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kari74 [83]
3 years ago
12

Dribbling a basketball harder into the floor makes it bounce higher is an example of what newton's law?

Chemistry
1 answer:
vodomira [7]3 years ago
8 0
Newton’s third law, because a person(a) is acting upon the ball(b) by dribbling the ball on the floor
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What is the half-life of a pharmaceutical if the initial dose is 500 mg and only 31 mg remains after 6 hours?
Sergeu [11.5K]

Answer:

\large \boxed{\text{b. 1.5 h}}

Explanation:

1. Calculate the rate constant

The integrated rate law for first order decay is

\ln \left (\dfrac{A_{0}}{A_{t}}\right ) = kt

where

A₀ and A_t are the amounts at t = 0 and t

k is the rate constant

\begin{array}{rcl}\ln \left (\dfrac{500}{31}\right) & = & k \times 6\\\\\ln 16.1 & = & 6k\\2.78& =& 6k\\k & = & \dfrac{2.78}{6}\\\\& = & 0.463 \text{ h}^{-1}\\\end{array}

2. Calculate the half-life

t_{\frac{1}{2}} = \dfrac{\ln2}{k} = \dfrac{\ln2}{\text{0.463  h}^{-1}} = \textbf{1.5 h}\\\\ \text{The half-life is $\large \boxed{\textbf{1.5 h}}$}

4 0
4 years ago
Calculate the formula/molecular mass of following
ddd [48]

Answer:

Refer to the period table. Use the atomic mass for calculation. Multiply the atomic mass with the number of atoms. For example, Al2(SO4)3. (27 × 2) + (32 x 3) + (16 x 12) = molecular mass

8 0
3 years ago
If a car speeds up from 80km/hr to 110km/hr to pass another vehicle in 1.5 min (0.025hrs), what is their acceleration?
Xelga [282]

Answer: uhhhh im pretty sure the answer is 69m/hr

Explanation:

8 0
3 years ago
Light used to produce food in plants is chemical or physical change?
Nitella [24]
This is Chemical change
4 0
3 years ago
370 cm3 of water at 80°C is mixed with 120 cm3 of water at 4°C. Calculate the final equilibrium temperature, assuming no heat is
NikAS [45]

Answer : The final temperature of the mixture is 61.4^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)

where,

c_1 = c_2 = specific heat of water = same

m_1 = m_2 = mass of water  =  same

\rho_1 = \rho_2 = density of water = 1.0 g/mL

V_1 = volume of water at 80.0^oC  = 370cm^3=370mL

V_2 = volume of water at 4^oC  = 120cm^3=120mL

T_f = final temperature of mixture = ?

T_1 = initial temperature of water = 80.0^oC

T_2 = initial temperature of water = 4^oC

Now put all the given values in the above formula, we get:

(\rho_1\times V_1)\times (T_f-T_1)=-(\rho_2\times V_2)\times (T_f-T_2)

(1.0g/mL\times 370mL)\times (T_f-80.0)^oC=-(1.0g/mL\times 120mL)\times (T_f-4)^oC

T_f=61.4^oC

Therefore, the final temperature of the mixture is 61.4^oC

6 0
4 years ago
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