1. A car begins at a speed of 3 m/s and accelerates at 2 m/s2

Two identical objects, A and B, are sitting on a table. If the net force on object A is 5 N and the net force on object B is 10

sveta [45]

If the net force on object A is 5 N and the net force on object B is 10 N, then object B will accelerate more quickly than object A provided the mass of both objects are same.

Answer: Option C

<u>Explanation: </u>

According to Newton’s second law of motion, any external force applied on an object is directly proportional to the mass and acceleration of the object. In order to state this law in terms of acceleration, it is stated that acceleration exhibited by any object is directly proportional to the net force applied on the object and inversely proportional to the mass of the object as shown below:

$\text {Acceleration of the object } \propto \frac{\text {Net force on the object}}{\text {Mass of the object}}$

So if two objects A and B are identical which means they have same mass, then the acceleration attained by the object will be directly proportionate to the net forces exerted on the objects only.

Thus if the force applied is more for one object, then the object will be exhibiting more acceleration compared to the other one. So as object B is experiencing a net force of 10 N which is greater than the net force experiences by object A, then the object B will be accelerating more quickly compared to the object A's acceleration.

7 0

3 years ago

Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in

sattari [20]

Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, $P_{atm}$ = 101 Kpa

$P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}$

$P_{gauge} = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3$

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

$P_{gauge} = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge} = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge} = 110,298.26 pa$

Therefore, the gauge pressure of air is 110 kpa

4 0

3 years ago

URGENT!!

irinina [24]

Answer:

Explanation:La ecuación de Van der Waals es una ecuación de estado de un fluido compuesto de partículas con un tamaño no despreciable y con fuerzas intermoleculares, como las fuerzas de Van der Waals. La ecuación, cuyo origen se remonta a 1873, debe su nombre a Johannes van der Waals, quien recibió el premio Nobel en 1910 por su trabajo en la ecuación de estado para gases y líquidos, la cual está basada en una modificación de la ley de los gases ideales para que se aproxime de manera más precisa al comportamiento de los gases reales al tener en cuenta su tamaño no nulo y la atracción entre sus partículas.

3 0

3 years ago

I NEED HELP!! PLEASE HELP ME

vesna_86 [32]

Answer:

but I can tomorrow if you have time can you come to the meeting tonight but yyyy the person who is this and what

Explanation:

gyyyyyyyyyyyyyyyyy the number of the person who is this and what is the

8 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

so

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

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0

3 years ago