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2 years ago

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On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

finds that, under the same circumstances, the stone returns to his hand in 2T. In both cases, he throws the stone with the same initial velocity and it feels negligible air resistance. The acceleration due to gravity on planet X (in terms of g) is _______.a) g/. b) g/4. c) 2g. d) g/2. e) g.

1 answer:

Liula [17]2 years ago

4 0

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

v = v₀ - a t

a = (v₀- v) / T

On planet X

v = v₀ - a' t’

a ’= (v₀-v) / 2T

Let's substitute the land values in plot X

a’= a / 2

Now let's use Newton's second law

W = ma

m g = m a

a = g

We substitute

a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

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Quinn is testing the motion of two projectiles x and y by shooting them from a sling shot. What can we say best describes the mo

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Explanation:

A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path subjected to gravity force only.

Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.

In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.

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Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

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2 years ago

A force of 200 N is applied on small piston of a pascal press. What would be the

VladimirAG [237]

Answer:

The force applied on the big piston is 1306.67 N

Explanation:

Given;

force applied on small piston, F₁ = 200 N

diameter of the small piston, d₁ = 4.37 cm

radius of the small piston, r₁ = d₁/2 = 2.185 cm

Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²

Area of the big piston, A₂ = 98 cm²

The pressure of the piston is given by;

$P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}$

Where;

F₂ is the force on big piston

$F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N$

Therefore, the force applied on the big piston is 1306.67 N

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1 year ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

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2 years ago

A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t

arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vector $n=-0.60\hat{i}-0.080\hat{j}$

We need to calculate the torque on the loop

Using formula of torque

$\tau=NIAB\sin\theta$

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

$\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}$

$\tau=4.7\times10^{-4}\ N-m$

Hence, The magnitude of the torque on the loop due to the magnetic field is $4.7\times10^{-4}\ N-m$ .

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2 years ago