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4 years ago

____________________ when objects collide, the total initial momentum equals the total final momentum

Physics

2 answers:

Karo-lina-s [1.5K]4 years ago

8 0

Its called law of conservation momentum

Alexus [3.1K]4 years ago

7 0

The law of conservation of linear momentum says that when objects collide, the total initial momentum equals the total final momentum.
The momentum of a body is given by the product of mass and velocity of the object.

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Zigmanuir [339]

In energy point of view, the larger stone had more potential energy before dropping. impacting the water, the larger one, having more kinetic energy which changed from potential energy, tranfered energy to the water and formed wave. the amplitude of the wave indicate the energy of the wave. more energy more amplitude.

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3 years ago

An experimenter using a gas thermometer found the pressure at the triple point of water (0.01°C) to be 4.80 × 10⁴ Pa and the pre

irakobra [83]

Answer:

$T = -282.33^o C$

Explanation:

As we know that the relation between temperature and pressure is a linear relation

so we have

$P - P_o = \frac{P_1 - P_o}{T_1 - T_o} (T - T_o)$

here we know that

$P_1 = 6.50 \times 10^4$

$P_o = 4.80 \times 10^4$

$T_1 = 100^o C$

$T_o = 0.01^o C$

now we will have

$P - 4.80 \times 10^4 = \frac{(6.50 - 4.80)\times 10^4}{100 - 0.01}(T - 0.01)$

$P = 4.80 \times 10^4 + 170.02(T - 0.01)$

now if P = 0

then we will have

$0 = 4.80 \times 10^4 + 170.02(T - 0.01)$

$T = -282.33^o C$

7 0

3 years ago

NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant

Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8 / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²