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ICE Princess25 [194]
2 years ago
5

1. Which science will discuss how matter is created throughout the universe?

Physics
1 answer:
Len [333]2 years ago
4 0

Answer:

physics....................

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How to convert 250 newton to pounds? It is likely to be difficult to move?
maxonik [38]
I have no idea I need the answer too
3 0
3 years ago
A mass m attached to a horizontal massless spring with spring constant k, is set into simple harmonic motion. its maximum displa
Lesechka [4]
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:
U_i= \frac{1}{2} ka^2
while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:
K_i =0
And the total energy of the system is
E_i = U_i+K= \frac{1}{2}ka^2

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:
U_f = 0
while the mass is moving at speed v, and therefore the kinetic energy is
K_f =  \frac{1}{2} mv^2
And the total energy is
E_f = U_f + K_f =  \frac{1}{2} mv^2

For the law of conservation of energy, the total energy must be conserved, therefore E_i = E_f. So we  can write
\frac{1}{2} ka^2 =  \frac{1}{2}mv^2
that we can solve to find an expression for v:
v= \sqrt{ \frac{ka^2}{m} }
6 0
3 years ago
if the kinetic and potential energy in a system or equal to the potential energy increases what happens as a result
8090 [49]

Answer:If the kinetic and potential energy in a system are equal, then the potential energy increases. ... Stored energy decreases. Energy of motion decreases. Total energy decreases

Explanation:

8 0
3 years ago
A 500 gram mass attached to a horizontal spring
Daniel [21]
Is that a question? .
..............................
8 0
3 years ago
Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,
klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble=2\hat{i}-3\hat{k} N

Position vector=r=0.5\hat{j}-2\hat{k} m

a.We have to find the resulting torque on the pebble about origin.

Torque=r\times F

Substitute the values then we get

Torque= (0.5j-2k)\times (2i-3k)

Torque=-k-1.5i-4jN-m

By using i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0

b.r=2i-3k

r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k

Torque about point (2,0,-3)

\tau=(-2i+0.5j+k)\times (2i-3k)

\tau=-6j-k-1.5i+2j=-1.5i-4j-kN-m

6 0
3 years ago
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