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3 years ago

_____ is the nonpersonal, paid communication a company places to influence consumer purchases.

Physics

1 answer:

GarryVolchara [31]3 years ago

6 0

Well publicity would be the socially accepted by standard prospect

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At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown strai

zhannawk [14.2K]

Answer:

The center of mass of the two-ball system is 7.05 m above ground.

Explanation:

Motion of 0.50 kg ball:

Initial speed, u = 0 m/s

Time = 2 s

Acceleration = 9.81 m/s²

Initial height = 25 m

Substituting in equation s = ut + 0.5 at²

s = 0 x 2 + 0.5 x 9.81 x 2² = 19.62 m

Height above ground = 25 - 19.62 = 5.38 m

Motion of 0.25 kg ball:

Initial speed, u = 15 m/s

Time = 2 s

Acceleration = -9.81 m/s²

Substituting in equation s = ut + 0.5 at²

s = 15 x 2 - 0.5 x 9.81 x 2² = 10.38 m

Height above ground = 10.38 m

We have equation for center of gravity

$\bar{x}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

m₁ = 0.50 kg

x₁ = 5.38 m

m₂ = 0.25 kg

x₂ = 10.38 m

Substituting

$\bar{x}=\frac{0.50\times 5.38+0.25\times 10.38}{0.50+0.25}=7.05m$

The center of mass of the two-ball system is 7.05 m above ground.

8 0

3 years ago

At what temperature do the fahrenheit and celsius scales give the same reading?

anygoal [31]

At -40.

-40 gives the same reading for Fahrenheit and Celsius scale.

7 0

3 years ago

Um ladrão tenta fugir sozinho carregando em suas mãos uma mala cheia de barras de ouro. A densidade do ouro é igual a 20 g/cm³,

Roman55 [17]

I believe the answer to your question is A

4 0

3 years ago

What momentum does a 70 kg person running 10m/s (a fast sprint) have ?

gogolik [260]

Momentum = (mass) x (speed)

Momentum = (70 kg) x (10 m/s)

Momentum = 700 kg-m/s

8 0

2 years ago

12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be

Nat2105 [25]

Answer:

$81:256$ .

Explanation:

Let $T$ denote the absolute temperature of this object.

Calculate the value of $T$ before and after heating:

$T(\text{before}) = 27 + 273 = 300\; \rm K$ .

$T(\text{after}) = 127 + 273 = 400\; \rm K$ .

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to $T^4$ .

Ratio between the absolute temperature of this object before and after heating:

$\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}$ .

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

$\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}$ .

4 0

2 years ago