WILL GIVE BRAINLIEST AND 100 POINTS! NEED THIS ASAP!

You need to determine the density of an unknown liquid and decide to perform an experiment. You notice that a wooden block float

Allushta [10]

Answer:

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

Explanation:

The density of a liquid is inversely proportional to the volume (height) of object submerged in it.

High density liquid possess higher buoyant force preventing objects from submerging.

p ∝ 1/V ∝ 1/h

since V = Ah

pu/pw = hw/hu

pu = pwhw/hu

Where;

p = density

h = height submerged

pu and pw is the density of unknown liquid and water respectively

hu and hw is the height of object submerged in unknown liquid and water respectively

pw = 1000kg/m^3

hu = 4.6cm = 0.046m

hw = 5.8cm = 0.058m

Substituting the given values;

pu = 1000×0.058/0.046

pu = 1260.9kg/m^3

the density of the unknown liquid is 1260.9kg/m^3

5 0

2 years ago

You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m

Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

$v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y$

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

$|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}$ ≈ $6.32 m/s$

And the angle with respect to the east should be with:

$arctan(\frac{2}{6} )=18.43 \°$

8 0

2 years ago

Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

3 0

2 years ago

An alloy is made of a material of specific gravity 7.87 and another material of specific gravity 4.50. The alloy of mass 750g ha

julsineya [31]

Answer:

13.9

Explanation:

Apparent weight is the normal force. Sum of the forces on the alloy when it is submerged:

∑F = ma

N + B − W = 0

N + ρVg − mg = 0

6.6 + (0.78 × 1000) V (9.8) − (0.750) (9.8) = 0

V = 9.81×10⁻⁵

If x is the volume of the first material, and y is the volume of the second material, then:

x + y = 9.81×10⁻⁵

(7.87×1000) x + (4.50×1000) y = 0.750

Two equations, two variables. Solve with substitution:

7870 (9.81×10⁻⁵ − y) + 4500 y = 0.750

0.772 − 7870 y + 4500 y = 0.750

0.0222 = 3370 y

y = 6.58×10⁻⁶

x = 9.15×10⁻⁵

The ratio of the volumes is:

x/y = 13.9

8 0

2 years ago

An electric power plant uses energy from burning coal to generate steam at 450∘C. The plant is cooled by 20∘C water from a nearb

Goryan [66]

Answer:

40 MJ (D)

Explanation:

Quantity of heat (Qh) = 100 MJ

temperature of steam (Th) = 450°c = 450 + 273 = 723 K

emperature of water (TI) = 20 °c = 20 + 273 = 293 k

efficiency = (Qh-Qi)/Qh = (Th-Ti)/Th

$\frac{100x10x^{6}-Qi }{100x10^{6}} = \frac{723-293}{723}$

$100x10^{6}$ - Qi= 0.5947 x $100 x 10 ^{6}$

$100x10^{6}$ - (0.5947 x $100x10^{6}$ ) = Qi

Qi = 40.5 MJ equivalent to 40 MJ (D)

6 0

2 years ago