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uranmaximum [27]

3 years ago

11

In a golf ball game, a person hits the golf ball with a club. The club is in contact with the ball, which is initially at rest,

for 2.45 ms. The ball has a mass of 0.0550 kg and leaves the club with a speed of 1.80 X 102 m/s. What is the average force exerted on the ball by the club

1 answer:

Anon25 [30]3 years ago

6 0

Answer:

F=4040.81 N

Explanation:

Given that

Time ,t= 2.45 ms

Mass ,m= 0.055 kg

v= 1.8 x 10² m/s = 180 m/s

We know that rate of change in the linear momentum is known as force.

Momentum P = m v

$F=\dfrac{dP}{dt}$

Therefore force F

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{0.055\times 180}{2.45\times 10^{-3}}\ N$

F=4040.81 N

Therefore force on the ball will be 4040.81 N

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A spring gun is made by compressing a spring in a tube and then latching the spring at

inn [45]

Answer:

a) $v=13.2171\,m.s^{-1}$

b) $H=8.9605\,m$

Explanation:

Given:

mass of bullet, $m=4.97\times 10^{-3}\,kg$

compression of the spring, $\Delta x=0.0476\,m$

force required for the given compression, $F=9.12 \,N$

(a)

We know

$F=m.a$

where:

a= acceleration

$9.12=4.97\times 10^{-3}\times a$

$a\approx 1835\,m.s^{-2}\\$

we have:

initial velocity, $u=0\,m.s^{-1}$

Using the eq. of motion:

$v^2=u^2+2a.\Delta x$

where:

v= final velocity after the separation of spring with the bullet.

$v^2= 0^2+2\times 1835\times 0.0476$

$v=13.2171\,m.s^{-1}$

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

$u=13.2171\,m.s^{-1}$

∵At maximum height the final velocity will be zero

$v=0\,m.s^{-1}$

Using the equation of motion:

$v^2=u^2-2g.h$

where:

h= height

g= acceleration due to gravity

$0^2=13.2171^2-2\times 9.8\times h$

$h=8.9129\,m$

is the height from the release position of the spring.

So, the height from the latched position be:

$H=h+\Delta x$

$H=8.9129+0.0476$

$H=8.9605\,m$

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3 years ago

Transfer of thermal or heat energy through "touch"

MaRussiya [10]

It is conduction. Rhejrjrjejehrhrhvrbrgr

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3 years ago

Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of

EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

$E_{n}$ = k e² / 2a₀ (1 /n²)

ao = h'² / k m e² h' = h/2πi

For another atom with a single electron in the last layer

a₀ ’= h’² / k m (Ze)²

a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

$E_{n}$ = - Z² Eo/n²

E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

$E_{n}$ - $E_{m}$ = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

c = λ f

E = h f

E = h c /λ

h c / λ = Z² ΔE

λ = 1 / Z² (hc / ΔE)

λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

$E_{n}$ = - 9 Eo / n²

40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

40.5 10-9 = 1/9 λ_hydrogen

tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

40.5 10⁻⁹ = 1/9 (16 $\lambda_{Be}$ )

tex]\lambda_{Be}[/tex] = 40.5 9/16

tex]\lambda_{Be}[/tex] = 22.78 nm

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3 years ago

Hello everyone. This is a question about Dimensional Analysis and I came across this question but I am unable to wrap my head ar

omeli [17]

Answer:

2. [B] = [L]/[T] and [C] = [L]/[T]

Explanation:

I assume you mean this:

A = B² + 2B⁴/C²

Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.

B² = [L]²/[T]²

B = [L]/[T]

B⁴/C² = [L]²/[T]²

C²/B⁴ = [T]²/[L]²

C² = B⁴ [T]²/[L]²

C² = ([L]/[T])⁴ [T]²/[L]²

C² = [L]²/[T]²

C = [L]/[T]

Notice we ignore the 2 coefficient, which is unitless.

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Water and h20 delivery have been a very meaningful company to the community

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