Any fracture or system of fractures along which Earth moves is known as a

A hydrometer is made of a tube of diameter 2.3cm.The mass of the tube and it's content is 80g. If it floats in a liquid density

iris [78.8K]

Answer:

The depth to which the hydrometer sinks is approximately 24.07 cm

Explanation:

The given parameters are;

The diameter of the hydrometer tube, d = 2.3 cm

The mass of the content of the tube, m = 80 g

The density of the liquid in which the tube floats, ρ = 800 kg/m³

By Archimedes' principle, the up thrust (buoyancy) force acting on the hydrometer = The weight of the displaced liquid

When the hydrometer floats, the up-thrust is equal to the weight of the hydrometer which by Archimedes' principle, is equal to the weight of the volume of the liquid displaced by the hydrometer

Therefore;

The weight of the liquid displaced = The weight of the hydrometer, W = m·g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ W = 80 g × g

The volume of the liquid that has a mass of 80 g (0.08 kg), V = m/ρ

V = 0.08 kg/(800 kg/m³) = 0.0001 m³ = 0.0001 m³ × 1 × 10⁶ cm³/m³ = 100 cm³

The volume of the liquid displaced = 100 cm³ = The volume of the hydrometer submerged, $V_h$

$V_h$ = A × h

Where;

A = The cross-sectional area of the tube = π·d²/4

h = The depth to which the hydrometer sinks

h = $V_h$ /A

∴ h = 100 cm³/( π × 2.3²/4 cm²) ≈ 24.07 cm

The depth to which the tube sinks, h ≈ 24.07 cm.

3 0

2 years ago

a woman is swimming across a cold lake. her body temperature is 98 degrees fahrenheit , and the lake water is at 60 degrees fahr

sergiy2304 [10]

Answer:

the woman has more thermal energy becuase of her body temp.

Explanation:

none

5 0

3 years ago

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f

amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//) → a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂) → a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).

6 0

2 years ago

Second class lever short note

seraphim [82]

Answer:

wow

Explanation:

6 0

3 years ago

Read 2 more answers

Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration

eimsori [14]

The complete text of the problem is:

"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "

Solution:

1) Acceleration: $-2336 m/s^2$ on the hardwood floor, $-382 m/s^2$ on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

$v^2 - u^2 = 2ad$

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s$

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

$d = 1.8 mm = 0.0018 m$

So we can find the acceleration by using again the equation

$v'^2 - v^2 = 2ad$

Solving for a,

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2$

For the carpeted floor instead,

$d=1.1 cm = 0.011 m$

therefore the acceleration is

$a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2$

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

$a=\frac{v'-v}{t}$

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

$a=-2336 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms$

For the carpeted floor,

$a=-382 m/s^2$

So the duration of the collision is

$t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms$

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

8 0

2 years ago