What branch of science that deals with landforms,features,inhabitants and phenomena on earth

Please write ONE complete sentence describing the relationship between Kinetic Energy and Gravitational Potential Energy in a sy

Scilla [17]

Answer:

Explanation:

The sum of kinetic and potential energy is the mechanical energy of the system. Assuming the energy of the system is conserved, which it is in your case, then the initial mechanical energy will equal the final mechanical energy. This relationship is very useful in mechanics.

3 0

3 years ago

Photosynthesis uses all of the following except _____ to make food.

Mashcka [7]

It is Carbon dioxide

8 0

3 years ago

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Now assume that the frictional force f is not at its maximum value. What is the relation between the torque Ï„ applied to each w

leva [86]

Answer:

a)The direction the frictional force will acts is in the positive x direction.

Explanation:

a)The direction the frictional force will acts is in the positive x direction

b)in the horizontal direction, the total force F(total) is equal to 4times the frictional force in the wheel.

F(total)=4f

''f'' is taken as the frictional force.

c)4times the normal force on each wheel minus the acceleration equals zero i.e 4N(wheel)-a=0

=4N(wheel)-mg=0

d) torque is the force that tends to bend rotation

ζ=rf

but acceleration=4×frictional force

cross multiply

f=ζ/r

f=ma/4

ma/4=ζ/r

a=4ζ/r

5 0

3 years ago

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A baseball (m=140g) traveling 32m/s moves a fielders

posledela

Answer:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average force exerted by the ball on the glove. It is solved using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

F = mg

$\dfrac{1}{2}mv^2=Fh$

$F=\dfrac{mv^2}{2h}$

$F=\dfrac{0.14\times (32)^2}{2\times 0.25}$

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

7 0

3 years ago

A spring has a natural length of 0.5 m and was stretched by 0.02 m. if the spring had a resultant energy of 0.5 j what is the sp

Anna71 [15]

$\textbf{2500 }\dfrac{\textbf{kg}}{\textbf{s}^{\textbf{2}}}$

Explanation:

Natural length of a spring is $0.5\text{ }m$ . The spring is streched by $0.02\text{ }m$ . The resultant energy of the spring is $0.5\text{ }J$ .

The potential energy of an ideal spring with spring constant $k$ and elongation $x$ is given by $\dfrac{1}{2}kx^{2}$ .

So, in the current problem, the natural length of the spring is not required to find the spring constant $k$ .

$\text{Potential Energy in the spring = }\dfrac{1}{2}kx^{2}\\0.5\text{ }J\text{ }=\text{ }\dfrac{1}{2}k(0.02\text{ }m)^{2}\\k\times0.0004\text{ }m^{2}\text{ }=\text{ }1\text{ }J\text{ }=\text{ }1\text{ }kg\frac{m^{2}}{s^{2}}\\k\text{ }=\text{ }\dfrac{1\text{ }kg\dfrac{m^{2}}{s^{2}}}{0.0004\text{ }m^{2}}\text{ }=\text{ }2500\text{ }\frac{kg}{s^{2}}$

∴ The spring constant of the spring = $2500\text{ }\frac{kg}{s^{2}}$

4 0

3 years ago