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anzhelika [568]
2 years ago
7

The wind blows a jay bird south with a force of 300 Newtons. The

Physics
1 answer:
adoni [48]2 years ago
6 0

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The  jay bird flies north, against the wind, with a force of 100  newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

F=\sqrt{300^2+100^2} \\\\F=316.22\ N

Hence, the net force on the jay bird is 316.22 N.

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evaluate the numerical value of the vertical velocity of the car at time t=0.25 s using the expression from part d, where y0=0.7
likoan [24]

Given :

Displacement , y = 0.75 m .

Angular acceleration , \alpha=0.95\ s^{-2} .

Initial angular velocity , \omega_o=6.3\ s^{-1} .

To Find :

The value of vertical velocity after time t = 0.25 s .

Solution :

By equation of circular motion is given by :

\omega=\omega_o+\alpha t

Putting all given values we get :

\omega=6.3+0.95\times 0.25\\\\\omega= $$6.5375\ s^{-1}

Now , vertical velocity is given by :

v=y\omega\\\\v=0.75\times 6.5375\ m/s\\\\v=4.90\ m/s

Therefore , the numerical value of the vertical velocity of the car at time t=0.25 s is 4.90 m/s .

Hence , this is the required solution .

8 0
2 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
2 years ago
Nisha was making a list of things that are obtained from plants and Amol was making a list of things that can dissolve in water.
Allisa [31]

Answer:

Sugar

Explanation:

Nisha's list:                               Amol's list

  Jute                                          Vinegar

  Wood                                       Lemon juice

  Rubber                                   Cooking soda

Nisha's list is made up of things that can be derived from plants. From the list given, oil, sugar and cotton can also be obtained from plants.

Most plant materials are organic matter.

Amol's list is made up of things that can dissolve in water. Sugar and table salt can also be dissolved in water.

Water is able to dissolve these materials because they are polar compounds. One rule of solubility is that like dissolves like.

New list:

Nisha's list:                               Amol's list

  Jute                                          Vinegar

  Wood                                       Lemon juice

  Rubber                                   Cooking soda

  Sugar                                       Sugar

                                   

Only sugar from the list can be added to both lists. It can be obtained from plant and can also dissolve in water.

                                             

6 0
2 years ago
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
SCORPION-xisa [38]

Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

7 0
2 years ago
How do you do projectile motion problems
PtichkaEL [24]
The key  projectile motion is that gravity allows downward only
7 0
2 years ago
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