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Genrish500 [490]
3 years ago
6

World class swimmers can swim the 100 meter in about 55 seconds what is their approximate average speed

Physics
2 answers:
Mice21 [21]3 years ago
3 0

About 1.8181 meters per second

goldfiish [28.3K]3 years ago
3 0

Answer:

v = 1.82 m/s

Explanation:

As we know that average speed is defined as the total distance covered in total interval of time.

So here given that

total distance = 100 m

total time = 55 s

so here from above formula

v_{avg} = \frac{d}{t}

v_{avg} = \frac{100}{55}

v_{avg} = 1.82 m/s

so the average speed of the swimmer will be 1.82 m/s

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What are the tree types of thermal transfer
avanturin [10]

Convection, conduction, and radiation.

4 0
3 years ago
Read 2 more answers
An electron that has a velocity with x component 2.4 x 106 m/s and y component 3.6 x 106 m/s moves through a uniform magnetic fi
likoan [24]

Answer:

(a) 7.315 x 10^(-14) N

(b) - 7.315 x 10^(-14) N

Explanation:

As you referred at the final remark, the electron and proton undergo a magnetic force of same magnitude but opposite direction. Using the definition of magnetic force,  a cross product must be done. One technique is either calculate the magnitude of the velocity and magnetic field and multiplying by sin (90°), but it is necessary to assure both vectors are perpendicular between each other ( which is not the case) or do directly the cross product dealing with a determinant (which is the most convenient approach), thus,

(a) The electron has a velocity defined as:  \overrightarrow{v}=(2.4x10^{6} i + 3.6x10^{6} j) \frac{[m]}{[s]}\\\\

In respect to the magnetic field; \overrightarrow{B}=(0.027 i - 0.15 j) [T]

The magnetic force can be written as;

\overrightarrow{F} = q(\overrightarrow{v} x \overrightarrow{B})\\ \\\\\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]

Bear in mind q =-1.6021x10^{-19} [C]  

thus,

\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= -1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=7.3152x10^{-14} [N]

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, k^{2}=k\cdot k = 1

(b) Considering the proton charge has the same magnitude as electron does, but the sign is positive, thus

\overrightarrow{F}= q \left[\begin{array}{ccc}i&j&k\\2.4x10^{6}&3.6x10^{6}&0\\0.027&-0.15&0\end{array}\right]\\\\\\\overrightarrow{F}= q(2.4x10^{6}* (-0.15)- (0.027*3.6x10^{6}))\\\\\\\overrightarrow{F}= 1.6021x10^{-19} [C](-457200) [T]\frac{m}{s}\\\\\overrightarrow{F}=(-7.3152x10^{-14}) k [\frac{N*m/s}{C*m/s}]\\\\|F|= \sqrt{ (-7.3152x10^{-14})^{2}[N]^2 *k^{2}}\\\\F=-7.3152x10^{-14} [N]

Note: The cross product is operated as a determinant. Likewise, the product of the unit vector k is squared and that is operated as dot product whose value is equal to one, i.e, k^{2}=k\cdot k = 1

Final remarks: The cross product was performed in R3 due to the geometrical conditions of the problem.  

6 0
3 years ago
What is the energy of an electromagnetic wave that has a frequency of
sukhopar [10]

Answer:

Energy, \; E = 2.6504 * 10^{-34} \; Joules

Explanation:

Given the following data;

Frequency = 4.0 x 10⁹ Hz

Planck's constant, h = 6.626 x 10-34 J·s.

To find the energy of the electromagnetic wave;

Mathematically, the energy of an electromagnetic wave is given by the formula;

E = hf

Where;

E is the energy possessed by a wave.

h represents Planck's constant.

f is the frequency of a wave.

Substituting the values into the formula, we have;

Energy, \; E = 4.0 x 10^{9} * 6.626 x 10^{-34}

Energy, \; E = 2.6504 * 10^{-34} \; Joules

8 0
2 years ago
The plates of a spherical capacitor have radii 6.25 cm and 15.0 crn. The space between the two spheres is filled with a material
aleksklad [387]

Answer:

Capacitance is 0.572×10⁻¹⁰ Farad

Explanation:

Radius = R₁ = 6.25 cm = 6.25×10⁻² m

Radius = R₂ = 15 cm = 15×10⁻² m

Dielectric constant = k = 4.8

Electric constant = ε₀ = 8.854×10⁻¹² F/m

ε/ε₀=k

ε=kε₀

Capacitance\ (C)=\frac{4\pi k\epsilon_0 R_1\times R_2}{R_2-R_1}\\\Rightarrow C=\frac{4\pi 4.8\times 8.854\times 10^{-12}\times 15\times 10^{-2}\times 6.25\times 10^{-2}}{15\times 10^{-2}-6.25\times 10^{-2}}\\\Rightarrow C=0.572\times 10^{-10}\ Farad

∴ Capacitance is 0.572×10⁻¹⁰ Farad

3 0
2 years ago
By what percent must one increase the tension in a guitar string to change the speed of waves on the string from 301 m/s to 343
garri49 [273]

Answer:

29.8 %

Explanation:

7 0
3 years ago
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