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Lelechka [254]
3 years ago
5

What is not a raquet sport

Physics
1 answer:
uysha [10]3 years ago
7 0

Answer:

what

Explanation:

Racket sports include tennis, badminton, squash or any other sport where you use rackets to hit a ball or shuttlecock to play. They can be played competitively or just for fun and are a great form of physical activity.

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A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the wate
Vilka [71]

Answer:

   f = 5.3 Hz

Explanation:

To solve this problem, let's find the equation that describes the process, using Newton's second law

        ∑ F = ma

where the acceleration is

         a = \frac{d^2 y}{dt^2 }

        B- W = m \frac{d^2 y}{dt^2 }

To solve this problem we create a change in the reference system, we place the zero at the equilibrium point

           B = W

In this frame of reference, the variable y'  when it is oscillating is positive and negative, therefore Newton's equation remains

         B’= m \frac{d^2 y'}{dt^2 }

           

the thrust is given by the Archimedes relation

         B = ρ_liquid g V_liquid

     

the volume is

        V = π r² y'

     

we substitute

          - ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }

          \frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi  r^2/m ) y' \ =0

this differential equation has a solution of type

         y = A cos (wt + Ф)

where

         w² = ρ_liquid g π r² /m

angular velocity and frequency are related

         w = 2π f

         

we substitute

          4π² f² = ρ_liquid g π r² / m

          f = \frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \  \pi  r^2 \ g}{m } }

calculate

         f = \frac{1}{2 \pi }  \sqrt{ \frac{ 1000 \ \pi  \ 0.03^2 \ 9.8 }{0.025}  }

         f = 5.3 Hz

6 0
3 years ago
According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the
slega [8]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

A=\dfrac{\pi}{4}\times d^2

Put the value into the formula

A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2

A=1.96\times10^{-9}\ m^2

We need to calculate the magnitude of the force

Using formula of force

F=\sigma A

Put the value into the formula

F=196\times10^{6}\times1.96\times10^{-9}

F=0.38416\ N

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

strain=\dfrac{\Delta l}{l_{0}}

\Delta l=strain\times l_{0}

Put the value into the formula

\Delta l=0.380\times l_{0}

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

l=l_{0}+\Delta l

Put the value into the formula

I=l_{0}+0.380\times l_{0}

l=1.38l_{0}

l_{0}=\dfrac{l}{1.38}

l_{0}=\dfrac{12\times10^{-2}}{1.38}

l_{0}=0.0869\ m

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

4 0
4 years ago
When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for
sveta [45]

Answer:

F = -4567.40 N

Explanation:

Given that,

The power developed by the engine, P = 196 hp

1 hp = 746 W

196 hp = 146157 W

Speed of the car, v = 32 m/s

Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :

P=-F\times v

F=\dfrac{-P}{v}

F=\dfrac{-146157\ W}{32\ m/s}

F = -4567.40 N

So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.

7 0
3 years ago
At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at
love history [14]

The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block <em>A</em>, and block <em>B</em>  after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>

<em />

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block <em>A</em> = 0 m/s

Speed of block <em>B</em>, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)

The location of block <em>B</em>, = v₂ × t

The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}

x_{cm} = \dfrac{8  \times0 + 16 \times  30}{8 + 16} = 20

The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)

Learn more about the center of mass here:

brainly.com/question/18557256

brainly.com/question/20714030

brainly.com/question/17088562

4 0
3 years ago
How do you solve 0.004 dm + 0.12508 dm?
Effectus [21]
0.004 of something added to 0.12508 of the same thing
adds up to 0.12908 of it.  

The thing could be a glass of water, a sheet of paper,
a pound of ground beef, a gallon of gas, or a snowball.  
In this problem, it just happens to be a dm. 
7 0
3 years ago
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