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2 years ago

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A piece of wood that floats on water has a mass of 0.0175 kg. A lead weight is tied to the wood, and the apparent mass with the

wood in air and the lead weight submerged in water is 0.0765 kg. The apparent mass with the wood and the lead weight both submerged in water is 0.0452 kg. What is the specific gravity of the wood

1 answer:

crimeas [40]2 years ago

5 0

Answer:

Specific gravity is 0.56

Explanation:

We know that

mass of water displaced by the wood is, m1( apparent mass when wood in air and lead is submerged in water) - m2(the apparent mass when wood and lead both are submerged in water)

= 0.0765 - 0.0452 = 0.0313 Kg

So the specific gravity of the wood is, = mass of wood / mass of water displaced by the wood

= 0.0175/0.0313

=0.56

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A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

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a) 219.8 rad/s

b) $20.0 rad/s^2$

c) $2.9 m/s^2$

d) $7005 m/s^2$

e) Towards the axis of rotation

f) $0 m/s^2$

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

$\omega_{max}=35 rev/s$

The angle covered during 1 revolution is

$\theta=2\pi$ rad

Therefore, the maximum angular velocity is:

$\omega_{max}=35 \cdot 2\pi = 219.8 rad/s$

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

$\alpha = \frac{\Delta \omega}{t}$

where

$\Delta \omega$ is the change in angular velocity

t is the time elapsed

Here we have:

$\omega_0 = 0$ is the initial angular velocity

$\omega_{max}=219.8 rad/s$ is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

$\alpha = \frac{219.8-0}{11}=20.0 rad/s^2$

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

Here we have

$\alpha =20.0 rad/s^2$

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

$a_t=(20.0)(0.145)=2.9 m/s^2$

d)

The magnitude of the radial (centripetal) acceleration is given by

$a_c = \omega^2 r$

where

$\omega$ is the angular velocity

r is the radius of the circle

Here we have:

$\omega_{max}=219.8 rad/s$ is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

$a_c=(219.8)^2(0.145)=7005 m/s^2$

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

$a_t = \alpha r$

where

$\alpha$ is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

$\alpha=0$ , since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

$a_t=(0)(0.145)=0 m/s^2$

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

$\omega=219.8 rad/s$

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

$v=(219.8)(0.145)=31.9 m/s$

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2 years ago

Hippos spend much of their lives in water, but amazingly, they don’t swim. manatees, They have, like little very body fat. The d

kenny6666 [7]

Answer:

428.59 N

Explanation:

Buoyant force, $B=Vg\rho$ where V is volume, g is gravitational constant and \rho is density

$B+F_{upward}=mg$ where $F_{upward}$ is upward force

$Vg\rho_{w}+F_{upward}=mg$

$F_{upward}=mg- Vg\rho_{w}$

$F_{upward}=g(mg- V\rho_{w})=g(m-m\frac {\rho_{w}{\rho_{hippo}}$ where $\rho_{hippo}$ is the density of hippo

$F_{upward}=mg(1-\frac {\rho_{w}}{\rho_{hippo}})$

Using g as 9.81

$F_{upward}=1500*9.81*(1-1000/1030)= 428.5922 N$

Therefore, the upward force=428.59 N

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2 years ago

Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s

vichka [17]

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances

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2 years ago

In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r

scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W = 2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

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zhuklara [117]

Lab safety equipment prevents damage from accidents and helps keep the people working in the lab safe. The equipment goes hand in hand with the clothing of the person. The first step would be to wear closed shoes and a lab coat.
The equipment that must be worn are goggles to protect the eyes from irritants and latex gloves to protect the skin on the hands.

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2 years ago

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