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labwork [276]
3 years ago
9

A boy im50kg at rest on a skateboard is pushed by another boy who exerts a force of 200 N on him. If the first boy's

Physics
1 answer:
sergey [27]3 years ago
5 0

Answer:

Time, t = 2 seconds

Explanation:

Given the following data;

Mass, m = 50 kg

Initial velocity, u = 0 m/s (since it's starting from rest).

Final velocity, v = 8 m/s

Force, F = 200 N

To find the time, we would use the following formula;

F = \frac {m(v - u)}{t}

Making time, t the subject of formula, we have;

t = \frac {m(v - u)}{F}

Substituting into the formula, we have;

t = \frac {50(8 - 0)}{200}

t = \frac {50*(8)}{200}

t = \frac {400}{200}

Time, t = 2 seconds

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It took a crew 9 h 36 min to row 8 km upstream and back again. If the rate of flow of the stream was 2 km/h, what was the rowing
babunello [35]

Answer:

3 km/h

Explanation:

Let's call the rowing speed in still water x, in km/h.

Rowing speed in upstream is: x - 2 km/h

Rowing speed in downstream is: x + 2 km/h

It took a crew 9 h 36 min ( = 9 3/5 = 48/5) to row 8 km upstream and back again. Therefore:

8/(x - 2) + 8/(x + 2) = 48/5      (notice that: time = distance/speed)

Multiplying by x² - 2², which is equivalent to (x-2)*(x+2)

8*(x+2) + 8*(x-2) =  (48/5)*(x² - 4)

Dividing  by 8

(x+2) + (x-2) = (6/5)*(x² - 4)

2*x = (6/5)*x² - 24/5

0 =  (6/5)*x² - 2*x - 24/5

Using quadratic formula

x = \frac{2 \pm \sqrt{(-2)^2 - 4(6/5)(-24/5)}}{2(6/5)}

x = \frac{2 \pm 5.2}{2.4}

x_1 = \frac{2 + 5.2}{2.4}

x_1 = 3

x_2 = \frac{2 - 5.2}{2.4}

x_2 = -1\; 1/3

A negative result has no sense, therefore the rowing speed in still water was 3 km/h

7 0
3 years ago
A light wave moves from diamond (n= 2.4) into water (n= 1.33) at an angleof 24°. what angle does it have in water?
worty [1.4K]

Answer:

n1 sin θ1 = n2 sin θ2      Snell's Law       (θ1 is the angle of incidence)

sin θ2 = n1 / n2 * sin θ1

sin θ2 = 2.4 / 1.33 * sin θ1

sin θ2 = 1.80 * .407 = .734

θ2 = 47.2 deg

3 0
2 years ago
A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the
Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)  

v₀ is the initial voltage on the cap  

v is the voltage after time t  

R is resistance in ohms,  

C is capacitance in farads  

t is time in seconds  

RC = τ = time constant  

τ = 20µ x 1.5k = 30 ms  

v = v₀e^(t/τ)  

0.5 = 1.5e^(t/30ms)  

e^(t/30ms) = 10/3  

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0
3 years ago
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