There are several actions that can be used to increase the action force of a canoe that is moving through water. One can increase the rate of striking or strike harder than before with a bigger paddle. Using a lighter canoe can also be helpful for increasing the speed. I hope the answer comes to your help.

4 0

3 years ago

Learning Goal: To practice Problem-Solving Strategy 7.2 Problems Using Mechanical Energy II. The Great Sandini is a 60.0-kg circ

andreev551 [17]

Answer:

v = 15.8 m/s

Explanation:

Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force

$W_{fr}$ = ΔEm = $Em_{f}$ -Em₀

Let's write the mechanical energy at each point

Initial

Em₀ = Ke = ½ k x²

Final

$Em_{f}$ = K + U = ½ m v² + mg y

Let's use Hooke's law to find compression

F = - k x

x = -F / k

x = 4400/1100

x = - 4 m

Let's write the energy equation

fr d = ½ m v² + mgy - ½ k x²

Let's clear the speed

v² = (fr d + ½ kx² - mg y) 2 / m

v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0

v² = (160 + 8800 - 1470) / 30

v = √ (229.66)

v = 15.8 m/s

5 0

3 years ago

Read 2 more answers

Iodine 131 half life is 8.0 days. Ten percent of the original sample o his isotope remains after (a) 22.7 days (b) 24.9 days (c)

Artyom0805 [142]

Answer:

option (c) is correct

Explanation:

Half life of a substance is the time in which the element becomes half of is initial value.

half life, T = 8 days

Amount remaining, N = 10 % of original value

Let the original value is No.

N = 10% of No

N = 0.1 No

Let the time taken is t and the decay constant is λ.

The relation between the decay constant and the half life is given by

$\lambda =\frac{0.6931}{T}=\frac{0.6931}{8}=0.08664 per day$

Us the equation of radioactivity

$N=N_{0}e^{-\lambda t}$

$0.1N_{0}=N_{0}e^{-0.08664 t}$

$e^{0.08664 t}=10$

Taking natural log on both the sides, we get

0.08664 t = 2.303

t = 26.6 days

4 0

3 years ago

Se dispone de dos vasos con agua a 20 grados y queremos calentarlo hasta alcanzar 50 grados. Si el primero contiene 0,5l y el se

BigorU [14]

Answer:

The cup with 0.5L

Explanation:

To know what amount of water you take into account the specific heat of the water. The specific heat of water is:

$c_{water}=4186\frac{J}{kg\°C}$

Thus, 4186 J of energy are needed to icrease the temperature of 1 kg water in 1°C. Then, more grams of water will need more energy.

You have that one cup has 0.5 L and the other one has 750mL = 0.75L

The second cup of water will need more heat because the amount of water contained in the second cup is greater than in the first cup with 0.5L

4 0

2 years ago

What is electrical energy​

What is electrical energy