Any followers for me please i have solved 25 question thats too much please

5. During the annual shuffleboard competition, Renee gives her puck an initial speed of 9.32 m/s. Once leaving her stick, the pu

abruzzese [7]

V=0 v²=0, A=v-u/t. T=v-u/a. T= 0-9.32/-4.06 therefore time = 2.296 seconds

3 0

3 years ago

A person throws a ball straight up in the air. The ball rises to a maximum height and then falls back down so that the person ca

Lana71 [14]

Answer:

The acceleration is about 9.8 m/s2 (down) when the ball is falling.

Explanation:

The ball at maximum height has velocity zero

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)

$v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow u=9.8t$

The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero

5 0

4 years ago

What is a disadvantage of using moving water to produce electricity

klemol [59]

Most of the time the towns get flooded and many people die and lose their homes and their jobs, businesses by the floods and the natural environment is destroyed ,air pollution is produced, the ecosystem could be disrupted and dams are known to fall over and dams are really expensive to build, The dams also cause geological damage, example earthquakes and spills happen.

4 0

3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

How can a 1kg ball have more kinetic energy than a 100kg ball? Explain both using words and by providing a numerical example

MariettaO [177]

1 kg ball can have more kinetic energy than a 100 kg ball as increase in velocity is having greater impact on K.E than increase in mass.

<u>Explanation</u>:

We know kinetic energy can be judged or calculated by two parameters only which is mass and velocity. As kinetic energy is directly proportional to the $(velocity)^2$ and increase in velocity leads to greater effect on translational Kinetic Energy. Here formula of Kinetic Energy suggests that doubling the mass will double its K.E but doubling velocity will quadruple its velocity: