the kinetic energy of a bowling ball is 25 (kg-m^2/sec^2). if the mass is 2 kg, the what is the speed of the bowling ball

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441

Tasya [4]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
$g= \frac{4 \pi^2}{T^2}L$ (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
$f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz$
And its period is the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s$

So now we can use eq.(1) to find the gravitational acceleration of the planet:
$g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2$

3 0

3 years ago

What does the diagram represent?

solong [7]

Answer:

Atoms in a element on the periodic table I dont know which one tho so try google that sorry I am not much of a help

Explanation:

4 0

3 years ago

A body with the inertial

Andrews [41]

Answer:

Explanation:

Hi there,

To get started, recall the kinematic equations from either a textbook, equation sheet, etc. Kinematic equations are used when acceleration is <em>constant,</em> as stated in the prompt.

Best way to use kinematic equations is to see which variable you are looking for, then which variable is unknown to you and is not needed for that equation.

a) average velocity

Takes the form of:

$v_a_v_g=\frac{d_t_o_t_a_l}{t}=\frac{v+v_0}{2}$ this is the literal definition of average velocity; initial plus final divided by 2.

We know total displacement and total time elapsed, so we will use the middle form of the equation:

$v_a_v_g=\frac{1640m}{40s}=41 \ m/s$

b) the final velocity

We can still use the average velocity formula, as the other two equations that include final velocity have acceleration variable which is unknown as of now.

Solve for final velocity:

$v=(2v_a_v_g)-v_o = 2(41 \ m/s) - (8 m/s) = 74 m/s\\$ this makes sense, since a velocity later in time is higher than a velocity earlier in time. It is increasing with increasing time because of acceleration.

c) the acceleration

There are two equations that can be used to solve this, but we will use the less time-consuming one, but both produce same answer:

$a = \frac{v-v_0}{t_t_o_t_a_l} = \frac{(74-8)m/s}{40s} =1.65 m/s^{2}$

Notice, change in velocity over change in time, and acceleration is constant. When acceleration is constant, it models a linear function, and acc. is just slope!

Study well and persevere. If you liked this solution, hit Thanks or give a rating!

thanks,

3 0

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

6 0

3 years ago

Two pans of water are on different burners of a stove. One pan of water is boiling vigorously, while the other is boiling gently

vichka [17]

Answer:

The correct answer is C.

Explanation:

3 0

3 years ago