speed of the car is given on the road
reaction time is given as
now we can find the distance that it will move during this reaction time
now the deceleration of the car is 10 m/s^2
so the distance that it will move before stop is given by
so the total distance that it require to stop is given as
while the deer is standing at distance 38 m
so the car will stop 2.8 m before the position of deer
Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)
Answer:
a) fr = 224.3 N
, b) fr = 224.3 N
, c) v = 198.0 m/s
Explanation:
a) For this exercise let's start by calculating the acceleration in the fall
v² = v₀² - 2 a (y-y₀)
When it jumps the initial vertical speed is zero
a = -v² / 2 (y-y₀)
a = -68 2/2 (1000-2000)
a = 2,312 m / s²
Let's use the second net law to enter the average friction force
fr = m a
fr = 97 2,312
fr = 224.3 N
b) let's look for acceleration
v² = v₀² - 2 a y
a = (v² –v₀²) / 2 (y-y₀)
a = (4² - 68²) / 2 (0-1000)
a = 2,304 m / s²
fr = m a
fr = 97 2,304
fr = 223.5 N
c) the speed of the wallet is searched with kinematics
v² = v₀² - 2 g (y-y₀)
v = √ (0-2 9.8 (0-2000))
v = 198.0 m/s
Answer:
R = 98304.75 m = 98.3 km
Explanation:
The density of an object is given as the ratio between the mass of that object and the volume occupied by that object.
Density = Mass/Volume
Now, it is given that the density of Earth has become:
Density = 1 x 10⁹ kg/m³
Mass = Mass of Earth (Constant) = 5.97 x 10²⁴ kg
Volume = 4/3πR³ (Volume of Sphere)
R = Radius of Earth = ?
Therefore,
1 x 10⁹ kg/m³ = (5.97 x 10²⁴ kg)/[4/3πR³]
4/3πR³ = (5.97 x 10²⁴ kg)/(1 x 10⁹ kg/m³)
R³ = (3/4)(5.97 x 10¹⁵ m³)/π
R = ∛[0.95 x 10¹⁵ m³]
<u>R = 98304.75 m = 98.3 km</u>
Answer:
Explanation:
To solve this problem, we can use the following suvat equation:
where
is the vertical displacement of the frog
is the initial vertical velocity
t is the time
a is the acceleration
We have chosen this formula because apart from , all the other quantities are known. In fact:
is the vertical displacement
t = 2 s is the total time of flight
is the acceleration due to gravity (negative because it is downward)
Therefore, solving for , we find the initial velocity of the frog:
