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Vikentia [17]
3 years ago
9

A robot probe drops a camera off the rim of a 278 m high cliff on Mars, where the free-fall acceleration is 3.7 m/s2 . Find the

velocity with which it hits the ground. Answer in units of m/s.
Physics
1 answer:
FromTheMoon [43]3 years ago
4 0
S = u + at u = 0 278 = 3.7t t = 278/3.7 = 75.135.. v = ut + 0.5at^2 u = 0 v = 0.5 * 3.7 * 75.135^2 = 10,443 m/sec
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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

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5 0
3 years ago
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Answer:

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Explanation:

8 0
3 years ago
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A cow’s mass is 410 kg and a car’s mass is 565 kg. What is the difference between their weights?
solmaris [256]
B. 1520 is the difference between their weights.
5 0
4 years ago
What cell processes occour during interphase
Feliz [49]
The cell cycle has two main phases, interphase and mitosis. Mitosis is the process during which one cell divides into two. Interphase is the time during which preparations for mitosis are made. Interphase itself is made up of three phases -- G1 phase, S phase, and G2 phase -- along with a special phase called G0.
7 0
3 years ago
A piece of amber is charged by rubbing with a piece of fur. If the net excess charge on the fur is 8.9 nC ( 8.9 10-9 C), how man
Alika [10]

Answer:

Number of electrons, n=5.56\times 10^{10}

Explanation:

Given that,

Charge on the fur, q=8.9\ nC=8.9\times 10^{-9}\ C

A piece of amber is charged by rubbing with a piece of fur. We need to find the electrons were added to the amber. It can be calculated using the quantization of charge as

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e is the charge on electron

n=\dfrac{q}{e}

n=\dfrac{8.9\times 10^{-9}}{1.6\times 10^{-19}}

n=5.56\times 10^{10}\ electrons

So, n=5.56\times 10^{10}\ electrons number of electrons are added to the amber. Therefore, this is the required solution.

7 0
4 years ago
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