If a bird lands on a tree, is that an example of speed, velocity, or acceleration?

Transfer of thermal or heat energy through "touch"

MaRussiya [10]

It is conduction. Rhejrjrjejehrhrhvrbrgr

4 0

2 years ago

An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t

Vlad1618 [11]

Answer:5

Explanation:

Given

speed of object $v=1\ m/s$

radius of circle $r=1\ m$

Force towards the center $F=1\ N$

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

$W=F\cdot s\cos 90$

$W=0$

4 0

3 years ago

Capacitor 2 has half the capacitance and twice the potential difference as capacitor 1. What is the ratio (U_{\rm C})_1/\,(U_{\r

IrinaK [193]

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V is the potential difference

Calling $C_1$ the capacitance of capacitor 1 and $V_1$ its potential difference, the energy stored in capacitor 1 is

$U=\frac{1}{2}C_1 V_1^2$

For capacitor 2, we have:

- The capacitance is half that of capacitor 1: $C_2 = \frac{C_1}{2}$

- The voltage is twice the voltage of capacitor 1: $V_2 = 2 V_1$

so the energy stored in capacitor 2 is

$U_2 = \frac{1}{2}C_2 V_2^2 = \frac{1}{2}\frac{C_1}{2}(2V_1)^2 = C_1 V_1^2$

So the ratio between the two energies is

$\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}$

4 0

2 years ago

What best describes the motion of particles in the solid state

dimulka [17.4K]

The particles vibrate against each other in a soild state........i think

3 0

2 years ago

slader A transcontinental flight of 4890 km is scheduled to take 40 min longer westward than eastward. The airspeed of the airpl

11111nata11111 [884]

Answer:

$v_{s}=65.2km/h$

Explanation:

Given data

Flight distance S=4890 km

Time difference Δt=t₂-t₁=40 min

Air speed of plane=980 km/h

To find

Speed of jet stream

Solution

When moving in the same direction as the jet stream time taken as t₁=d/(v+vs),v is velocity of plane and vs is velocity of plane

While moving in opposite direction t₂=d/(v+vs)

So

$t_{2}-t_{1}=\frac{d}{(v-v_{s}) } - \frac{d}{(v+v_{s}) }\\t_{2}-t_{1}=\frac{d(v+v_{s})-d(v-v_{s})}{(v-v_{s})(v+v_{s})} \\t_{2}-t_{1}=\frac{2dv_{s}}{(v)^{2} -(v_{s})^{2} }\\0.666667h=\frac{2(4890km)v_{s}}{(980km/h)^{2} -(v_{s})^{2} }\\0.666667((980km/h)^{2} -(v_{s})^{2})=9780v_{s}\\640267-0.666667(v_{s})^{2}-9780v_{s}=0\\0.666667(v_{s})^{2}+9780v_{s}-640267=0$

Apply quadratic formula to solve for vs

So

$v_{s}=65.2km/h$

4 0

3 years ago