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3 years ago

You attach a cart with a mass of 0.92 kg to a spring and compress the cart-

Physics

1 answer:

Artyom0805 [142]3 years ago

8 0

Answer:

F = - K X force exerted by compressed spring

F X = 1/2 M V^2 energy (work) required to produce a speed V for mass M

K X * X = 1/2 M V^2 direction of F is the same as that of V

X^2 = M V^2 / K combining terms

X = (M / K)^1/2 V

X = (.92 / 130)^1/2 * 1.4 = .118 m

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An athlete stretches a spring an extra 40.0 cm beyond its initial length. How much energy has he transferred to the spring, if t

notsponge [240]

Answer:

Explanation:

Step one:

given data

extension= 40cm

Spring constant K= 52.9N/cm

Step two:

Required

the Kinetic Energy KE

the expression to find the kinetic energy is

KE= 1/2ke^2

substituting our data we have

KE= 1/2*52.9*40^2

KE=0.5*52.9*1600

KE= 42320Joules

<u>The answer is b) 4230 J </u>

4 0

3 years ago

How would life on earth be different if earth’s axis were not tilted with respect to its orbit?

Vladimir [108]

If Earth's axis was "straight up and down" instead of tilted, then ...

<span>-- There would be no seasons.

-- The climate at any one place would be the same all year around.

-- The days would be the same length, everywhere,
    and all year around.

-- So would the nights.

-- The sun would be up a little more than 12 hours every day.
    It would be down a little less than 12 hours every day.

-- At the middle of the day, the sun would be at the same height
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   in others.

-- The equator would be the only place on Earth where the sun
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-- If you were at the north pole or the south pole, the sun would be
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4 0

3 years ago

Read 2 more answers

two electrons are an angstrom (1x10^-10m) apart. What electrostatic force do they exert on one another?

Irina-Kira [14]

Answer:

2.30 × 10⁻⁸ N if the two electrons are in a vacuum.

Explanation:

The Coulomb's Law gives the size of the electrostatic force $F$ between two charged objects:

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}}$ ,

where

$k$ is coulomb's constant. $k = 8.99\times 10^{8}\;\text{N}\cdot\text{m}^{2}\cdot\text{C}^{-2}$ in vacuum.
$q_1$ and $q_2$ are the signed charge of the objects.
$r$ is the distance between the two objects.

For the two electrons:

$q_1 = q_2 = 1.60\times 10^{-19}\;\text{C}$ .
$r = 1\times 10^{-10}\;\text{m}$ .

$\displaystyle F = -\frac{k\cdot q_1 \cdot q_2}{r^{2}} = -\frac{8.99\times 10^{8}\times (1.60\times 10^{-19})^{2}}{(1\times 10^{-10})^{2}} = 2.30\times 10^{-8}\;\text{N}$ .