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 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

Answers

Answer 1
Answer:

Answer:

The speed of the arrow immediately after it leaves the bow is 38.73 m/s

Explanation:

given information:

force,  F = 150 N

x = 50 cm = 0.5 m

mass of arrow, m = 50 g = 0.05 kg

We start from the force of the spring

F = kx

k = (F)/(x)

  = (150)/(0.5)

  = 300 N/m

The potential energy, EP of the spring is

EP = (1)/(2) kx^(2)

the kinetic energy, EK of the spring

EK = (1)/(2) mv^(2)

According to conservative energy,

EP = EK

(1)/(2) kx^(2) = (1)/(2) mv^(2)

kx^(2) = mv^(2)

v^(2) = (kx^(2) )/(m)

v = x\sqrt{(k)/(m) }

  = 0.5\sqrt{(300)/(0.05) }

  = 38.73 m/s

Answer 2
Answer:

Final answer:

Using Hooke's Law, we can determine the speed of the arrow. The speed of the arrow immediately after it leaves the bow is approximately 38.7 m/s.

Explanation:

In this problem, we can use Hooke's Law to determine the speed of the arrow. Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring:

F = -kx

Where F is the force, x is the displacement, and k is the spring constant.

In this case, the force exerted by the bow on the arrow is acting like a spring force. The force of the bow is 150N, and the displacement is 50cm (which is equivalent to 0.5m). So we can set up the equation as:

150N = -k * 0.5m

Now we can solve for k:

k = -150N / 0.5m = -300 N/m

Now that we have the spring constant, we can use it to find the potential energy stored in the bow:

PE = 0.5kx^2 = 0.5*(-300N/m)*(0.5m)^2 = 37.5 J

Next, we can use the conservation of energy to find the kinetic energy of the arrow right after it leaves the bow. The potential energy stored in the bow is converted into kinetic energy:

KE = PE = 37.5 J

The kinetic energy is given by the equation:

KE = 0.5mv^2

Where m is the mass of the arrow and v is its velocity. Rearranging the equation, we can solve for v:

v = sqrt(2KE/m) = sqrt(2*37.5 J / 0.05 kg) = sqrt(1500) m/s ≈ 38.7 m/s.

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Answers

Answer:

Explanation:

a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final =  1/2 I₂ ω₂²

= 1/2 (I₁ / 2)  (2ω)²

=  I₁ ω²

c )

Change in rotational kinetic energy

=  I₁ ω² -  1/2 I₁ ω²

=  +  1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force

Answers

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

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Therefore, the correct answer is option C - the frictional force.

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.(a) What is the heat capacity of this object?
(b) What is the object's specific heat?

Answers

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT  = 15 ^0C = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

  So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

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A commercial aircraft is flying westbound east of the Sierra Nevada Mountains in California. The pilot observes billow clouds near the same altitude as the aircraft to the south, and immediately turns on the "fasten seat belt" sign. Explain why the aircraft experiences an abrupt loss of 500 meters of altitude a short time later.

Answers

Answer:

Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.

Explanation:

Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.

What is the magnitude of the force you must exert on the rope in order to accelerate upward at 1.4 m/s2 , assuming your inertia is 63 kg ? Express your answer with the appropriate units.

Answers

Answer:

The magnitude of the force you must exert on the rope in order to accelerate upward is 705.6 N

Explanation:

The magnitude of force, you must exert can be estimated as follows;

Since it is upward motion, we must consider acceleration due to gravity which opposes the upward motion.

F = m(a+g)

where;

F is the magnitude of the upward force

m is your mass, which is the measure of inertia = 63kg

a is the acceleration of the rope = 1.4 m/s²

F = 63(1.4 + 9.8)

F = 63(11.2)

F = 705.6 N

Therefore, the magnitude of the force you must exert on the rope in order to accelerate upward is 705.6 N

Answer:

705.6 N

Explanation:

Force: This can be defined as the product of mass a acceleration.

The S.I unit of force is Newton.

The expression for the force on the rope in order to accelerate upward is given as,

F-W = ma .......................... Equation 1

Where F = Force exerted on the rope, W = weight of the rope, m = mass of the rope, a = acceleration.

But,

W = mg........................ Equation 2

Where g = acceleration due to gravity

substitute equation 2 into equation 1

F-mg = ma

F = ma+mg

F = m(a+g).............. Equation 3

Given:  m = 63 kg, a = 1.4 m/s²

Constant: g = 9.8 m/s²

Substitute into equation 3

F = 63(1.4+9.8)

F = 63(11.2)

F = 705.6 N

The magnitude of the force exerted on the rope = 705.6 N

Every few years, winds in Boulder, Colorado, attain sustained speeds of 45.0 m/s (about 100 mi/h) when the jet stream descends during early spring. show answer No Attempt Approximately what is the force due to the Bernoulli effect on a roof having an area of 205 m2? Typical air density in Boulder is 1.14 kg/m3 , and the corresponding atmospheric pressure is 8.89 × 104 N/m2 . (Bernoulli’s principle assumes a laminar flow. Using the principle here produces only an approximate result, because there is significant turbulence.)

Answers

Answer:

The force exerted on the roof is F =2.37*10^(5)N

Explanation:

From the question we are told that

      The speed of the wind is v = 45.0 m/s

       The area of the roof is A = 205 m^2

       The air density of Boulder is \rho = 1.14 kg / m^3

        The atmospheric pressure is P_(atm) = 8.89 * 10^(4) N/ m^2

For a laminar flow the Bernoulli’s principle is  mathematically represented as

            P_1 + (1)/(2) \rho v_a ^2 + \rho g h_a = P_2 + (1)/(2)  \rho v_b ^2 + \rho h_b

Where  v_1 is the  speed of air in  the building

             v_b is the speed of air outside the building

             P_1 \ and \ P_2 are the pressure of inside and outside the house

             h_a \ and \ h_b are the height above and  below the roof

Now for  h_a = h_b

            The above equation becomes

                 P_1 + (1)/(2) \rho v_a ^2 = P_2 + (1)/(2) \rho v_b ^2

                 P_1 - P_2 = (1)/(2) \rho (v_b^2 - v_a^2)

Since pressure is mathematically represented as

           P = (F)/(A )

The above equation can be written as

             F  = (1)/(2) \rho ( v_b^2 - v_a ^2 ) A

The initial velocity is 0

    Substituting value  

                F = (1)/(2)  (1.14) [(45^2 - 0^2 ) ](205)

                F =2.37*10^(5)N