All categories

3 years ago

What is the name of the spacecraft that has travels the farthest away from earth

Physics

2 answers:

luda_lava [24]3 years ago

8 0

NASA Voyager 1 space craft

Hitman42 [59]3 years ago

5 0

N.A.S.A voyager 1 space craft

You might be interested in

A helicopter is hovering above the ground. Jim reaches out of the copter (with a safety harness on) at 180 m above the ground. A

vesna_86 [32]

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

6 0

3 years ago

A 7.00-cm-long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wi

Anvisha [2.4K]

Answer:

(a) 1.075 N

(b) 4.254 T

Explanation:

(a)

From the question,

Power = Force×Velocity.

P = Fv................... Equation 1

Where P = Power dissipated in the circuit, F = Pulling force of the wire, v = speed of the wire.

make F the subject of the equation

F = P/v................. Equation 2

Given: P = 4.30 W, v = 4.0 m/s.

Substitute into equation 2

F = 4.30/4

F = 1.075 N.

(b)

Applying,

F = BILsinФ.............. Equation 2

Where B = Strength of the magnetic field, I = current in the wire, L = Length of the wire, Ф = angle between the wire and the magnetic field.

make B the subject of the equation

B = F/ILsinФ................... Equation 3

But,

P = I²R...................... Equation 4

Where R = resistance of the wire.

make I the subject of the equation

I = √(P/R)............... Equation 5

Given: P = 4.30 W, R = 0.330 Ω

Substitute into equation 5

I = √(4.3/0.33)

I = √13.03

I = 3.61 A.

Also given: L = 7 cm = 0.07 m, Ф = 90°

Substitute into equation 3

B = 1.075/(0.07×3.61×sin90)

B = 1.075/0.2527

B = 4.254 T.

4 0

3 years ago

A 12 cm diameter piston-cylinder device contains air at a pressure of 100 kPa at 24oC. The piston is initially 20 cm from the ba

lina2011 [118]

Answer:

ΔQ = 0.1 kJ

$\mathbf{v_f = 1.445*10^{-3} m^3}$

$\mathbf{P_f = 156.5 \ kPa}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

Explanation:

GIven that:

Diameter of the piston-cylinder = 12 cm

Pressure of the piston-cylinder = 100 kPa

Temperature =24 °C

Length of the piston = 20 cm

Boundary work ΔW = 0.1 kJ

The gas is compressed and The temperature of the gas remains constant during this process.

We are to find ;

a. How much heat was transferred to/from the gas?

According to the first law of thermodynamics ;

ΔQ = ΔU + ΔW

Given that the temperature of the gas remains constant during this process; the isothermal process at this condition ΔU = 0.

Now

ΔQ = ΔU + ΔW

ΔQ = 0 + 0.1 kJ

ΔQ = 0.1 kJ

Thus; the amount of heat that was transferred to/from the gas is : 0.1 kJ

b. What is the final volume and pressure in the cylinder?

In an isothermal process;

Workdone W = $\int dW$

$W = \int pdV \\ \\ \\W = \int \dfrac{nRT}{V}dv \\ \\ \\ W = nRt \int \dfrac{dv}{V} \\ \\ \\ W = nRT In V |^{V_f} __{V_i}} \\ \\ \\ W = nRT \ In \dfrac{V_f}{V_i}$

Since the gas is compressed ; then $v_f< v_i$

However;

$W =- nRT \ In \dfrac{V_f}{V_i}$

$W =- P_1V_1 \ In \dfrac{V_f}{V_i}$

The initial volume for the cylinder is calculated as ;

$v_1 = \pi r^2 h \\ \\ v_1 = \pi r^2 L \\ \\ v_1 = 3.14*(6*10^{-2})^2*(20*10^{-2}) \\ \\ v_1 = 2.261*10^{-3} \ m^3$

Replacing over values into the above equation; we have :

$100 = - ( 100*10^3 *2.261*10^{_3}) In (\dfrac{v_f}{v_i}) \\ \\ - In (\dfrac{v_f}{v_i})= \dfrac{100}{(100*10^3*2.261*10^{-3})} \\ \\ - In \ v_f + In \ v_i = \dfrac{100}{226.1} \\ \\ - In \ v_f = - In \ v_i + \dfrac{100}{226.1} \\ \\ - In \ v_f = - In (2.261*10^{-3} + \dfrac{100}{226.1 } \\ \\ - In \ v_f = 6.1 + 0.44 \\ \\ - In \ v_f = 6.54 \\ \\ - In \ v_f = -6.54 \\ \\ v_f = e^{-6.54} \\ \\ \mathbf{v_f = 1.445*10^{-3} m^3}$

The final pressure can be calculated by using :

$P_1V_1 = P_2V_2 \\ \\ P_iV_i = P_fV_f$

$P_f =\dfrac{P_iV_i}{V_f}$

$P_f =\dfrac{100*2.261*10^{-3}}{1.445*10^{-3}}$

$P_f = 1.565*10^2 \ kPa$

$\mathbf{P_f = 156.5 \ kPa}$

c. Find the change in entropy of the gas. Why is this value negative if entropy always increases in actual processes?

The change in entropy of the gas is given by the formula:

$\Delta S=\dfrac{\Delta Q}{T}$

where

T = 24 °C = (24+273)K

T = 297 K

$\Delta S=\dfrac{-100 \ J}{297 \ K}$

ΔS = -0.337 J/K

The value negative is due to the fact that there is need to be the same amount of positive change in surrounding as a result of compression.

4 0

3 years ago

he summit of Mount Everest is 8850 m above sea level. (a) How much energy would a 78 kg climber expend against the gravitational

natali 33 [55]

Answer:

The energy which is expend by the person is $U=6.7\times 10^{6}J$ .

Explanation:

The formula for the potential energy due to the height of the object is as follows as;

U= mgh

Here, h is the height of the person, m is the mass of the object and g is the acceleration due to gravity.

It is given in the problem that a person is 8850 m above the sea level. 78 kg climber would expend energy against the gravitational force in climbing to the summit from sea level. Here, the person has potential energy as he is above the sea level.

Put h= 8850 m, m= 78 kg and $g = 9.8 m s^{-2}$ and in the above expression to get the value of energy.

U= (78)(9.8)(8850)

U= 6764940 J

$U=6.7\times 10^{6}J$

Therefore, the energy which is expend by the person is $U=6.7\times 10^{6}J$ .

8 0

3 years ago

Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station 100 m diameter rotatin

Zielflug [23.3K]

Given Information:

Diameter = d = 100 m

Required Information:

Period = T = ?

Answer:

Period = T = 14.2 seconds

Explanation:

We know that a station revolving at an angular velocity ω, have an acceleration given by

α = ω²r

Where ω is angular velocity and r is the radius of cylindrical space station.

Normal gravity means α = g = 9.8 m/s²

ω² = α/r

ω = √(α/r)

The radius is given by

r = d/2

r = 100/2

r = 50 m

ω = √(9.8/50)

ω = 0.4427 rad/sec

We also know that

ω = 2πf

f = ω/2π

f = 0.4427/2π

f = 0.0704 rev/sec

Finally time period is given by

T = 1/f

T = 1/0.0704

T = 14.2 sec

Therefore, the rotation period is 14.2 seconds.

3 0

3 years ago