Answer:
73.6 minutes
Explanation:
relative time = time interval / √(1 - observer velocity² / speed of light²)
we have relative time. we want time interval.
rearrange
time interval = relative time x √(1 - observer velocity² / speed of light²)
convert 85 mins into seconds
85 x 60 = 5100
1.5 x 10⁸ as a number is 150000000
for c = 299 792 458
time interval = 5100 x √(1 - 150 000 000² / 299 792 458²)
for c = 3 x 10⁸
time interval = 5100 x √(1 - 150 000 000² / 300 000 000²)
time interval = 5100 x 0.866
time interval = 4415.71
divide by 60 for back into minutes
time = 73.6 minutes
Answer:
0.098 N
Explanation:
From the question,
Spring scale reading = W-U............... Equation 1
Where W = weight of the cube, U = upthrust.
W = mg
Where m = mass of the cube, g = acceleration due to gravity.
Given: m = 11 g = 0.011 kg, g = 9.8 m/s².
W = 0.011(9.8)
W = 0.1078 N.
From Archimedes principle,
Upthrust = weight of water displaced.
U = (Density of water×volume of metal cube)×acceleration due to gravity.
U = (D×V)g
Given: D = 1000 kg/m², V = 1 cm³ = (1/1000000) = 1×10⁻⁶ m³, g - 9.8 m/s²
U = 1000(9.8)(10⁻⁶)
U = 0.0098 N.
Substitute the value of W and U into equation 1
Reading of the spring scale = 0.1078-0.0098
Reading of the spring scale = 0.098 N
the fossil informed us about the climatic and geographical changes that occurs in the environment over time.
<h3>What the mystery fossil dig site informed us about environment?</h3>
We can infer about the climate and geographical changes in the environment over time at the mystery fossil dig site because the fossil provides us information about the climatic condition as well as the geography of land when this fossil organism were alive.
So we can conclude that the fossil informed us about the climatic and geographical changes that occurs in the environment over time.
Learn more about fossil here: brainly.com/question/11829803
#SPJ1
Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Answer:
it is he sound of source it uses ur ears to hear something unectpected can u give me brainly plz
Explanation:
