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diamong [38]
3 years ago
9

What do the planets Saturn, Neptune, and Uranus have in common?

Physics
1 answer:
NemiM [27]3 years ago
5 0

Answer:

they are all outer planets

You might be interested in
A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat
Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg  

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

m g =\dfrac{mv^2}{r}

9.81 =\dfrac{v^2}{0.675}

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)

where, I =\dfrac{2}{5}mr^2

\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)

0.7 u^2 + g H = 0.7 v^2 + g(2R)

0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial  speed is 3.35 m/s

8 0
2 years ago
A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th
kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

P_1, V_1, T_1 & P_2, V_2, T_2

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container  therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

-Q=W+U_1----1

Q=-W+U_2-----2

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

U_1 & U_2 change in internal Energy of gas

Thus from 1 & 2 we can say that

U_1+U_2=0

n_1c_v(T-T_1)+n_2c_v(T-T_2)=0

T(n_1+n_2)=n_1T_1+n_2T_2

T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}

where n_1=\frac{P_1V_1}{RT_1}

n_2=\frac{P_2V_2}{RT_2}

T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}

T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}

and P=\frac{P_1V_1+P_2V_2}{V_1+V_2}

6 0
3 years ago
What does this equation mean FeDe=FrDr
rjkz [21]
Not exactly the best way to describe it but, it is used to calculate resistance of a lever as in the use of a pry bar or pulley. Technology used to increase output with little input.
3 0
3 years ago
Caleb is filling up water balloons for the Physics Olympics balloon tosscompetition. Caleb sets a 0.50-kg spherical water balloo
Mashcka [7]

a)

• P = F/A

P = pressure = 630 N/m^2

F = force

A = area

F = mg = 0.50 kg x 9.8 m/s^2 = 4.9 N

m= mass

g= gravity

P = F/A

A = F/P

A = 4.9 N / 630 N/m^2 = 7.778 x 10^-3 m^2

b)

• Area of a circle = pi* radius ^2

7.778 x 10^-3 m^2 = pi* radius ^2

√(7.778 x 10^-3 m^2 / pi ) = radius

radius = 0.04976 m

Answers:

a ) 7.778 x 10^-3 m^2

b) 0.04976 m

8 0
1 year ago
Two identical tiny spheres of mass m =2g and charge q hang from a non-conducting strings, each of length L = 10cm. At equilibriu
Citrus2011 [14]

Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

F_y:  T_y - W = 0\\T_y = m*g = 0.002 kg *9.81m/s^2 = 0.01962 N

T_y = T_*cos(50)\\T = \frac{T_y}{cos(50)} = 0.0305 N

T_x = T*sin(50) = 0.0234 N

The electric force is given by the expression:

F = k*\frac{q_1*q_2}{r^2}

In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

And k is the coulomb constan equal to 9 *10^9 N*m^2/C^2. q1 y q2 is the charge of each particle, in this case, they are equal.

F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

8 0
3 years ago
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