Which two men developed the steam engine?

Plastic foam is about 0.10 times as dense as water. What weight of bricks could you stack on a 1m x 1m x 0.10m slab of foam, so

goblinko [34]

Answer: Weight = 98.1N

Explanation:

Density of water = 1000 kg/m^3

Given that the Plastic foam is about 0.10 times as dense as water. That is,

Density of plastic foam = 0.1 × 1000 = 100kg/m^3

The volume V = 1 ×1×0.1 = 0.1 m^3

Density is the ratio of mass to volume

Density = mass/volume

Let us substitute for density and volume to get mass.

100 = M/0.1

Make M the subject of formula

M = 100 × 0.1 = 10 kg

Weight = mg

Where g = 9.81 m/s

Substitute the M and g into the formula

Weight = 10 × 9.81 = 98.1 N

Therefore, the weight of the brick is 98.1 N

4 0

4 years ago

The transmission coefficient of the hypothetical disease dumas fever is 0.2. if the combined death and recovery rate is 0.4, wha

borishaifa [10]

NT = r / β
where NT is the threshold density, r is the recovery and death rate and β is the transmission coefficient.

= 0.4 / 0.2
= 2

7 0

3 years ago

For a new TV series "Stupidity Factor contestants are dropped into the ocean (p 1030 kg/ms) along with a Styrofoam soo kg/ms blo

luda_lava [24]

Answer:

Explanation:

Heres the correct and full question:

For a new TV series "Stupidity Factor contestants are dropped into the ocean (p=1030 kg/m³) along with a Styrofoam p=3oo kg/m³ block that is 2m x 3 m X 20 cm thick. If too many contestants climb aboard the block and it sinks below the water surface, they are declared "stupid" and abandoned at sea. Find the maximum number of 7o kg contestants that the block can hold. (No fractional contestants, please. They hate it when that happens.

answer:

volume of styrofoam block=V=2m x 3m x 0.20m =1.2m³

density of styrofoam=ρs=300kg/m³

mass of styrofoam=ms=v(ρs)=1.2 x 300=360kg

weight of styrofoam = ws=(ms)g=360 x 9.8=3528N

consider number of contestants= n

toatal weight of ccontestants=W=n(70 x 9.8)=n(686N)

since, styrofoam fully submerged into water, then bat force,

B=ρ(vs)g=1030 x 1.2 x 9.8 = 12112.8N

At equilibrium,

B - W - Ws = 0

12112.8 - 686n - 3528 = 0

$n=\frac{12112.8-3528}{686}=12.5$

n=12 person(contestants)

7 0

4 years ago

Waves on water in a ripple tank are often used as a model to help students to understand sound waves

lesya692 [45]

Answer:

I'm sorry I

Explanation:

I don't know I'm sorry I will tell you another answer

3 0

3 years ago

The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final

Alex787 [66]

Answer:

Final temperature Water = 20.99-degree celsius

Final temperature Concrete = 24.98 degree celsius

Final temperature Steel = 50.1 degree celsius

Final temperature Mercury = 29.26 degree celsius

Explanation:

Given the mass of each substance = 1.50 kg

Ti = 15

Q = 1.5 kcal = 6276 joule

We have to use the heat capacity of each object so find the heat capacity from the table.

Heat capacity of water = 4186 J/kg degree celsius.

Heat capacity of concrete = 840 J/kg degree celsius.

Heat capacity of steel = 452 J/kg degree celsius.

Heat capacity of mercury = 139 J/kg degree celsius.

Use the below formula to find the final temperature.

$T_f = T_i + \frac{Q}{mc_w} \\$

$\text{Temperature in the case of water.} \\= 20 + \frac{6276}{1.5 \times 4186 } \\= 20.99 \ degree \ celsius \\\text{Temperature in the case of concrete.} \\= 20 + \frac{6276}{1.5 \times 840 } \\= 24.98 \ degree \ celsius \\\text{Temperature in the case of steel.} \\= 20 + \frac{6276}{1.5 \times 452 } \\= 29.26 \ degree \ celsius \\\text{Temperature in the case of mercury.} \\= 20 + \frac{6276}{1.5 \times 139 } \\= 50.1 \ degree \ celsius \\$

5 0

4 years ago