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Ira Lisetskai [31]
3 years ago
15

A force of 10 newtons toward the right is exerted on a

Physics
1 answer:
weeeeeb [17]3 years ago
6 0

Answer:

Explanation:

coefficient of kinetic friction of wooden floor μ = .4

force of friction = μ R , R is reaction force of floor

R = mg = weight of body

R = 25 N

force of friction = .4 x 25 = 10 N

Net force on the crate = 10 - 10 = zero .

Net force on the body will be nil.

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A train moving with a velocity of 42.9 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s^2 North until
ankoles [38]

Answer:

3658.24m

Explanation:

Hello!

the first thing that we must be clear about is that the train moves with constant acceleration

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

\frac {Vf^{2}-Vo^2}{2.a} =X

Vf = final speed =160km/h=44.4m/s

Vo = Initial speed =42.9km/h=11.92m/s

A = acceleration =0.25m/s^2

X = displacement

solving

\frac {44.4^{2}-11.92^2}{2.(0.25)} =X\\X=3658.24m

the distance traveled by the train is 3658.24m

5 0
3 years ago
In order to prevent injury in a car crash, it is recommended that you _______. Increase the initial velocity of the collision.
Vlad1618 [11]
Increase the change in momentum of the collision
5 0
3 years ago
Alexander calders mobiles, like untitled, move when air currents move through them, making them_____________
yanalaym [24]
Alexander Calders mobiles, like untitled, move when air currents move through them, making them kinetic. Alexander Calder was an American artist, famous for his abstract sculpture, hanging mobiles, and Kinetic art. Kinetic art is the <span>art that depends on motion for its effect. The kinetic art's works of Alexander Calders were called "mobiles".</span>
4 0
3 years ago
Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 590 Hz. The peg of one string slips
Slav-nsk [51]

To solve this problem it is necessary to apply the concepts related to frequency and vibration of strings. Mathematically the frequency can be expressed as

f = \frac{v}{\lambda}

Then the relation between two different frequencies with same wavelength would be

\frac{f'}{f} = \frac{v'/\wavelength}{v/\wavelength}

\frac{f'}{f} = \frac{v'}{v}

The beat frequency heard when the two strings are sounded simultaneously is

f_{beat} = f-f'

f_{beat} = f(1-\frac{f'}{f})

f_{beat} = f(1-\frac{v'}{v})

We have the velocity of the transverse waves in stretched string as

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{200N}{\mu}}

And,

v' = \sqrt{\frac{196N}{\mu}}

Therefore the relation between the two is,

\frac{v'}{v} = \sqrt{\frac{192}{200}}

\frac{v'}{v} = \sqrt{0.96}

Finally substituting this value at the frequency beat equation we have

f_{beat} = 590(1-\sqrt{0-96})

f_{beat} = 11.92Hz

Therefore the beats per second are 11.92Hz

4 0
3 years ago
A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may
12345 [234]

Answer:

The tension on the string is  T  =  43.302 \ N

Explanation:

From the question we are told that

    The mass of the rock is m_r = 5.00 \ kg =  5000 \ g

       The density of the rock is \rho  =  4300 \ kg/m^3 =  4.3 g/dm^3

       

Generally the volume of the rock is mathematically evaluated as

          V    =  \frac{m_r}{\rho}

substituting values

        V    =  \frac{5000}{4.3}

       V    =  1162.7 \  dm^3

The volume of the rock immersed in water is

      V_w = \frac{V}{2}  

substituting values

     V_w = \frac{1162.7 }{2}

     V_w = 581.4 \ dm^3

mass of water been displaced by the this volume is

     m_w  = V_w     According to Archimedes principle

=>   m_w =  581.4 \ g

     m_w =  0.5814 \ kg

The weight of the water displace is  

      W _w =  m_w  * g

      W _w =  0.5814  * 9.8

      W _w = 5.698 \ N

The actual weight of the rock is  

      W_r  =  m_r * g

     W_r  =  5.0 *  9.8

     W_r  =  49.0 \ N

The tension on the string is

       T  = W_r - W_w

substituting values

       T  = 49.0 -  5.698

       T  =  43.302 \ N

4 0
3 years ago
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