Answer:
0.45 miles
Explanation:
since, car will be running on the edge of track, distance covered in one lap by car will be equal to circumference of track.\
we will use value of pi as 22/7
Circumference for any circular shape is given by 

Hence, in one lap
distance will be covered
in 7 laps
distance will be covered.
but is given that 7 laps should be equal to 10 miles
so 10 miles will be equal to 

Thus, diameter of track will be 0.45 miles to meet the requirement.
To compute for the heat,

needed to be absorbed or released, we need

where

is the mass of the alligator,

is the change in temperature, and

is the specific heat of the alligator's body. Plugging in all the information we have,

Recall that 1 Watt = 1 J/s, thus time needed to absorb radiation from the sun is

That means it takes 4250 seconds for the alligator to warm up to be able to absorb the radiation from the sun.
Answer: 4250 seconds
Answer:
Explanation:
Let the required velocity of rocket be v .
We shall use the formula of time dilation to find the velocity of rocket .
t =
t = 430
t' = 38


= .0078
=.9922
= .996
v = .996 x 3 x 10⁸ m /s
= 2.988 x 10⁸ m /s
B )
Kinetic energy of rocket
= 1/2 m v²
= .5 x 20000 x (2.988 x 10⁸ )²
= 8.9 x 10²⁰ J .
C )
This energy is 8.9 times the energy requirement of United states in the year 2005 .
Answer:
Vd = 2.42 ×10⁻⁴ m/s
Explanation:
Given: A = 3.00×10⁻⁶ m², I = 7.00 A, ρ = 2.70 g/cm³
To find Drift Velocity Vd=?
Sol
the formula is Vd = I/nqA (n is the number of charge per unit volume)
n = No. of electron in a mole ( Avogadro's No.) / Volume
Volume = Molar mass / density ( molar mass of Al =27 g)
V = 27 g / 2.70 g/cm³ = 10 cm³ = 1 × 10 ⁻⁵ m³
n= (6.02 × 10 ²³) / (1 × 10 ⁻⁵ m³)
n= 6.02 × 10 ²⁸
Now
Vd = (7A) / ( 6.02 × 10 ²⁸ × 1.6 × 10⁻¹⁹ C × 3.00×10⁻⁶ m²)
Vd = 2.42 ×10⁻⁴ m/s