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Trava [24]
2 years ago
8

An archer pulls her bowstring back 0.410 m by exerting a force that increases uniformly from zero to 270 N. How much work is don

e in pulling the bow?
Physics
1 answer:
Nitella [24]2 years ago
6 0

Answer:

<em>110.7Joules</em>

Explanation:

<em>Work is said to be done when the force applied to a body cause the body to move through a distance.</em> Mathematically:

Work done = Force * Distance

Given the following

Force = 270 -0  = 270N

Distance moved = 0.410m

Required

The work done

Substitute the given parameters into the formula

Workdone = 270 * 0.41

Workdone = 110.7Joules

<em>Hence the work done in pulling the bow is 110.7Joules</em>

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A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an
kupik [55]

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

  • <em>mass of the car, m = 1000 kg</em>
  • <em>speed of the car, v = 50 km/h = 13.89 m/s</em>
  • <em>banking angle, θ = 20⁰</em>

The normal force on the car due to banking curve is calculated as follows;

Fcos(\theta) = mg

The horizontal force on the car due to the banking curve is calculated as follows;

Fsin(\theta) = \frac{mv^2}{r}

<em>Divide </em><em>the second equation by the first;</em>

\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m

Thus, the radius of the curved road at the given condition is 54.1 m.

Learn more about banking angle here: brainly.com/question/8169892

3 0
2 years ago
As an astronaunt travels from the surface of the earth to a postion that is four times
allochka39001 [22]

Answer:

Explanation:

4 0
3 years ago
Read 2 more answers
Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a
In-s [12.5K]

Answer:

T = 1.108\,s

Explanation:

The period of a physical pendulum is:

T = \sqrt{\frac{I_{O}}{m\cdot g \cdot L} }

T=2\cdot \pi \sqrt{\frac{\frac{1}{3}\cdot m \cdot L^{2} }{m\cdot g\cdot L} }

T=2\cdot \pi \sqrt{\frac{L }{3\cdot g} }

The length of the leg is approximately the height of the person:

L = 0.915\,m

The period is:

T = 2\cdot \pi \sqrt{\frac{0.915\,m}{3\cdot (9.807\,\frac{m}{s^{2}} )} }

T = 1.108\,s

4 0
3 years ago
Someone please help me!
kirill [66]

I think it’s the first one

4 0
3 years ago
an object moving at 10. km/hr has a kinetic energy of 10. J. what is the kinetic energy of the same object if it is moving at 20
Schach [20]
Kinetic energy is related to velocity by:
KE = (1/2)mv^2

solve for mass m
10 = (1/2)m(10)^2
10 = (1/2)m(100)
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10/50 = m
1/5 = m

at 20 km/hr

KE = (1/2)(1/5)(20)^2
KE = (1/10)(400)
KE = 40 J
5 0
3 years ago
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