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2 years ago

.A 0.6 kg basketball is thrown straight up at a velocity of 4 m/s. What is the initial kinetic energy of the basketball?

Physics

1 answer:

motikmotik2 years ago

7 0

Answer:

the initial kinetic energy of the basket ball is 4.8 J

Explanation:

The computation of the initial kinetic energy of the basket ball is shown below:

As we know that

Initial kinetic energy is

$= \frac{1}{2}mv^2 \\\\= \frac{1}{2}\times 0.6 \times (4m/s)^2\\\\= \frac{1}{2}\times 0.6 \times 16m/s^2\\\\= 4.8 J$

Hence, the initial kinetic energy of the basket ball is 4.8 J

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What type of light does this light bulb produce most (i.e. at what wavelength does the spectrum have maximum intensity)?

lesantik [10]

Answer: The light bulb produces the continuous light. At minimum wavelength the spectrum have maximum intensity.

Explanation:

According to Wein's displacement law, the wavelength is inversely proportional to the temperature.

The intensity depends on the frequency. The frequency is inversely proportional to the wavelength.

Therefore, when the temperature of the light bulb will be maximum then the wavelength will be minimum. At minimum wavelength the spectrum have maximum intensity.

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2 years ago

Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air

Nata [24]

The frequency of a sound is whatever frequency leaves the source. It doesn't change.

Voiced of swimmers at the pool don't change frequency in or out of the water. Only their speed and wavelength change.

5 0

2 years ago

Read 2 more answers

A neutron at rest decays (breaks up) to a proton and an electron. Energy is released in the decay and appears as kinetic energy

SashulF [63]

Answer:

$5.444\times 10^{-4}$

Explanation:

The momentum of the neutron before and after the decay is the same since there's no external force.

$P_{sys}=const\\\\P=mv\\\\K=0.5mv^2$

#The neutron is initially at rest, so after the decay:

$P_A+P_B=0\\\\P_A=-P_B$

#After decay, the proton has +ve direction with a velocity $v_A$ while the electron moves in a negative direction with a velocity $v_B$

Therefore:

$P_A=m_Av_A, P_B=m_Bv_B\\\\\therefore m_Av_A,=m_Bv_B$

Let the energy released during the decay be Q:

$Q=K_{tot}=K_A+K_B\\\\Q=K_A+0.5m_Bv_B^2\\\\Q=K_A+0.5m_B(\frac{m_A}{m_B})^2v_A^2\\\\\ But \ K_A=0.5m_Av_A^2\\\\\therefore Q=K_A+\frac{m_A}{m_B}K_A=K_A(1+\frac{m_A}{m_B})\\\\=Q=\frac{m_A+m_B}{m_B}K_A\\\\m_A=1836m_B\\\\\frac{K_A}{Q}=\frac{m_B}{1836m_B+m_B}=\frac{1}{1837}\\\\\frac{K_A}{Q}=5.444\times10^{-4}$

Hence,Kp/Ktot is 5.444x10^(-4)

4 1

2 years ago

How many significant figures are in 0.0069

posledela

Two significant figures, the 6 and the 9

7 0

2 years ago

A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a

aleksley [76]

Answer:

T₂ =602 °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

$\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\$

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602 °C

5 0

2 years ago