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 =

=

= 300 N/m

The potential energy, EP of the spring is

EP =

the kinetic energy, EK of the spring

EK =

According to conservative energy,

EP = EK

=

=

=

v =

=

= **38.73 m/s**

Answer:
### Final answer:

### Explanation:

### Learn more about speed of the arrow here:

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.

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**.

#SPJ3

g During a collision with a wall, the velocity of a 0.4 KgKg ball changes from 25 m/sm/s towards the vall to 12 m/sm/s away from the wall. If the time the ball was in contact with the call was 0.5 secsec , what was the magnitude of the avarage force applied to the ball

Solve for x–30 = 5(x + 1)

If Jim could drive a Jetson's flying car at a constant speed of 490 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 4.5 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days). unanswered

a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?

Solve for x–30 = 5(x + 1)

If Jim could drive a Jetson's flying car at a constant speed of 490 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 4.5 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days). unanswered

a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?

**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.

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.

Since the road is banked, and 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, for cornering the banked road, the car will not rely only on the frictional force.

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

(b) What is the object's specific heat?

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 = 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

c =

So object's specific heat = 911.11 J/kgK

**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.

**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**

**Answer:**

The force exerted on the roof is

**Explanation:**

From the question we are told that

The speed of the wind is

The area of the roof is

The air density of Boulder is

The atmospheric pressure is

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

Where is the speed of air in the building

is the speed of air outside the building

are the pressure of inside and outside the house

are the height above and below the roof

Now for

The above equation becomes

Since pressure is mathematically represented as

The above equation can be written as

The initial velocity is 0

Substituting value