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3 years ago

How does the elbow medical and lateral epicondylitis?

1 answer:

4 0

Lateral epicondylitis, or “tennis elbow,” is an inflammation of the tendons that join the forearm muscles on the outside of the elbow. The bony bump on the outside (lateral side) of the elbow is called the lateral epicondyle. The ECRB muscle and tendon is usually involved in tennis elbow. 
Medial epicondylitis, or “golfer’s elbow,” is an inflammation of the tendons that attach your forearm muscles to the inside of the bone at your elbow. It's identified by pain from the elbow to the wrist on the inside (medial side) of the elbow. The pain is caused by damage to the tendons that bend the wrist toward the palm.

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Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m. What is the force (in N) of

Lostsunrise [7]

Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = $\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}$

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = $\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}$

(|r₂₁|)² = 148.25

$F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }$

= 0.050938(0.19107i + 0.54933j) N

= (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

7 0

3 years ago

A bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes.

zaharov [31]

Given that,

Frequency emitted by the bat, f = 47.6 kHz

The speed off sound in air, v = 413 m/s

We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{413}{47.6\times 10^3}\\\\\lambda=0.00867\ m$

$\lambda=8.67\ mm$

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.

3 0

3 years ago

How much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate

Nuetrik [128]

Answer:

<h3>2479 Newton</h3>

Solution,

Mass=100 kg

Acceleration due to gravity(g)=24.79 m/s^2

Now,.

 $weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton$ 

hope this helps ..

Good luck on your assignment..

5 0

3 years ago

1 Give reasons:

lorasvet [3.4K]

Answer:

a) The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. ... It allows hydrophobic molecules and small polar molecules diffuse through the lipid layer, but does not allow ions and large polar molecules cannot diffuse through the membrane

b) Plastids are present in the cells of plants. They are characterised by the presence of pigments. ... Chloroplasts contain chlorophyll and carotenoid pigments responsible for capturing the light energy that is necessary for photosynthesis. The chloroplasts are therefore known as the kitchen of the cell.

c) Lysosomes are known as the suicidal bag of the cell because it is capable of destroying its own cell in which it is present. It contains many hydrolytic enzymes which are responsible for the destruction process. This happens when either the cell is aged or gets infected by foreign agents like any bacteria or virus.

d) Mitochondria are often called the “powerhouses” or “energy factories” of a cell because they are responsible for making adenosine triphosphate (ATP), the cell's main energy-carrying molecule. ... In mitochondria, this process uses oxygen and produces carbon dioxide as a waste product.

e) In Hydra, the cells are arranged in two germinal layers—outer ectoderm and inner endoderm. Between these two layers is a layer of undifferentiated cells called mesoglea. Such kind of pattern of embryonic layers is seen in diploblastic animals. Hence, Hydra is a diploblastic animal.

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4 0

2 years ago

A car accelerates at a rate of 13m/s^2[S]. If the car's initial velocity is 120km/h[N]. What will its final velocity be in m/s,

Delvig [45]

Answer:

the final velocity of the car is 59.33 m/s [N]

Explanation:

Given;

acceleration of the car, a = 13 m/s²

initial velocity of the car, u = 120 km/h = 33.33 m/s

duration of the car motion, t = 2 s

The final velocity of the car in the same direction is calculated as follows;

v = u + at

where;

v is the final velocity of the car

v = 33.33 + (13 x 2)

v = 59.33 m/s [N]

Therefore, the final velocity of the car is 59.33 m/s [N]

6 0

3 years ago