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2 years ago

NEED ASAP PLEASE

Physics

1 answer:

miskamm [114]2 years ago

6 0

Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.

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What conditions would have to exist in order for a space station to support life

amid [387]

Answer:

Mechanical life support system

Explanation:

Mechanical life support system is one which allows humans to survive on places or conditions which are not natural to their survival so by the help of mechanical devices humans can survive in places like underwater or in space.

The life support system may not only provide water, air and food but it should also keep a proper check on the maintenance of body pressure, temperature, management of the waste product and the absorption or the radiations from the body itself.

Thus mechanical life support system are designed with the hep of high engineering techniques for the safety and security as these are life-critical.

4 0

3 years ago

A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga

pashok25 [27]

A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
$p_i V_i = nRT_i$
where
$p_i=1.0 atm=1.01 \cdot 10^5 Pa$ is the initial pressure of the gas
$V_i$ is the initial volume of the gas
$n=2.3 mol$ is the number of moles
$R=8.31 J/K mol$ is the gas constant
$T_i=240^{\circ}C=513 K$ is the initial temperature of the gas

By re-arranging this equation, we can find $V_i$ :
$V_i = \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3$

2) Now the gas cools down to a temperature of
$T_f = 14^{\circ}C=287 K$
while the pressure is kept constant: $p_f = p_i = 1.01 \cdot 10^5 Pa$ , so we can use again the ideal gas law to find the new volume of the gas
$V_f = \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3$

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
$W=p \Delta V= p(V_f -V_i)$
by using the data we found at point 1) and 2), we find
$W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J$
where the negative sign means the work is done by the surrounding on the gas.

5 0

3 years ago

A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards

Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

$Y = 1.6 X - 0.032x^2$

we know that maximum height occurs is given as

$x = -\frac{b}{2a}$

$y =- \frac{1.6}{2(-0.032)} = 25$

and maximum height is

$y = 1.6\times 25 - 0.032\times 25^2$

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

8 0

3 years ago

Read 2 more answers

A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine

Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = $\frac{4}{3}\pi(\frac{d}{2})^3$

Volume of the balloon = $\frac{4}{3}\pi(\frac{0.296}{2})^3$

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = $\frac{3}{2}\times K_BT$

where,

Boltzmann constant, $K_B=1.3807\times10^{-23}J/K$

Average kinetic energy = $\frac{3}{2}\times1.3807\times10^{-23}\times292$

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = $\frac{3RT}{m}$

where, m is the molar mass of the Helium = 0.004 Kg

rms speed = $\frac{3\times8.314\times292}{0.004}$

rms speed = 1349.35 m/s

5 0

3 years ago

A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio

Fed [463]

Answer:

Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g

Explanation:

The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.

And sum of upward forces = sum of downward forces.

N = mg cos θ

m = 100 g = 0.10 kg

g = acceleration due to gravity = 9.8 m/s²

θ = 15°

N = (0.1×9.8×cos 15°) = 0.946582 N

The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).

For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.

Frictional force = Tension + F

Frictional force = μN

where μ = coefficient of kinetic friction = 0.60

N = normal reaction = 0.9466 N

Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N

The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ

F = (0.10 × 9.8 × sin 15°) = 0.253624 N

Tension in the rope = T = ?

Fr = F + T

T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N

But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.

2T = mg

2 × 0.3144 = 9.8m

m = 0.06416 kg = 64.16 g.

Mass of the container = 30 g

So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g

Hope this Helps!!!

8 0

3 years ago