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4 years ago

List five exothermic reactions that are going on around you

Physics

1 answer:

Stells [14]4 years ago

7 0

1) exhalation

2)neutralization reaction of antacids

3)washing clothes with detergents release a small amount of heat

4) degradation of biowaste

5) adding water to calcium oxide

Hope this answer helps you...

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A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

3 0

4 years ago

Read 2 more answers

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

Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thru

suter [353]

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

Determine the Thrust developed

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D ) = 2.6 m

first step : calculate the area of the duct

A = π/4 * D^2

= π/4 * ( 2.6)^2 = 5.3092 m^2

next : calculate the velocity of propeller

Punit = (A*v*β ) / 2 * V^2 ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

= ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

Finally determine the thrust developed

F = Punit / V2

= (1900 * 10^3) / ( 8.9480)

= 212.3373 kN

8 0

3 years ago

FONTE: JBC w our tree to eat por Pirage B097 VO DED expleyo 2. How do you feel about it? (Are you happy? A little bit sad? Conte

xxTIMURxx [149]

Answer:

sorry kilqngqn kolqng ng point

Explanation:

sorry tlaga please pa heart nlang please

5 0

3 years ago

How much work does a student (m = 60 kg) do when he Climb a tower 80 m high?

gtnhenbr [62]

Answer:

W = 47040 J

Explanation:

Given that,

The mass of a student, m = 60 kg

Height of the tower, h = 80 m

We need to find the work done in climbing the tower. The work done is given by :

W = mgh

So,

W = 60 × 9.8 × 80

W = 47040 J

So, the required work done is 47040 J.

7 0

3 years ago