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2 years ago

Which undergoes greater acceleration: an airplane that goes from 1000 km/h to 1005 km/h in 10 seconds or a skateboard that

Physics

1 answer:

Alex_Xolod [135]2 years ago

6 0

Answer:

Skateboard

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

The airplane's acceleration is:

a = (1005 km/h − 1000 km/h) / 10 s

a = 0.5 km/h/s

The skateboard's acceleration is:

a = (5 km/h − 0 km/h) / 1 s

a = 5 km/h/s

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A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri

enyata [817]

Answer:

$T=8.1N$

Explanation:

From the question we are told that:

Mass m=0.40

Radius r=1.8m

Angle Beneath the Horizontal \theta =40 \textdegree

Speed v=5.0m/s

The Tension Angle

$\alpha=90-\theta\\\\\alpha=90-40$

$\alpha=50 \textdegree$

Generally the equation for Tension is is mathematically given by

$T=\frac{mv^2}{r}+mgcos \alpha$

$T=\frac{0.40*5^2}{1.8}+0.40*5cos50$

$T=8.1N$

7 0

2 years ago

A parallel-plate capacitor has plates of area 0.30 m2 and a separation of 2.10 cm. A battery charges the plates to a potential d

Dima020 [189]

Answer:

a) 1.26e^-10F

b) 1.47e^-10F

c) 2.39e^-8C 2.89e^-8C

d) E=4500.94N/C

e) E'=5254.23N/C

f) 100.68V

g) 1.65e^-10J

Explanation:

To compute the capacitance we can use the formula:

$C=\frac{k\epsilon_o A}{d}$

where k is the dielectric constant of the material between the plates. d is the distance between plates and A is the area.

(a) Before the material with dielectric constant is inserted we have that k(air)=1. Hence, we have:

$k=1\\A=0.30m^2\\d=0.021m\\e_o=8.85*10^{-12}C^2/(Nm^2)\\\\C=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{0.021m}=1.26*10^{-10}F$

(b) With the slab we have that k=4.8 and the thickness is 4mm=4*10^{-3}m. In this case due to the thickness of the slab is not the same as d, we have to consider the equivalent capacitance of the series of capacitances:

$C=(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_1})\\\\\\C_1=\frac{(1)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{8.5*10^{-3}m}=3.1*10^{-10}F\\\\C_2=\frac{(4.8)(8.85*10^{-12}C^2/(Nm^2))(0.30m^2)}{4*10^{-3}m}=3.186*10^{-9}F\\\\C=1.47*10^{-10}F$

(c)

The charge between the plates for both cases, with the slab is given by:

Q : without the slab

Q': with the slab

$Q=CV=(1.26*10^{-10}F)(190V)=2.39*10^{-8}C\\\\Q'=C'V=(1.47*10^{-10F})(190V)=2.79*10^{-8}C\\$

(d) The electric field between the plate is given by:

$E=\frac{Q}{2\epsilon_o A}$

E: without the slab

E': with the slab

$E=\frac{2.39*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=4500.94N/C\\\\E'=\frac{2.79*10^{-8}C}{2(8.85*10^{-12}C^2/Nm^2)(0.30m^2)}=5254.23N/C\\$

(f) We can assume the system as composed by V=V1+V'+V1 as in (c). By using the equation V=Ed we obtain:

$V=2(4500.94)(8.85*10^{-3}m)+(5254.23)(4*10^{-3}m)=100.68V$

(g) External work is the difference between the energies of the capacitor before and after the slab is placed between the parallels:

$\Delta E=\frac{1}{2}[(1.26*10^{-10}F)(120V)-(1.47*10^{-10})(100.6V)]=1.65*10^{-10}J$

5 0

2 years ago

PLEASE HELPPP!!! D:

pogonyaev

Answer:

a is 4.6 x 10-2m

b is 1x8204m

Explanation:

The fire net is actually going past zero into a negative vertical displacement. Essentially because again, we're saying that why initial equals thirty eight meters. The person is falling down to zero meters. Zero meters is where the fire that is. And then after the person stretches the fire. Yet it has a certain vertical displacement. But this particles placement is negative. So here we're going to choose the negative value. so trying to see how much of the fire net would stretch if the person was lying on tough it and if the person jumped from a height of thirty eight meters. So for here Mass is going to be equal to sixty two kilograms and why initial for party is going to be twenty meters and then why final is going to be negative one point four meters And so in order to find in order to find this is an instance, essentially, this is an instance perspective in the sense that we need this data in order to figure out this during constant so we can see that energy initial equals energy final and then we consider it mg y initial equals mg.

7 0

2 years ago

A race car travels on a straight racetrack with

Julli [10]

Answer:968 m/s

Explanation:

6 0

2 years ago

a ball is thrown upward at 25 m/s from the ground. what is the average velocity and average speed of the ball after 5 seconds?

Phoenix [80]

When the ball reaches the highest point, its velocity will be zero. So final velocity is v=0m/s
Here, the time to reach highest point is equal to time to hit ground from highest point.
Consider upward motion, the acceleration of the ball is a=−g=−9.8m/s
2
(minus for motion opposite to gravity)
If t be the required time.
Using v=u+at,
0=25+(−9.8)t
⟹ t=2.5s
Thus the ball hits the ground after 2.5 seconds after reaching its highest point.

8 0

2 years ago