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Oduvanchick [21]
3 years ago
10

Which ball has the greater average speed during the 1-s interval after release (assuming neither hits the ground during that tim

e)? View Available Hint(s) Which ball has the greater average speed during the 1-s interval after release (assuming neither hits the ground during that time)? the ball thrown upward the ball thrown downward Neither; the average speeds of both balls are the same.
Physics
1 answer:
notka56 [123]3 years ago
3 0

Answer:

The ball thrown downward

Explanation:

When the ball is thrown vertically, the acceleration of it is the gravity acceleration independent if it is thrown downward or upward. However, the acceleration is a vector, so, when the ball is thrown upward, the movement is against the gravity, so the acceleration is negative, and so, the velocity decreases during time; and when the ball is thrown downward, the movement goes to the gravity, so the acceleration is positive, so the velocity increase after time passes.

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Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² %  = 0,036 %

4 0
3 years ago
If a machine will do 50 J of work in 35 seconds how much power did it produce?
grandymaker [24]
Power is the amount of work done over a period of time. If you will put that into an equation, the formula of power will be:

P = W/t
Where:
P = power
W=work
t = time

Your problem already provides you with work and time so all you need to do is divide:

P = W/t
P = 50J/30s
P = 1.67 W
6 0
3 years ago
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0
3 years ago
What is true about high-level nuclear waste?
fenix001 [56]

the answer is b. it's commercially reprocessed

3 0
3 years ago
Read 2 more answers
As the scattering angle of the photon increases, what happens to the wavelength associated with the photon?
frez [133]

As the scattering angle of the photon increases, the wavelength associated with the photon increases.

<h3><u>Explanation:</u></h3>

The particle with quantum mechanical property is known as Compton wavelength. The wavelength of a photon increases during collision. When the scattering angle of the photon is 0 degree then the photon's wavelength increases by 0 and when the scattering angle  is 180 degree then the wavelength of  the photon will become double. This is known as Compton wavelength.

When a photon undergoes collision process, the photo loses its energy and this energy is transferred to the electrons. This causes energy of the photon to decrease and thus the frequency also decreases. Thus, the wavelength of the photon will increase.

6 0
3 years ago
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