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4 years ago

What are two examples of an accelerating object ?

Physics

1 answer:

givi [52]4 years ago

3 0

-- Steven is in his car, rolling slowly but with his foot on the gas and speeding up.

-- Sharon is in her car, rolling fastly but with her foot on the brake and slowing down.

-- Josh is in his car, keeping at a steady 20 miles per hour as he turns the corner.

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Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.80

sergiy2304 [10]

Answer:

1.3 rev/s

Explanation:

$I_{o}$ = Moment of inertia when arms and one leg of the skater is out = 3.5 kgm²

$I_{i}$ = Moment of inertia when arms and legs of the skater are in = 0.80 kgm²

$w_{i}$ = Angular speed of skater when arms and legs of the skater are in = 5.5 rev/s

$w_{o}$ = Angular speed of skater when arms and legs of the skater are out = ?

Using conservation of angular momentum

$I_{i}$ $w_{i}$ = $I_{o}$ $w_{o}$

(0.80) (5.5) = (3.5) $w_{o}$

$w_{o}$ = 1.3 rev/s

8 0

3 years ago

A toy cannon uses a spring to project a 5.35-g soft rubber ball. The spring is originally compressed by 5.08 cm and has a force

Elenna [48]

Answer:

a) the velocity is v=1.385 m/s

b) the ball has its maximum speed at 4.68 cm away from its compressed position

c) the maximum speed is 1.78 m/s

Explanation:

if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore

we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball

W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki

therefore

Ui + Ki = Uf + Kf + W fr

where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.

since

U= 1/2 k x² , K= 1/2 m v² , W fr = F*L

X= compression length , L= horizontal distance covered

therefore

Ui + Ki = Uf + Kf + W fr

1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L

a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0

1/2 k X² + 0 = 0 + 1/2 m v² + F*L

1/2 m v² = 1/2 k X² - F*L

v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s

b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.

1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]

1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d

taking the derivative

dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm

at x m = -0.4 cm the velocity is maximum

therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position

c) the maximum speed is

1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0

v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s

4 0

3 years ago

The first law of thermodynamics invention that spurred its development is

Ket [755]

Energy and matter can neither be created or destroyed.

5 0

3 years ago

A solid sphere of radius R carries a fixed, uniformly distributed charge q. Obtain an expression for the magnitude of the electr

NISA [10]

Answer:

The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

Explanation:

Given that,

Radius of solid sphere = R

Charge = q

According to figure,

Suppose r is the distance between the point P and center of sphere.

If $\rho$ be the volume charge density,

Then, the charge will be,

$q=\rho\times\dfrac{4}{3}\pi R^3$ .....(I)

Consider a Gaussian surface of radius r.

We need to calculate the electric field outside the sphere

Using formula of electric field

$\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}$

$E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}$

Put the value from equation (I)

$E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}$

$E=\dfrac{qr}{4\pi\epsilon_{0}R^3}$

Hence, The electric field outside the sphere will be $\dfrac{qr}{4\pi\epsilon_{0}R^3}$ .

4 0

3 years ago

Matter is anything that has mass and takes up space. Which units of measurement can you use to describe these two properties?

Papessa [141]

Answer: Matter is anything that has mass and takes up space . Mass is amount of mass that an object contains. Mass is measured by scale . Volume is amount of space that matter takes up .

Explanation:

4 0

3 years ago