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vitfil [10]
3 years ago
11

Suppose a flagellum is rotating at a constant rps (revolution per second). What is the angular displacement of the flagellum in

seconds?
Physics
1 answer:
IgorLugansk [536]3 years ago
5 0

Answer:

1794.47 radian.

Explanation:

Suppose a flagellum is rotating at a constant 952 rps (revolution per second). What is the angular displacement of the flagellum in 0.3seconds? (report your answer in SI units)

Let us suppose that a flagellum is rotating at a constant 952 revolution per second.

Firstly, we will convert 952 revolution per second to radian per second as follows :

952 revolution per second = 2π × 952 rad/s

=5981.592 rad/s

The angular displacement is given by the expression as follows :

\theta=\omega t\\\\\theta=5981.592\times 0.3\\\\\theta=1794.47\ \text{rad}

So, the angular displacement of the flagellum in 0.3 seconds is 1794.47 radian.

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The velocity of blood flow is ________. A) in direct proportion to the total cross-sectional area of the blood vessels B) slower
Leto [7]

Answer:

Option (D)

Explanation:

The velocity at which blood flows in the blood vessels is inversely proportional to the total cross-sectional area of the blood vessels present in the body. This means that if the cross sectional area of the vessels low, then there will be high rate of blood flow, and vice versa. This blood flow is minimum in the case of capillaries, where it gets enough time for the exchanging of essential nutrients as well as gases.

Thus, the correct answer is option (D).

5 0
3 years ago
Before the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 5J. After the exper
finlep [7]

Answer:

Option (b) is correct.

Explanation:

Elastic collision is defined as a collision where the kinetic energy of the system remains same. Both linear momentum and kinetic energy are conserved in case of an elastic collision.

Inelastic collision is defined as a collision where kinetic energy of the system is not conserved whereas the linear momentum is conserved. This loss of kinetic energy may due to the conversion to thermal energy or sound energy or may be due to the deformation of the materials colliding with each other.

As given in the problem, before the collision, total momentum of the system is 2.5~Kg~m~s^{-1} and the kinetic energy is 5~J. After the collision, the total momentum of the system is  2.5~Kg~m~s^{-1}, but the kinetic energy is reduced to 4~J. So some amount of kinetic energy is lost during the collision.

Therefor the situation describes an inelastic collision (and it could NOT be elastic).

5 0
3 years ago
Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,
makkiz [27]

Answer: \theta=cos^{-1}0.991=7.69^o

The following vectors have been given: \vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}

The angle between these two vectors can be found by:

cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\
||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}

\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54

||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}

cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o

7 0
4 years ago
explain how many minimum number of geostationary satellites are required for global coverage of T.V transmission
kobusy [5.1K]

Answer:I honestly don't know

Explanation:

4 0
4 years ago
The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ν0. find the minimum en
shusha [124]
The energy carried by the incident light is
E=hf
where h is the Planck constant and f is the frequency of the light. The threshold frequency is the frequency that corresponds to the minimum energy needed to eject the electrons from the metal, so if we substitute the threshold frequency in the formula, we get the minimum energy the light must have to eject the electrons:
E=hf=(6.63 \cdot 10^{-34}Js)(5.64 \cdot 10^{14}s^{-1})=3.74 \cdot 10^{-19}J
4 0
4 years ago
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