If you were told an atom was an ion, you would know the atom must have a _____

Two mechanical devices typically used in laboratories to accurately measure small objects or distances are the _____ and _____.

Lera25 [3.4K]

Micrometers and calipers

3 0

3 years ago

Cylindrical beaker of height 0.100 mm and negligible weight is filled to the brim with a fluid of density rhorhorho = 890 kg/m3k

Nesterboy [21]

Incomplete part of the question

A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 . What is the weight Wb of the ball? Express your answer numerically in newtons.

What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale. Calculate your answer from the quantities given in the problem and express it numerically in newtons.

What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative). Express your answer numerically in newtons.

The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.

What weight W3 does the scale now show?

Answer:

(a) 2.94 N

(b) 1 N

(c) 2.42 N

(d) 3.42 N

Explanation:

(a)

From the definition of density, it's mass per unit volume hence mass is a product of density and volume. To get weight, we multiply mass by acceleration due to gravity

The weight of the ball is $W=\rho g V$

Where $\rho$ is the density, V is volume and g is acceleration due to gravity

Substituting density for 5000 Kg/m3 and g for 9.8 m/s2 and v for 0.00006 m3 then

$W= 5000 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3}=2.94 N$

(b)

Because the ball is being held up mostly by the rod, the fluid pressure on the bottom of the cylinder is just the same as before.

The scale does not "know" the ball is there at all.

That's why it still reads 1 N.

Therefore, the reading is 1 N

(c)

The buoyant force of the fluid on the ball is equal to the weight of the displaced fluid, namely,

$890 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3} = 0.52 N$

so the force needed for the rod to hold up the ball is 2.94 N - 0.52 N = 2.42 N.

(d)

Now the scale "feels" the weight of the ball,

so the scale reads the weight of the ball

PLUS the weight of the original fluid

MINUS the weight of the fluid that was displaced

= 2.94 N + 1.00 N - 0.52 N = 3.42 N

6 0

3 years ago

3. ¿Cuál es la distancia focal de un espejo CONVERo sabiendo que un objeto colocado

lions [1.4K]

i dont know for real

8 0

3 years ago

A horizontal beam of electrons initially moving at 4.0×10^7 m/s is deflected vertically by the vertical electric field between o

givi [52]

Answer:

$1.77\times 10^{-7}\ C/m^2$

0.000439077936334 m

Explanation:

q = Charge of electron = $1.6\times 10^{-19}\ C$

E = Electric field = $2\times 10^{4}\ N/C$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

d = Distance between plates = 2 cm (assumed)

m = Mass of electron = $9.11\times 10^{-31}\ kg$

The beam consists of electrons which means it has negative charge this means the upper plates will be positive and the lower plate will be negative.

The direction is upper to lower lower plate.

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

Electric flux is given by

$\phi=\epsilon_0E\\\Rightarrow \phi=8.85\times 10^{-12}\times 2\times 10^{4}\\\Rightarrow \phi=1.77\times 10^{-7}\ C/m^2$

The charge per unit area on the plates is $1.77\times 10^{-7}\ C/m^2$

Deflection is given by

$s=\dfrac{1}{2}\dfrac{qE}{m}(\dfrac{d}{v})^2\\\Rightarrow s=\dfrac{1}{2}\dfrac{1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}(\dfrac{0.02}{4\times 10^7})^2\\\Rightarrow s=0.000439077936334\ m$

The deflection is 0.000439077936334 m

7 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. You reaction time

bearhunter [10]

Answer:

Explanation:

Discount the time here; it's not important. It doesn't tell you how long it takes the car to stop, it only refers to reaction time, which means nothing in the scheme of things.

The useful info is as follows:

initial velocity = 20 m/s

final velocity = 0 m/s

a = -10 m/s/s

and we are looking for the displacement. Use the following equation:

$v^2=v_0^2+2a$ Δx

where v is the final velocity, v₀ is the initial velocity, a is the deceleration (since it's negative), and Δx is displacement. Filling in:

$0^2=(20)^2+2(-10)$ Δx and

0 = 400 - 20Δx and

-400 = -20Δx so

Δ = 20 meters

3 0

3 years ago

If you were told an atom was an ion, you would know the atom must have a ______?