Which of the following is transferred in order for static electricity to occur?

A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of

dimulka [17.4K]

Answer:

True

Explanation:

When a satellite is orbiting the earth, the centripetal force is balanced by the gravitational force as :

$\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{GM}{r}}$ ...........(1)

Where

M is the mass of the earth

m is the mass of the planet

From equation (1), the speed of the satellite depends only on the mass of the earth and the orbital radius.

So, If a payload of material is added until it doubles the satellite's mass, the earth's pull of gravity on this satellite will double but the satellite's orbit will not be affected. It is true.

6 0

3 years ago

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You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12-m-hi

Vlada [557]

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

$E_A} =E_{B}$

Spring potential energy =potential energy due to elevation

0.5*k*x²= mg $(h_{B}-h_{A} )$ =mgh

0.5*k*2.3²= 430*9.81*6

k=9568.92 N/m

For safety reason

k"=1.13 *k= 1.13*9568.92

k"=10812.88 N/m

agsin from conservation of energy

$E_A} =E_{C}$

spring potential energy=change in kinetic energy

0.5*k"*x²=0.5*m* $V_{max}^{2}$

10812.88 *2.3²=430* $V_{max}^{2}$

$V_{max}$ =11.53 m/s

5 0

4 years ago

According to Newton, there were two things needed for an object to fall around or orbit the earth. Label the diagram below with

iVinArrow [24]

Centripetal force and centrifugal force

4 0

3 years ago

What is the acceleration of a 4,000 kg car pushed with a<br> force of 12,000 N?

jek_recluse [69]

Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

8 0

3 years ago

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A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and

cricket20 [7]

Answer:

The frequency $f$ = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = $\frac{d}{2}$ = $\frac{44}{2}$ = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = $\frac{d}{2}$ = $\frac{3.0}{2}$ = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = $\frac{d}{2}$ = $\frac{2.0}{2}$ = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

$P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2$

$pgy_1=\frac{1}{2}pv^2_2 +pgy_2$

$v_2=\sqrt{2g(y_1-y_2)}$

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = $\sqrt{2*9.8*(0.35-0)}$

v₂ = $\sqrt {6.86}$

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

$v_3 = \frac{v_2r_2^2}{v_3^2}$

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = $(2.62 m/s)*$ $(\frac{1.5mm^2}{1.0mm^2} )$

v₃ = 0.0589 m/s

∴ velocity of the fluid in the cylinder = 0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

$f=\frac{v_s}{4(h-v_3t)}$

where;

$v_s$ = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

$f=\frac{343}{4(0.40-(0.0589)(0.4)}$

$f= \frac{343}{0.6576}$

$f$ = 521.59 Hz

∴ The frequency $f$ = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

$\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})$

$\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)$

$\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}$

where;

$v_s$ (velocity of sound) = 343 m/s

v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

$\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}$

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.

8 0

3 years ago