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masya89 [10]
3 years ago
8

1. What is an antigen? Give an example​

Physics
2 answers:
jasenka [17]3 years ago
5 0

A toxin or other foreign substance which induces an immune response in the body, especially the production of antibodies.
An example would be a common cold virus which causes the bodies to make antibodies which prevents the person from getting sick.
Marizza181 [45]3 years ago
4 0
A toxin or foreign substance
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When a molecule absorbs an infrared photon it contributes to the ________energy of that molecule.
Tpy6a [65]

Answer: INFRARED ENERGY

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4 years ago
Calculate the induced electric field in a 54-turn coil with a diameter of 19.5 cm that is placed in a spatially uniform magnetic
storchak [24]

Answer:

E = 3.049 N/C

Explanation:

Induced electric field = e/2 *(pi) *r

Induced electric field = e /(3.14 *19.5)  - Eq (1)

e = (pi)r^2*B/t

= 3.14 * (19.5/2*100)^2 * 0.50 T/ 0.1

= 1.867 V

Substituting this value in equation 1, we get –  

E = 1.867 V/(3.14 *19.5/100)  

E = 3.049 N/C

4 0
3 years ago
Which shows one example of physical change and the one example of a chemical change?
atroni [7]

Answer:

D. Freezing water and burning coal

Explanation:

8 0
3 years ago
The density of table sugar is 1.59 g/cm3. What is the volume of 7.85 g of sugar?
Likurg_2 [28]
\frac{7.85g }{1.59 \frac{g}{ cm^{3} } } = 4.937 cm^{3}
4 0
4 years ago
A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how
Montano1993 [528]
<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity V_{o} has two components, because the brick was thrown at an angle \alpha=61\º:

V_{ox}=V_{o}cos\alpha   (1)

V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}  (2)

V_{oy}=V_{o}sin\alpha   (3)

V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}   (4)

As this is a projectile motion, we have two principal equations related:

<h2>In the x-axis: </h2>

X=V_{ox}.t  (5)

Where:

X=10.1m is the distance where the brick landed

t is the time in seconds

If we already know X and V_{ox}, we have to find the time (we will need it for the following equation):

t= \frac{X}{ V_{ox}}  (6)

t=2.42s  (7)

<h2>In the y-axis: </h2>

-y=V_{oy}.t+\frac{1}{2}g.t^{2}   (8)

Where:

y is the height of the building (<u>in this case it has a negative sign because of the reference system we chose)</u>

g=-9.8\frac{m}{s^{2}} is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}   (9)

-y=-10.52m   (10)

Multiplying by -1 each side of the equation:

y=10.52m >>>>This is the height of the building

3 0
3 years ago
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