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2 years ago

How many cranial nerves do we have

1 answer:

8 0

12.

The twelve cranial nerves are: olfactory nerve, optic nerve, oculomotor nerve, trochlear nerve, trigeminal nerve, abducens nerve, facial nerve, vestibulocochlear nerve, glossopharengeal nerve, vagus nerve, spinal accessory nerve, and hypoglossal nerve.
:)

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One way to monitor global warming is to measure the average temperature of the ocean. Researchers are doing this by measuring th

Marat540 [252]

Answer:

0.07°C

Explanation:

solution:

the speed of a sound in water is:

v(T)=1480+4(T-4°C)

at 4°C the travel time is:

t(4◦C) = ( 7600 × 103 m ) / (1480 m/s) = 5202.7 s

5°C, the travel time is:

t(5◦C) = ( 7600 × 103 m ) / (1484 m/s) = 5188.7 s

one degree C corresponds to a ∆t of 14 s so temperature difference is:

ΔT=1 s/14 s=0.07◦C

5 0

2 years ago

Jorge hits a drum as hard as he can, and it makes a loud sound. Then he hits the same drum with less strength, and it makes a qu

Leviafan [203]

Answer:

when he uses more force to hit the drum then its makes a louder sound so the more force you hit the drum with then the louder it is.

8 0

3 years ago

Name and explain two sleeping disorders

vfiekz [6]

Insomnia and night terrors

4 0

3 years ago

A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,

Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a) Kinetic energy of block = potential energy in spring

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg. Spring constant k is unknown, but you can find it from given data:

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.

From the energy equation above, solve for v,

v = v √(k/m)

= 0.15 √(300/1)

= 2.598 m/s.

b) Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.

This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0

3 years ago

A 13561 N car traveling at 51.1 km/h rounds

Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

$a_{c} = \frac{v^{2}}{r}$

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

3 0

3 years ago