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2 years ago

How many meters will a person run during a 5-mile race

Physics

2 answers:

Rashid [163]2 years ago

8 0

Answer:

8046.72 meters pretty sure

IgorLugansk [536]2 years ago

8 0

I think the same 8046.72 ^^^^

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What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Volgvan

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

4 0

3 years ago

A mango fruit drops from the top of it's tree which is 40 m. How long does it takes to reach the ground

IrinaVladis [17]

T=distance over speed
T=40m over 9.8ms
T=answer

4 0

3 years ago

How does a mathematical model help you understand the science concepts

Dmitrij [34]

It can help with measurements and when you want to add measurements to a cylinder or a beaker so ya

7 0

2 years ago

A circular window of 30 cm diameter in a submarine can withstand a maximum force of 5.20 × 105 N. The maximum depth in a lake to

Svetach [21]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

F = $5.20 \times 10^{5}$ N

g = 9.8 m/s

radius = $\frac{diameter}{2}$

= $\frac{30 cm}{2}$ = 15 cm = 0.15 m (as 1 m = 100 cm)

Formula to calculate depth is as follows.

F = $\rho \times g \times h \times A$

or, h = $\frac{F}{\rho \times g \times A}$

h = $\frac{5.2 \times 10^{5}}{1000 \times 9.8 \times (3.1416 \times (0.15 m^{2})}$

= 751 m

Thus, we can conclude that the maximum depth in a lake to which the submarine can go without damaging the window is closest 750 m.

3 0

3 years ago

A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving downward at 30 cm/s just before l dith t

viktelen [127]

Answer:

The speed of the cart and clay after the collision is 50 cm/s .

Explanation:

Given :

Mass of lump , m = 500 g = 0.5 kg .

Velocity of lump , v = 30 cm/s .

Mass of cart , M = 1 kg .

Velocity of cart , V = 60 cm/s .

We know by conservation of momentum :

$mv+MV=(m+M)v'$

Here , $v'$ is the speed of the cart and clay after the collision .

Putting all value in above equation .

We get :

$0.5\times 30+1\times 60=(0.5+1)\times v'\\\\v'=\dfrac{15+60}{1.5}\\\\v'=50\ cm/s$

Hence , this is the required solution .

3 0

3 years ago