Answer:

**Answer:**

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

At the time of collision velocity of ball one is descending.

**Explanation:**

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

v² = u² + 2as

v² = 44.50² + 2 x (-9.81) x 71.32

v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

v = u + at

24.10 = 44.50 - 9.81 t , t = 2.07 s

-24.10 = 44.50 - 9.81 t , t = 6.99 s

Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

s = ut + 0.5 at²

71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s

Answer:

**Answer:**

**v2=139 ft**

**Explanation:**

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=**2.57 s, 7.44 s**

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, **when the first ball is descending**. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->**v2=139 ft**

The acrylic plastic rod is 20 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Eₚ = 2.70 GPa, vₚ = 0.4.

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.

What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

A force of 40 N is applied in a direction perpendicular to the end of a 9 m long bar that pivots about its other end. Find the torque that this force produces about the pivot point. magnitude

an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and the push lasts 0.8 seconds. The skater's mass is 55 kg. what is the skater's speed after she stops pushing on the wall

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its speed is decreased linearly from 60 mph to 30 mph in 10 seconds. Calculate the theoretical maximum energy in kWh that can be recovered during this interval. Ignore all losses.

What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

A force of 40 N is applied in a direction perpendicular to the end of a 9 m long bar that pivots about its other end. Find the torque that this force produces about the pivot point. magnitude

an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and the push lasts 0.8 seconds. The skater's mass is 55 kg. what is the skater's speed after she stops pushing on the wall

**Answer:**

the correct option is C

**Explanation:**

The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.

Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.

When we see the possibilities we see that the correct option is C

The **wavelength **of the **light **passing through the slit is **214** nm.

The **wavelength **is the **distance **between identical **points **in the adjacent cycles of a **waveform**.

Given that the **separation **between two slits **d **is **3.00 × 10^-5 **m and the **distance **from the **slit **to screen **r **is **2 **m. The **distance **from the **central** spot to fringe **s **is **10.0 **m and the **bright **bands of the spectrum **m **are **7 **for the seventh bright fringe.

The **wavelength **of the **light **passing through the slit is calculated as given below.

Hence we can conclude that the **wavelength **of the **light **passing through the slit is **214** nm.

To know more about the **wavelength**, follow the link given below.

To solve this problem, the concepts related to destructive and constructive Interference of light spot and dark spot are necessary.

By definition in the principle of superposition, light interference is defined as

Where,

d = Separation of the two slits

R = Distance from slit to screen

m= Any integer, which represents the repetition of the spectrum. The order of m equal to 1,2,3,4,5 represent bright bands and the order of m equal to 1.5,2.5,3.5 represent the dark bands.

Y = Distance from central spot to fringe.

Re-arrange the equation to find \lambda we have that

Our values are gives according the problem as,

m = 7 (The seventh bright fringe)

R = 2m

Therefore the wavelength of the light passing through the slits is 214nm

**Answer:**

a)

b)

**Explanation:**

Given:

- mass of the body,

- mass of the tyre,

- length of hanging of tyre,

- distance run by the body,

- acceleration of the body,

**(a)**

Using the equation of motion :

..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

putting the values in eq. (1)

Now, the momentum of the body just before the jump onto the tyre will be:

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

**(b)**

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

above the normal hanging position.

**Answer:**

**π*R²*E**

**Explanation:**

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒**Flux = E*π*R²**

**Answer:**

**ratio of the piccolo's length to the flute's length is 0.4916**

**Explanation:**

given data

frequency of piccolo = 522.5 Hz

frequency of flute = 256.9 Hz

to find out

ratio of the piccolo's length to the flute's length

solution

we get here length of tube that is express as

length of tube = velocity of sound ÷ fundamental frequency .......................1

so here ratio of Piccolo length to flute that is

= 0.4916

**so ratio of the piccolo's length to the flute's length is 0.4916**

**Answer:**

a=-2.77 m/s^2

**Explanation:**

Assuming constant acceleration,

v=at + v_0

where v_0 is the initial velocity.

At rest, v=0, so

0=at+v_0

So solving the equation for a:

a=(-v_0)/t

Inserting the numbers yields

a=-2.77 m/s^2