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mojhsa [17]
3 years ago
6

A motorcycle cop, parked at the side of a highway reading a magazine, is passed by a woman in a red Ferrari 308 GTS doing 120.0

km/h. After a few attempts to get his cycle started, the officer roars off 2.00 s later. At what average rate must he accelerate if 159 km/h is his top speed and he is to catch her just at the state line 3.04 km away?
Physics
1 answer:
jeyben [28]3 years ago
3 0

Answer:

XZXDHF

Explanation:ZDHF

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You walk east for 10m then turn north 15m what is your displacementm
Doss [256]
25m north east hope this helps
5 0
3 years ago
A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many
jekas [21]

The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration

Complete question:

A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?

Answer:

the total units of insulin admistered each morning

= 22 units of qam and humulin

Explanation:

given

44 units and Humnlin N

with concentration 50/100 = 1/2 = 0.5

∴ 44 × 0.5 ≈ 22 units in the morning

regular insulin administered each day

(22 + 35)units of qam and humulin

= 57units

5 0
3 years ago
HELP
astraxan [27]

Answer:

Ultraviolet light has shorter wavelengths than infrared light. The frequency of ultraviolet light is greater than that of infrared light. Ultraviolet light is higher in energy than infrared light.

7 0
3 years ago
Which way does wind blow,
ruslelena [56]
A. Horizontally - The wind blows parallel to the horizon.
7 0
3 years ago
A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.
victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A)  = N₀(B)

N = N₀ e^{-\lambda t}

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ e^{-\lambda_1 t}

For B

N(B) = N(B)₀ e^{-\lambda_2 t}

N(A) = N(B) , for t = 2 h

N(A)₀ e^{-\lambda_1 t} = N(B)₀ e^{-\lambda_2 t}

N(A)₀ e^{-\lambda_1 t} = 5 x N₀(A)  e^{-\lambda_2 t}

e^{-\lambda_1 t} = 5  e^{-\lambda_2 t}

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

half life = .693 / λ

For A

.77 =  .693 / λ₁

λ₁ = .9 h⁻¹

e^{\lambda_2 t} = 5  e^{\lambda_1 t}

Putting t = 2 h , λ₁ = .9 h⁻¹

e^{\lambda_2\times  2} = 5  e^{.9\times  2}

e^{\lambda_2\times  2} = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

6 0
3 years ago
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