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andriy
1 month ago
7

A paper in the journal Current Biology tells of some jellyfish-like animals that attack their prey by launching stinging cells i

n one of the animal kingdom's fastest movements. High-speed photography showed the cells were accelerated from rest for 700 ns at 5.30 ✕ 107 m/s2. Calculate the maximum speed reached by the cells and the distance traveled during the acceleration.
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
3 0
Given that, the starting speed of the cells is 0 since they were at rest. The cell's acceleration is specified, along with time t = 700 ns. We aim to calculate the peak speed achieved by the cells and the distance covered during the acceleration. Let v signify the final velocity. Let d represent the distance traversed. We'll apply the equations of motion to find the solution.
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The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Sav [3153]

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

           v represents the object's velocity

<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">

The KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

</psubstituting>
6 0
2 months ago
A 20.00-kg lead sphere is hanging from a hook by a thin wire 2.80 m long and is free to swing in a complete circle. Suddenly it
Keith_Richards [3271]
Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v = \sqrt{rg} = \sqrt{2.80 \times 9.8} = 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as

P.E_{highest} = mgh

, where h signifies the diameter of the circle (2r). Hence, the expression will be written as P.E_{highest} = mg(2r)

where u is the velocity at the lowest point. Consequently, the modified equation is

= \sqrt{v^{2} + 4gr}

= \sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}

= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v = \frac{(m_{1} + m_{2})u}{m_{2}}

= \frac{(20 + 5) \times 11.71}{5}

= \frac{292.75}{5}

= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.

3 0
1 month ago
Ability of the muscles to function effectively and efficiently without undue fatigue
Keith_Richards [3271]

Response:

Physical well-being

Clarification:

7 0
2 months ago
A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
inna [3103]

Answer:

At this position, the magnetic field equals ZERO

Explanation:

The magnetic field produced by a moving charge is described as

B = \frac{\mu_0 qv sin\theta}{4\pi r^2}

Here, we determine the direction of the magnetic field using

\hat B = \hat v \times \hat r

Thus, we find

\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

Leading to a magnetic field of ZERO

Consequently, when the charge moves in the same line as the given position vector, the magnetic field will be nonexistent

3 0
1 month ago
An astronaut stands by the rim of a crater on the Moon, where the acceleration of gravity is 1.62 m/s2 and there is no air. To d
Keith_Richards [3271]

Answer:

12.1 seconds

Explanation:

t = time duration

u = initial speed

v = final speed = 0

s = distance = 120 m

a = lunar gravity acceleration = 1.67 m/s²

Motion equation

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -1.67\times 120-0^2\\\Rightarrow u=\sqrt{2\times 1.67\times 120}\\\Rightarrow u=20.02\ m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-20.2}{-1.67}\\\Rightarrow t=12.1\ s

The rock takes 12.1 seconds to reach the bottom of the crater.

5 0
2 months ago
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