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Sergeu [11.5K]
2 years ago
8

50 N

Physics
1 answer:
Mariana [72]2 years ago
5 0

The work is done by the force pushing the mower is determined as 500 J.

<h3>Work done by the force</h3>

The work done by the force is calculated as follows;

W = Fsinθ x d

where;

  • Fsinθ = Fx  is the applied force parallel to the ground
  • d is the distance
  • θ is the angle of inclination

W = Fx(d)

W = 25 x 20

W = 500 J

Thus, the work is done by the force pushing the mower is determined as 500 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

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At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh
Svetradugi [14.3K]

Answer:

The time needed is T  = 16.8 s

Explanation:

From the question we are told that

      The magnitude of the stimulated acceleration due gravity is  a  =  0.5 g

        The diameter of the spaceship is  d =  35m

       

Generally the force acting on the spaceship is  

       F  =  ma

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

      F  = \frac{mv^2}{r}

Thus  

       ma  =  \frac{mv^2}{r}

=>    \frac{v^2}{r}  =  a

      Generally the speed of this spaceship is mathematically represented as

      v =  \frac{2 \pi}{T}

=>    v^2  =   [\frac{2\pi}{T}] ^2

=>     \frac{\frac{4\pi^2 r^2}{T^2} }{r}  = 0.5g

=>       \frac{4 \pi^2 r }{T^2} =  0.5 g

=>         T  = \sqrt{ \frac{4\pi^2 r}{0.5g}}

substituting values

          T  = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}

         T  = 16.8 s

4 0
3 years ago
Read 2 more answers
A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d
I am Lyosha [343]

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

6 0
3 years ago
Use the graph below to answer the following question: What is happening to the object's velocity?
hjlf

The object's velocity is decreasing.

Explanation:

From graph is the attached image, we can clearly point that the velocity of this motion is decreasing with time.

Velocity is a vector quantity.

  • The y-axis represent displacement.
  • The x-axis depicts time
  • Using the graph, we know that the slope of the line on the graph gives us the velocity as it denotes the change of displacement with time.
  • When we find the slope, it will give us a negative value which shows that the body is slowing down and not increasing speed.

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

4 0
3 years ago
2 Points
Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

:\implies  Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

Substituting the respective values in the Formula, we get :

:\implies  50 N = 10 kg × Acceleration

:\implies \mathsf{Acceleration = \dfrac{50\;N}{10\;kg}}

:\implies Acceleration of the Object = 5 m/s²

4 0
3 years ago
True or false? If two components are connected in series, the current through one component will
Molodets [167]

the answer is ( True ) .

the current is the same in series circuits .

8 0
3 years ago
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