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OlgaM077 [116]
2 years ago
8

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50.

cm and holds it in position with a force of 150 N. If the mass of the arrow is 50. g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?
Physics
1 answer:
Nady [450]2 years ago
8 0

Answer:

v = 38.73 m/s

Explanation:

Given

Extension of the bow, x = 50 cm = 0.5 m

Force of the arrow, F = 150 N

Mass of the arrow, m = 50 g = 0.05 kg

speed of arrow, v = ? m/s

We start by finding the spring constant

Remember, F = kx, so

k = F/x

k = 150 / 0.5

k = 300 N/m

the potential energy if the bow when pulled back is

E = 1/2kx²

E = 1/2 * 300 * 0.5²

E = 0.5 * 300 * 0.25

E = 37.5 J

The speed of the arrow will now be found by using the law of conservation of energy

1/2kx² = 1/2mv²

kx² = mv²

v² = kx²/m, on substituting, we have

v² = (300 * 0.5²) / 0.05

v² = 75 / 0.05

v² = 1500

v = √1500

v = 38.73 m/s

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Crank

Answer:

33.33 %

Explanation:

given,

error last year = 63

error this year = 42

percent decrease in the error = ?

to find the percentage difference in the error formula used is

   = \dfrac{difference}{original}\times 100

   = \dfrac{63-42}{63}\times 100

   = \dfrac{21}{63}\times 100

   = 33.33 %

Percentage decrease in the number of error is equal to 33.33%.

7 0
2 years ago
At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
dalvyx [7]

Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}

Part(b):

Also the angular displacement (\Delta \theta) can be written as

&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad

8 0
2 years ago
To exercise, a man attaches a 4.0 kg weight to the heel of his foot. When his leg is stretched out before him, what is the torqu
Masja [62]

Answer:

B. τ = 16 Nm

Explanation:

In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:

τ = Fd

here,

τ = Torque = ?

F = Force exerted by the weight = Weight = mg

F = mg = (4 kg)(10 m/s²) = 40 N

d = distance from knee to weight = 40 cm = 0.4 m

Therefore,

τ = (40 N)(0.4 m)

<u>B. τ = 16 Nm</u>

8 0
2 years ago
Imagine you derive the following expression by analyzing the physics of a particular system: a=gsinθ−μkgcosθ, where g=9.80meter/
BARSIC [14]

Answer:

Explanation:

Analysis of structure gives

a=gsinθ−μkgcosθ

Notice that all the expression are right but we want to know of we can simplify the expression further.

We want to analyse if we can still further simplify the expression,

Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.

So option a is wrong because the expression can be simplified further to

a=g(sinθ−μkcosθ)

Option b is right and the best option.

Since we are given that, g=9.8m/s²

We can as well substitute that to option a

So we will have

a=9.8metre/second²(sinθ−μkcosθ)

Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first

So it will have been the best option if it was written as

a=9.8metre/second²(sinθ−μkcosθ)

So the best option is B.

8 0
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pashok25 [27]

Answer:

U for the reaction is 15,048.58 kJ/mol of pentane.

Explanation:

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Moles of pentane = mass/MW

Mass = 0.1523 g

MW of pentane (C5H12) = 72 g/mol

Moles of pentane = 0.1523/72 = 0.00212 mol

U for the reaction = quantity of heat required ÷ moles of pentane = 31.903 kJ ÷ 0.00212 mol = 15,048.58 kJ/mol

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