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Pie
3 years ago
11

The speed of a boat in still water is 20 mph. it travels from one pier to another with the current in 4 hours and back against t

he current in 6 hours and 40 minutes. if the distance between piers is 100 miles, what is the speed of the current
Physics
1 answer:
saw5 [17]3 years ago
3 0
The answer is, "the speed of the current is 5 miles per hour."

To calculate the speed of the current,
let's assume speed of current =  xmph. Time taken to travel from one pier to another with the current = 100/(20+x)h


But the time taken to travel from one pier to another with the current, which is given is = 4 hours. So,  4=100/(20+x) 80+4x = 100

4x = 20

x = 5 Thus, the speed of the current is 5 miles per hour.
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Communicate is 25 a multiple of 2 or 5 ? how do you know?
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A body initially at 100°C cools to 60°C in minutes and to 40°C. The temperature of body at the end of 15 minutes will be​
zhuklara [117]

The question is incomplete, the complete question is;

A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?

Answer:

See explanation

Explanation:

From Newton's law of cooling;

θ1 - θ2/t = K(θ1 + θ2/2 - θo]

Where;

θ1 and θ2 are initial and final temperatures

θo is the temperature of the surroundings

K is the constant

t is the time taken

Hence;

100 - 60/5 = K(100 + 60/2 - θo)

100 - 40/10 = K(100 + 40/2 - θo)

8= (80 - θo)K -----(1)

6= (70 - θo)K -----(2)

Diving (1) by (2)

8/6 = (80 - θo)/(70 - θo)

8(70 - θo) = 6(80 - θo)

560 - 8θo = 480 - θo

560 - 480 = -θo + 8θo

80 = 7θo

θo = 11.4°

Again from Newton's law of cooling;

θ = θo + Ce^-kt

Where;

t= 0, θ = 60° and θo = 11.4°

60 = 11.4 + C e^-K(0)

60 - 11.4 = C

C= 48.6°

To obtain K

40 = 11.4 + 48.6e^-10k

40 -11.4 = 48.6e^-10k

28.6/48.6 = e^-10k

0.5585 = e^-10k

-10k = ln0.5585

k= ln0.5585/-10

K= 0.0583

Hence, the temperature in 15 minutes;

θ= 11.4 + 48.6e^(-0.0583 × 15)

θ= 31.7°

4 0
3 years ago
a proton travelling along is x-axis is lowed by a niform electric field E. at x = 20.0 cm, the proton has a speed of 3.5x10^6 m/
anzhelika [568]

Answer:

Magnitude of electric field is 1.06 x 10^5 V/m along negative X-direction

Explanation:

Given: initial velocity of proton = u = 3.5 x 10^6 m/s

final velocity of proton = v = 0 m/s

initial point l_i = 0.2 m and final point is l_f = 0.8 m

According to conservation of energy:

change in in kinetic energy = change in potential energy of proton

⇒\frac{m}{2}(v^2-u^2 ) = qE(l_i - l_f)

where q and m is the charge and mass of proton E is the electric field , l_i and l_f is the initial and final position of proton

on substituting the respected values we get,

1.023 x 10^-^1^4 = 9.6 x 10^-^2^0 x E

⇒ E = 1.06 x 10^5 V/m

external force is opposite to the motion as velocity of proton decreases with distance.

Therefore, magnitude of electric field is 1.06 x 10^5 V/m along negative X-direction

5 0
3 years ago
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