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2 years ago

10

A block of copper of density 8.9g/cm3 measures 5 cm by 3 cm by 2 cm. Given that the force of gravity is 10N/kg-1, determine. a)

The maximum pressure b) The minimum pressure exerted on a horizontal surface.

1 answer:

kow [346]2 years ago

8 0

Answer:

hajdbslalbsgsvzblalsgdgxbzhagsg

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Explain how a book can have energy even if it is not moving.

frutty [35]

Can something have energy even if it's not moving?
All moving objects have kinetic energy. When an object is in motion, it changes its position by moving in a direction: up, down, forward, or backward. ... Potential energy is stored energy. Even when an object is sitting still, it has energy stored inside that can be turned into kinetic energy (motion).

Does a book at rest have energy?
A World Civilization book at rest on the top shelf of a locker possesses mechanical energy due to its vertical position above the ground (gravitational potential energy).

Does a book lying on a table have energy?
The book lying on a desk has potential energy; the book falling off a desk has kinetic energy.

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2 years ago

Un joven pelotero llamado Saúl en su primer año de ser firmado en grandes ligas, batea varias veces de cuadrangular, el segundo

Brilliant_brown [7]

Answer:

135-15=120

120÷3=40

40+40+(10)+40+(5)=135

8 0

3 years ago

Read 2 more answers

A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.

Simora [160]

Answer:

$\mu_k=\frac{a}{g}$

Explanation:

The force of kinetic friction on the block is defined as:

$F_k=\mu_kN$

Where $\mu_k$ is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:

$N=mg\\F_k=F=ma$

Replacing this in the first equation and solving for $\mu_k$ :

$ma=\mu_k(mg)\\\mu_k=\frac{a}{g}$

6 0

3 years ago

Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m

bearhunter [10]

Answer:

(0, πR/4)

Explanation:

The linear mass density (mass per length) is λ = λ₀ sin θ.

A short segment of arc length is ds = R dθ.

The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

4 0

3 years ago

The total surface area of the human body is 1.20 m2 and the surface temperature is 30∘C=303∘K. If the surroundings are at a temp

zalisa [80]

To solve this problem we must consider the expressions of Stefan Boltzmann's law for which the rate of change of the radiation of energy H from a surface must be

$H = Ae\sigma T^4$

Where

A = Surface area

e = Emissivity that characterizes the emitting properties of the surface

$\sigma$ = Universal constant called the Stefan-Boltzmann constant $(5.67*10^{-8}m^{-2}K^{-4})$

T = Absolute temperature

The total heat loss would be then

$Q = H_2 -H_1$

$Q =Ae\sigma T_2^4-Ae\sigma T_1^4$

$Q = Ae\sigma (T_2^4-T_1^4)$

$Q = (1.2)(1)(5.67*10^{-8})(303^4-280^4)$

$Q = 155.29J$

Therefore the net rate of heat loss from the body by radiation is 155.29J

8 0

3 years ago