How much force is required to accelerate a 12 kg mass at m 5 m/s

Most ocean waves obtain their energy and motion from _____. the moon's gravitational attraction the sun plate movement the wind

IrinaVladis [17]

Hello!

Most ocean waves obtain their energy and motion from the wind.

Ocean waves are surface waves that move across the surface of the ocean. When wind touches the surface of the water, there is friction in the contact zone. This friction causes a drag effect, that makes wrinkles on the surface of the water. As the wrinkles get bigger, they transform into full-blown waves, and the taller the wave, the more energy it can extract from the wind, making them even bigger and to move longer distances.

Have a nice day!

3 0

3 years ago

A cart for hauling ore out of a gold mine has a mass of 413 kg, including its load. the cart runs along a straight stretch of tr

Andrei [34K]

Given from the problem :mass m = 413 kg;coefficient of friction u = 0.0163;acceleration due to gravity g = 9.8 m/s2;inclined angle @1 = 14.3;inclined angle @2 = 4.69;distance travelled d = 175 m;applied fore F = 410 N; the component of the force from the donkey in the direction of motion isF2 = F1 [email protected]= 397.2964498768165 N
Fy = N - mg [email protected] = 0N = mg [email protected] = 4037.964151113007 NFx = F2 - mg [email protected] - f = mahere f = u N=65.8188156631420141

F2 - mg [email protected] - f = maa = F2 - mg [email protected] - f/ m=0.31923412183075155 m/s^2
work done by donkeyW = F2 d=69526.8787284428875 J

8 0

3 years ago

6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction betw

USPshnik [31]

Answer:

$x=22.65m$

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

$v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}$

The acceleration can be calculated using Newton's second law:

$\sum F_x:F_f=ma\\\sum F_y:N=mg$

Recall that the maximum force of friction is defined as $F_f=\mu N$ . So, replacing this:

$\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}$

Now, we calculate the distance:

$x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m$

7 0

3 years ago

A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$

Where in this case:

$m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m$

Therefore:

$E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J$

8 0

3 years ago

How did Bourne believe that an object could be made to rise or sink at will?

Angelina_Jolie [31]

Bourne believed that an object would float or sink at will as long as he could <span>manipulate the effect's of buoyancy which control and object to sink or float. Hope this helps!

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4 0

3 years ago

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