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3 years ago

7

A certain sports car can accelerate from 0 m/s to 27.7 m/sec with an acceleration of 11.2 m/s². How long does this acceleration

take in seconds?

1 answer:

Otrada [13]3 years ago

8 0

Answer:

Δt = 2.47 s

Explanation:

The problem involves just the formula of the acceleration:

$a = \frac{Δv}{Δt}$

$Δv = (27.7 - 0) \: m/s$

You want to know Δt, knowing the time and the speed. So:

$Δt = Δv/a$

Δt = 2.47 s

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By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.

Rashid [163]

Answer:

λ = 102.78 nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

E = - 13.606 / n²

where 13.606 eV is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

ΔE = 13.606 (1 / $n_{f}^{2}$ - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

DE = 13.606 (1/1 - 1/3²)

DE = 12.094 eV

let's reduce the energy to the SI system

DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

E = h f

f = E / h

f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

c = λ f

λ = c / f

λ = 3 10⁸ / 2.9186 10¹⁵

λ = 1.0278 10⁻⁷ m

let's reduce to nm

λ = 102.78 nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

5 0

3 years ago

A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).

Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c. The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω² + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh = 1/2Iω² + 1/2mv² = 1/2Iω² + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω² + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω² + 1/2mr²ω²

mgh = mr²ω²/5 + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0

2 years ago

20 points+ Brainliest

Helen [10]

Answer:

The answer is B. hope this helps

Explanation:

3 0

3 years ago

ILL GIVE BRAINLY THING

Bogdan [553]

Answer:

About 3 trips

Explanation: if we do 2.5m*1.6m*0.75 it equals to 11000 then we divide that to 11m3 and it gives you 3.6 so it will be about 3 times

Thx

4 0

3 years ago

Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o

o-na [289]

Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

$\displaystyle V_1=\frac{Q_1}{C_1}$

$\displaystyle V_2=\frac{Q_2}{C_2}$

They are both the same after connecting them, thus

$\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}$

Or, equivalently

$\displaystyle Q_2=\frac{C_2Q_1}{C_1}$

The total charge of both capacitors is

$\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)$

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

$Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C$

Now we compute Q1 from the equation above

$\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C$

The final voltage of any of the capacitors is

$\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V$

7 0

3 years ago