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zalisa [80]
2 years ago
5

Some tropical butterflies can reach speeds of up to 11 m/s. Suppose a butterfly flies at a speed of 6.0m/s while another flying

insect some distance ahead flies in the same direction with a constant speed. The butterfly then increases its speed at a constant rate of 1.4m/s2 and catches up to the other insect 3.0s later. How far the butterfly travel during the race
Physics
1 answer:
ladessa [460]2 years ago
7 0

Answer:

24.3 m

Explanation:

Using the equation of motion

S=ut+0.5at^{2} where s is the distance, u is initial velocity, t is time and a is acceleration

Substituting u for 6 m/s, t for 3 s, a for 1.4 m/s2 we obtain

S=6*3+(0.5*1.4*3^{2}=18+6.3=24.3 m


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Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car
juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

a = \frac{-F}{m}

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity  because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion



5 0
3 years ago
Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the
BlackZzzverrR [31]

relation between potential difference and electric field is given as

E . d = \Delta V

so here we know that

d = 3 cm

\Delta V = 30 V

E \times 0.03 = 30

E = 1000 N/C

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0
3 years ago
FILL IN THE BLANKS!
Aleksandr [31]
1. First blank is A. Conductors
Second blank is D. Insulators

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8 0
2 years ago
 A beach ball moving with the speed of 1.27 m/s rows of a pier and hits the water 0.75 m from the end the pier how high is the
Vera_Pavlovna [14]

Answer:

75.8

Explanation:

because just divide 1.27 into 0.75 and there's your answer

8 0
2 years ago
Four capacitors with capacitance 3.0 pF, 2.0 pF, 5.0 pF and X pF are connected in series to each other. If the equivalent capaci
bagirrra123 [75]

Answer:

<h2>A. 6pF</h2>

Explanation:

If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

\frac{1}{C_t} = \frac{1}{C_1} +\frac{1}{C_2} +\frac{1}{C_3} +\frac{1}{C_4} \\

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

\frac{1}{0.83} = \frac{1}{3.0} +\frac{1}{2.0} +\frac{1}{5.0} +\frac{1}{C_4} \\\\\\1.205 = 0.333+0.5+0.2+\frac{1}{C_4} \\\\1.205 = 1.033 + \frac{1}{C_4} \\\\\frac{1}{C_4}  = 1.205-1.033\\\\\frac{1}{C_4}  = 0.172\\\\C_4 = \frac{1}{0.172}\\ \\C_4 = 5.8pF\\\\

C₄ ≈ 6pF

Hence the value of the X capacitor is approximately 6pF

8 0
3 years ago
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