Can someone help me in the 26th one pleaseeee

A frictionless cart attached to a spring vibrates with amplitude A. Part A Determine what fraction of the total energy (Etot) of

Komok [63]

Answer:

the fraction of the total energy (Etot) of the cart-spring system is elastic potential energy (Us) is 0.25

the fraction is kinetic energy (K) when the cart is at position x=A/2 is 0.75

Explanation:

the solution is in the attached Word file

Download docx

5 0

3 years ago

How do you calculate joules? someone please help me I beg

Gnoma [55]

Answer:

<h2>Joules = Watts × Seconds</h2>

Explanation:

If you multiply the number of watts by the number of seconds, you'll end up with joules.

5 0

3 years ago

If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch

lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

Φ = B*A*cos(θ)

ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

θ_1 = 90° - ∅

θ_1 = 90° - 20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

θ_2 = ∅

θ_2 = 20°

- Hence, the change in flux can be calculated:

ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

ΔΦ = -3.39*10^-6

8 0

4 years ago

A satellite is put in a circular orbit about Earth with a radius equal to 35% of the radius of the Moon's orbit. What is its per

Debora [2.8K]

Answer:

0.21 lunar month

Explanation:

the radius of moon = r₁

time period of the moon = T₁ = 1 lunar month

The radius of the satellite = 0.35 r₁

Time period of satellite

The relation between time period and radius

$T\ \alpha\ \sqrt{r^3}$

now,

$\dfrac{T_2}{T_1}=\dfrac{\sqrt{r_2^3}}{\sqrt{r_1^3}}$

$\dfrac{T_2}{T_1}=\dfrac{\sqrt{0.35^3r_1^3}}{\sqrt{r_1^3}}$

$\dfrac{T_2}{1}=\sqrt{0.35^3}$

T₂ = 0.21 lunar month

hence, the time period of revolution of satellite is equal to 0.21 lunar month

6 0

4 years ago

A high-speed train travels 480 km from Paris to Grenoble, France, at an average speed of 180 km/h. How long will it take to reac

KengaRu [80]

The number of hours it will take for the high-speed train to reach Grenoble is the ratio between the distance to the speed. For this item, the equation would be,
t = d / s
Substituting,
t = 480 km / 180 km/h = 2.7 h
Thus, it will take approximately 2.7 hours to reach Grenoble.

7 0

3 years ago

Can someone help me in the 26th one pleaseeee​

Can someone help me in the 26th one pleaseeee