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Phoenix [80]
2 years ago
9

7.

Physics
1 answer:
vodomira [7]2 years ago
8 0

Answer:

the true about n i b i n is they are Commissioner

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Jenny pushes a 40 N crate down the hall 2m. How much work did she do?
Shalnov [3]
Work done = force x distance = 40 x 2 = 80 Joules.
8 0
2 years ago
What is the amount of charge when 13.5a is flowing for 2 1/2 hours
abruzzese [7]

Answer:

Quick Answer:

a.  8.9Ω

b.  1.2×104 C

Explanation:

4 0
3 years ago
Read 2 more answers
The bohr shift on the oxygen-hemoglobin dissociation curve is produced by changes in
Alisiya [41]

Answer:

THE BOHR SHIFT ON THE OXYGEN-HEMOGLOBIN DISSOCIATION CURVE IS PRODUCED BY CHANGES IN THE CONCENTRATION OF CARBON IV OXIDE.

Explanation:

The oxygen-hemoglobin dissociation curve shows the relationship between the saturated hemoglobin concentration and oxygen. It shows how the blood hold on to and releases oxygen. The Bohr shift can occur as a result of changes in concentration of carbon iv oxide and other factors such as acidity or pH, 2,3-bisphosphoglycerate, exercise, also temperature of the body. These factors contributes to the right or left shift on the curve. Carbon iv oxide prevents the binding of oxygen to the hemoglobin. The is because hemoglobin has the same binding site for both oxygen and carbon iv oxide. Carbon iv oxide increase also leads to a change in the pH of the blood through the formation of bicarbonate ion. Bicarbonate ion formation causes reduced acidity and therefore lead a shift in the dissociation curve for more of the carbon iv oxide to be excreted as hemoglobin's affinity for oxygen reduces. And when the concentration of carbon iv oxide is low in the plasma, acidity increases and this provides more affinity for oxygen by the hemoglobin.

7 0
3 years ago
The spring constant for the spring in above Table is 20 N/m.
satela [25.4K]
The answer is 0.025J.

W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J
5 0
3 years ago
94. The diagram shows the orbit of a satellite
Valentin [98]

Answer:

7.65x10^3 m/s

Explanation:

The computation of the satellite's orbital speed is shown below:

Given that

Earth mass, M_e = 5.97 × 10^24 kg

Gravitational constant, G = 6.67 × 10^-11 N·m^2/kg

Orbital radius, r = 6.80 × 10^6m

Based on the above information

the satellite's orbital speed is

V_o = √GM_e ÷ √r

= √6.67 × 10^-11 × 5.97 × 10^24 ÷ √6.80 × 10^6

=  7.65x10^3 m/s

4 0
3 years ago
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