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sweet-ann [11.9K]
3 years ago
6

Although blood cells are contained within a special liquid called plasma, the cells themselves are___________.

Physics
1 answer:
valentina_108 [34]3 years ago
8 0

Answer:

Solid

Explanation:

The plasma is the liquid part of blood, it is 90% and accounts for 55% of blood volume. It is what red blood cells, white blood cells, and platelets move around in. These cells remain solid within the plasma. I hoped this helped!

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Longitudinal waves transfer energy ___________ to the direction of the wave motion.
viva [34]
Longitudinal waves transfer energy parallel to the direction of the wave motion
4 0
3 years ago
Read 2 more answers
A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju
stira [4]

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90

3 0
3 years ago
A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1
Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain d=2.46\times10^{3}\ m

Height of islandh=1.80\times10^{3}\ m

Distance of the enemy ship from the mountain d'=6.10\times10^{2}\ m

Initial velocity v=2.55\times10^{2}\ m/s

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=2.55\times10^{2}\cos74.9

v_{x}=66.42\ m/s

We need to calculate the vertical component of initial velocity

Using formula of vertical component

v_{y}=v\sin\theta

Put the value into the formula

v_{y}=2.55\times10^{2}\sin74.9

v_{y}=246.19\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{v_{x}}

t=\dfrac{2.46\times10^{3}}{66.42}

t=37.03\ sec

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

H= v_{y}t-\dfrac{1}{2}gt^2

Put the value in the equation

H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2

H=2397.4\ m

We need to calculate the distance close to the peak

Using formula of distance

H'=H-h

Put the value into the formula

H'=2397.4-1800

H'=597.4\ m

Hence, The distance close to the peak is 597.4 m.

6 0
3 years ago
which of the following locations has the largest electric field? A)0.4 cm from a +3.0 uC point charge B)0.2 cm frm a +1.5 uC poi
JulsSmile [24]
<span>Electric field is proportional to q/d^2, where q is the magnitude of the charge and d is the distance. Since all the given units are identical, we can just compare their relative magnitudes without calculating for the exact values.
A) 3/(0.4)^2 = 18.75
B) 1.5/(0.2)^2 = 37.5
C) 6/(0.4)^2 = 37.5
D) 3/(0.2)^2 = 75
Therefore, choice D has the largest electric field of all.
</span>
7 0
2 years ago
HELP PLEASE 15 POINTS !!!
Nookie1986 [14]
B(I)84.0*10^5J
B(ll)1400W
3 0
3 years ago
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