I want to answer these questions about vectors

Why is it advisable to wear long sleeves when a student works in a chemistry lab

love history [14]

So then if they do spill the chemical then it gets on their cloths and not on then it dosen"t harm them instead it ruins their shirt.

3 0

3 years ago

Read 2 more answers

PLZ Help Me I give you brainlist also NO LONKS OR I REPORT and Have a gr8 day my peeps

Sergio039 [100]

Answer:

the answer is a because I saw it in a syllabus

3 0

3 years ago

A pharmacist attempts to weigh 100 milligrams of codeine sulfate on a balance with a sensitivity requirement of 4 milligrams. Ca

ira [324]

Answer:

4 %

2 ) 3.42 %

Explanation:

Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .

Given , 4 milligram is the maximum error possible .

Measured weight = 100 milligram

So percentage maximum potential error

= (4 / 100) x 100

4 %

2 )

As per measurement

weight of 6 milliliters of water

= 48.540 - 42.745 = 5.795 gram

6 milliliters of water should measure 6 grams

Deviation = 6 - 5.795 = - 0.205 gram.

Percentage of error =(.205 / 6 )x 100

= 3.42 %

3 0

3 years ago

A 65.0-Ω resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 Ω. Calcula

Luda [366]

Answer:

a) 0.1832 A

b) 11.91 Volts

c) 2.18 Watt , 0.0168 Watt

Explanation:

(a)

R = external resistor connected to the terminals of the battery = 65 Ω

E = Emf of the battery = 12.0 Volts

r = internal resistance of the battery = 0.5 Ω

i = current flowing in the circuit

Using ohm's law

E = i (R + r)

12 = i (65 + 0.5)

i = 0.1832 A

(b)

Terminal voltage is given as

$V_{ab}$ = i R

$V_{ab}$ = (0.1832) (65)

$V_{ab}$ = 11.91 Volts

(c)

Power dissipated in the resister R is given as

$P_{R}$ = i²R

$P_{R}$ = (0.1832)²(65)

$P_{R}$ = 2.18 Watt

Power dissipated in the internal resistance is given as

$P_{r}$ = i²r

$P_{r}$ = (0.1832)²(0.5)

$P_{r}$ = 0.0168 Watt

5 0

3 years ago

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

6 0

3 years ago