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2 years ago

What objects have orbital paths? A. the sun and stars

Physics

1 answer:

suter [353]2 years ago

7 0

Answer:

The planets and moons.

Explanation:

Planets follow an elliptical path around the sun (kinda oval shaped). Moons do the same to planets.

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Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold

arsen [322]

Answer:

Resistance of gold wire, $R=1977 \times 10^3 ohm$

Explanation:

In this question we have given

Density of gold, $d=19.3\times 10^3 \frac{kg}{m^3}$

resistivity of gold, $r=2.44\times 10^{-8} ohm.m$

Length of wire, $L= 2.05 km$

Temperature, $T= 20^oC$

We know that relation between volume and density is given as

$Density= \frac{mass}{Volume}$

Therefore, volume occupied by one gram gold is given as,

$V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3$ .........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

$V=\pi \times r^2 \times L$ ......(2)

Here,

$A= \pi \times r^2$ ...........(3)

here A is the cross sectional area of cylendrical gold wire

From equation 2 and 3

we got

$V=A \times L$ ...............(4)

on comparing equation 1 and equation 4, we got,

$A \times L=5.181\times 10^{-8} m^3$

$A=\frac{5.181\times 10^{-8} m^3}{2050 m}$

$A=2.53\times 10^{-11}m^2$

we know that resistance and resistivity are related by following formula,

$Resistance = resistivity\times \frac{L}{A}$ ................(5)

Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

$R=1977 \times 10^3 ohm$

Therefore resistance of gold wire, $R=1977 \times 10^3 ohm$

7 0

3 years ago

One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by w

Valentin [98]

Answer:

ω = 10.75 rad/s

Explanation:

The stone leaves the circular path with a horizontal speed

v₀ = vt = R*ω

ω = v₀ / R

we are given that R = x / 25.7 so ω = 25.7*v₀ / x

Kinematics gives x = v₀*t

With this substitution for x the expression for ω becomes ω = 25.7 / t

Kinematics also gives for the vertical displacement y that

y = v₀y*t + 0.5*ay*t²

we know that v₀y = 0 m/s

since the stone is launched horizontally, so that

y = 0.5*ay*t² ⇒ t = √(2*y / ay)

Using this result for t in the expression for ω and assuming that upward is positive, we get

ω = 25.7 / √(2*y / ay)

⇒ ω = 25.7 / √(2*(-28) / (-9.8))

⇒ ω = 10.75 rad/s

5 0

3 years ago

In a thundercloud there may be an electric charge of 24 C near the top of the cloud and −24 C near the bottom of the cloud. If t

lidiya [134]

Answer:

Electric force, $F=1.29\times 10^6\ N$

Explanation:

Given that,

Charge 1, $q_1=24\ C$

Charge 2, $q_2=-24\ C$

Distance between charges, d = 2 km

The electric force is given by :

$F=k\dfrac{q_1q_2}{d^2}$

k is the electrostatic constant

$F=8.98755\times 10^9\times \dfrac{(24)^2}{(2\times 10^3)^2}$

F = 1294207.2 N

$F=1.29\times 10^6\ N$

Hence, this is the required solution.

5 0

4 years ago

An inquisitive physics student and mountian climber climbs a 43.6 m cliff that overhangs a calm pool of water. He throws two sto

USPshnik [31]

Answer:

Explanation:

What we are basically looking for here is how long it takes the first stone to hit the water. We have everything we need to figure that out. We will use the equation

Δx = . Filling in, we will solve for t, the time is takes the first stone to hit the water (which is the same for both since they both hit the water at the same time):

which is a quadratic that we will have to factor. Get it into standard form, setting it equal to 0:

and factor to get that

t = 3.2 s and t = -2.8 s

Since time can't ever be negative, it takes 3.2 s for the stones to hit the water.

4 0

3 years ago

18 of 25

anyanavicka [17]

Answer:

Explanation:

Cause its true

7 0

3 years ago

Read 2 more answers