All categories

3 years ago

Can I apply my homework questions about physic?

Physics

2 answers:

Anuta_ua [19.1K]3 years ago

5 0

I’m not sure how this app works but I want points lololololo but I think the answer is yes you can

docker41 [41]3 years ago

3 0

Answer:

mhm

Explanation:

You might be interested in

What exited kaveh pahlevan about his work?

USPshnik [31]

Answer:

i dont know

Explanation:

i dont know since you didn't provide something to base off of

7 0

2 years ago

An astronaut having mass 320 kg with equipment included is attempting an untethered space walk. The astronaut is initially at re

ExtremeBDS [4]

This can be solved using momentum balance, since momentum is conserved, the momentum at point 1 is equal to the momentum of point 2. momentum = mass x velocity
m1v1 = m2v2
(0.03kg x 900 m/s ) = 320(v2)
v2 = 27 / 320
v2 = 0.084 m/s is the speed of the astronaut

7 0

3 years ago

An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an

Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of $\gamma$

Using formula of relativistic energy

$E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

Put the value into the formula

$1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

$(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}$

Here, $\gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

$\gamma-1=0.01953$

$\gamma=0.01953+1$

$\gamma=1.01953$

We need to calculate the length

Using formula of length

$L'=\dfrac{L}{\gamma}$

Put the value into the formula

$L'=\dfrac{4}{1.01953}$

$L'=3.92\ m$

Hence, The length of the tube is 3.92 m.

8 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

6 0

3 years ago

~~~~~NEED HELP ASAP~~~~~

Romashka-Z-Leto [24]

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

8 0

2 years ago