Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
Answer:
The question is wrong Since if you apply Force on 0.0m²It would mean That the pressure exerted=F/A=F/0
An since we can't divide a number by 0, the question is wrong
<span>2002 seconds, or 33 minutes, 22 seconds.
First, let's calculate how many joules it will take to lift 78 kg against gravity for 1100 meters. So:
78 kg * 9.8 m/s^2 * 1100 m = 840840 kg*m^2/s^2
Now a watt is defined as kg*m^2/s^3, so a division of the required joules should give us a convenient value of seconds. So:
840840 kg*m^2/s^2 / 420 kg*m^2/s^3 = 2002 seconds.
And 2002 seconds is the same as 33 minutes, 22 seconds.</span>
Given that,
Energy 
Surface temperature = 11000 K
Emissivity e =1
(a). We need to calculate the radius of the star
Using formula of energy
Put the value into the formula
(b). Given that,
Radiates energy 
Temperature T = 10000 K
We need to calculate the radius of the star
Using formula of radius
Put the value into the formula
Hence, (a). The radius of the star is 
(b). The radius of the star is 
