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3 years ago

Give one chemical and physical property of wood.

1 answer:

6 0

Chemical 1. Wood is burnable , it can be burned , its most of the time brown

Physical 1. Wood is most of the time hard , sometimes soft

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A tiger runs one kilometer. The tiger does 15,000 J of work and has a total power output of 500 W. How many minutes does it take

defon

Power is the amount of energy consumed per unit time. Having no direction, it is a scalar quantity. As is implied by the equation for power, a unit of power is equivalent to a unit of work divided by a unit of time. The formula would be as follows:

P = W/t

We calculate as follows:

500 W = 15000 J / t
t = 30 s

3 0

3 years ago

What would happen when there is a collision between two objects having unequal masses

Mashcka [7]

Answer:

Newton's Laws Applied to Collisions. Newton's third law of motion is naturally applied to collisions between two objects. ... Thus, if the colliding objects have unequal mass, they will have unequal accelerations as a result of the contact force that results during the collision.

Explanation:

3 0

3 years ago

A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

igomit [66]

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

3 0

3 years ago

Saturated water vapor at 200 kPa is condensed into a saturated liquid via a constant-pressure process inside of a piston-cylinde

evablogger [386]

Answer:

The process is not possible

Explanation:

if we want to determine if the process is possible , we can check with the second law of thermodynamics

ΔS≥ ∫dQ/T

for a constant temperature process ( condensation)

ΔS≥ 1/T ∫dQ

and from the first law of thermodynamics

ΔH = Q - ∫VdP , but P=constant → dP=0 → ∫VdP=0

Q=ΔH

then

ΔS≥ΔH/T

from steam tables

at P= constant = 200 Kpa → T= 120°C = 393 K

at P= constant → H vapor = 2201.5 kJ/kg , H liquid = 1.5302 kJ/kg

, S vapor= 7.1269 kJ/kg , S liquid 1.7022 kJ/kg

therefore

ΔH = H vapor - H liquid = 2201.5 kJ/kg - 1.5302 kJ/kg = 2199.9698 kJ/kg

ΔS = S vapor - S liquid = 7.1269 kJ/kg - 1.7022 kJ/kg = 5.4247 kJ/kg

therefore since

ΔS required = ΔH/T = 2199.9698 kJ/kg/(393 K)= 5.597 kJ/kg K

and

ΔS= 5.4247 kJ/kg ≤ ΔS required=5.597 kJ/kg K

the process is not possible

5 0

3 years ago

Find the initial velocity of a ball that has a final velocity of 17.9 m/s, west after it accelerated at 63.7 m/s2, west for 0.21

Musya8 [376]

Explanation:

Taking west to be positive, given:

v = 17.9 m/s

a = 63.7 m/s²

t = 0.21 s

Find: v₀

v = at + v₀

17.9 m/s = (63.7 m/s²) (0.21 s) + v₀

v₀ = 4.52 m/s

The initial velocity is 4.52 m/s west.

5 0

3 years ago