{F}{m}

\\ \sf{:}\!\implies Acceleration=\dfrac{8}{0.25}

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5 0
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A 460 W heating unit is designed to operate with an applied potential difference of 120 V (a) By what percentage will its heat o
dybincka [34]

Answer:

(a) = -0.16%

(b) = smaller

Explanation:

given

power = 460 W

potential difference = 120 V

(a) what percentage will   its heat output drop if the applied potential difference drops to 110 V ?

we know p = \frac{v^2}{R} .....................(i)

we need to find change in power

\Delta P = \frac{\Delta (V^2)}{R}  

\Delta P = \frac{2 V \Delta V}{R}..............(ii)

from equations we get

\frac{\Delta P}{P} =  \frac{2 \Delta V}{V}

\frac{\Delta P}{P} = 2 \frac{110 -120}{120}

\frac{\Delta P}{P} =  -2(\frac{10}{120})

\frac{\Delta P}{P} = - 0.16 %

(b)

if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power

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