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3 years ago

When the ball on the table starts moving when the table is tilted

Physics

1 answer:

Elis [28]3 years ago

6 0

Answer:

When the ball is placed on tilted tale it starts rolling because of the force of gravity pulls it downward along the plane of table and frictional force produces torque which generate rotational motion as a result body will start rolling.

Explanation:

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components of the speed of the coin is given as

$v_x = v cos60$

$v_x = 6.4 cos60 = 3.2 m/s$

$v_y = vsin60$

$v_y = 6.4 sin60 = 5.54 m/s$

now the time taken by the coin to reach the plate is given by

$t = \frac{\delta x}{v_x}$

$t = \frac{2.1}{3.2}$

$t = 0.656 s$

now in order to find the height

$h = vy * t + \frac{1}{2} at^2$

$h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2$

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so it is placed at 1.52 m height

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3 years ago

People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p

Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

$g' = \omega^2 (\dfrac{r}{2})$

$g' =\dfrac{1}{2}(\omega^2 r)$

$g'=\dfrac{g}{2}$

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hence, the correct answer is option B

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3 years ago

astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating o

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Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

The mass of the string will be:

⇒ $T=2 \pi\sqrt{\frac{m}{k} }$

or,

⇒ $m=\frac{k T^2}{4 \pi^2}$

On substituting the values, we get

⇒ $=\frac{280\times (3)^2}{4 \pi^2}$

⇒ $=\frac{280\times 9}{4\times (3.14)^2}$

⇒ $=68.83 \ kg$

(b)

The speed of the string will be:

⇒ $\frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2$

then,

⇒ $v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }$

On substituting the values, we get

⇒ $=\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }$

⇒ $=\sqrt{\frac{280(0.16-0.04)}{63.83} }$

⇒ $=\sqrt{\frac{280\times 0.12}{63.83} }$

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