Ask question

Ask question

All categories

2 years ago

13

UDAY WAS TOLD TO PUT SOME CONTAINERS IN ONE OF THE COLD STORES AT WORK. THE LABLES ON THE CONTAINERS READ STORE BELOW -5 C.THERE

ARE TWO STORE ROOMS .ONE IS KEPT AT 15 F AND THE OTHER AT 25 F . WHICH ONE SHOULD HE CHOOSE?

1 answer:

Alja [10]2 years ago

3 0

He should choose the room that’s 15 F.

-5 C = 23 F
Meaning that 15 F is below -5 C and 25 F is not.

You might be interested in

Rachel is helping her younger brother replace a broken part in his toy ambulance. This part is responsible for converting electr

Studentka2010 [4]

The light bulb, it takes electrical energy and turns it into l<span>ight energy!</span>

5 0

2 years ago

Read 2 more answers

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = 14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

= 24.25 × 10⁶ N/m²

Volume = $\dfrac{4}{3} \pi r^3$

$\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}$

Bulk modulus is equal to

$B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }$

now

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }$

$B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }$

$(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}$

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0

2 years ago

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

garik1379 [7]

Answer:

T=2.94*10^-10 N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

l=length of the spider silk, 14cm

velocity of wave = √(T/μ)

where T = tension and

μ = mass per unit length)

λ/2=l

for fundamental frequency λ/2 =14cm

(λ= wavelength of standing wave; as there will be no node

except the endpoints of silk strand)

λ = 28 cm = 0.28 m

and since frequency * wavelength = speed of wave. we have,

150 * 0.28 = √(T/μ) ..................(#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

= [π * (10 * 10^(-6))²] * 1300 = 13π * 10^(-8).

now putting this in equation (#) we get

150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)² =

2.94*10^-10 N/m.

6 0

2 years ago

The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp

Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

3 0

1 year ago

James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than

Molodets [167]

Answer:

0.12 K

Explanation:

height, h = 51 m

let the mass of water is m.

Specific heat of water, c = 4190 J/kg K

According to the transformation of energy

Potential energy of water = thermal energy of water

m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h = c x ΔT

9.8 x 51 = 4190 x ΔT

ΔT = 0.12 K

Thus, the rise in temperature is 0.12 K.

7 0

2 years ago