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4 years ago

A 100 kg individual consumes 1200 kcal of food energy a day. Calculate

Physics

1 answer:

IrinaK [193]4 years ago

4 0

Answer:

(a) 5142.86 m

(b) 317.5 m/s

Explanation:

m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

(a) Let the altitude be h

Q = m x g x h

504 x 10^4 = 100 x 9.8 x h

h = 5142.86 m

(b) Let v be the speed

Q = 1/2 m v^2

504 x 10^4 = 1/2 x 100 x v^2

v = 317.5 m/s

Let the final temperature is T2.

Q = m x c x (T2 - T1)

504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

T2 = 49.3 degree C

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A vertical spring with k=490N/m is standing on the ground. You are holding a 5.0kg block just above the spring, not quite touchi

WARRIOR [948]

Answer:

Compression in the spring, x = 0.20 m

Explanation:

Given that,

Spring constant of the spring, k = 490 N/m

Mass of the block, m = 5 kg

To find,

Compression in the spring.

Solution,

Since the block is suddenly dropped on the spring gravitational potential energy of block converts into elastic potential energy of spring. Its expression is given by :

$mgx=\dfrac{1}{2}kx^2$

Where

x is the compression in the spring

$x=\dfrac{2mg}{k}$

$x=\dfrac{2\times 5\times 9.8}{490}$

x = 0.20 m

So, the compression in the spring due to block is 0.20 meters.

5 0

3 years ago

All living organisms reproduce by a live functioning baby leaving the mother's body.

algol13

Answer:

there is no question

Explanation:

4 0

3 years ago

In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radi

Vitek1552 [10]

Answer:

(a) 1,569.63 N

(b) 195,933.99 Pa

(d) 1,647.47 kg

Explanation:

Let the cross-sectional area (CSA) of small piston = A₁

Let the cross-sectional area (CSA) of the bigger piston = A₂

Let the Force applied at the smaller piston = F₁

Let the Force applied at the bigger piston = F₂

The principle of hydraulic lift assumes the that the fluid is in-compressible, resulting to a constant pressure system.

F₁/A₁ =F₂/A₂----------------------------------------------------------- (1)

(a) F₁= F₂ xA₁ /A₂

F₂ = 13,300 N

A₁ = π r₁²

=π x (0.0505)²

= 0.008011 m²

A₂= π r₂²

=π x (0.147)²

= 0.06788 m²

Substituting into (1)

F₁ = 13,300 x 0.008011/0.06788

= 1,569.6272

≈ 1,569.63 N

(b) Air pressure = Force/Area

= F₁/A₁

= 1,569.6272/ 0.008011

= 195,933.99 Pa

Workdone = force x distance----------------------------------------- (2)

force = pressure × area

distance = volume/area from

Substituting into (2)

Workdone = pressure × volume.

As pressure and volume are equal for each piston, work must also be equal

(d) F₂ = F₁ x A₂/ A₁---------------------------------------------------- (3)

A₁ = π r₁²

=π x (0.079)²

= 0.01960 m²

A₂= π r₂²

=π x (0.353)²

= 0.3914 m²

Substituting into (2)

F₂= 825 x 0.3914/ 0.01960

= 16,474.7448

≈ 16,474.74 N

= 1,647.47 kg

3 0

3 years ago

Why do blades come in different lengths in a jig saw sander?

trasher [3.6K]

Answer:

To determine the minimum blade length, add 1" to the workpiece thickness. One type of material, and some materials can be cut by more than one type of blade. No matter the material, there's likely a jigsaw blade designed specifically for. Armed with the right blade, follow these pointers to make your work go (and cut) .

Explanation:

6 0

3 years ago

What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?Use 1.67×

vazorg [7]

The electric force on the proton is:

F = Eq

F = electric force, E = electric field strength, q = proton charge

The gravitational force on the proton is:

F = mg

F = gravitational force, m = proton mass, g = gravitational acceleration

Since the electric force and gravitational force balance each other out, set their magnitudes equal to each other:

Eq = mg

Given values:

q = 1.60×10⁻¹⁹C, m = 1.67×10⁻²⁷kg, g = 9.81m/s²

Plug in and solve for E:

E(1.60×10⁻¹⁹) = 1.67×10⁻²⁷(9.81)

E = 1.02×10⁻⁷N/C

5 0

3 years ago