Answer:
17.54N in -x direction.
Explanation:
Amplitude (A) = 3.54m
Force constant (k) = 5N/m
Mass (m) = 2.13kg
Angular frequency ω = √(k/m)
ω = √(5/2.13)
ω = 1.53 rad/s
The force acting on the object F(t) = ?
F(t) = -mAω²cos(ωt)
F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)
F(t) = -17.65 * cos (5.355)
F(t) = -17.57N
The force is 17.57 in -x direction
Answer:
Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.
Initial speed of the projectile is v and final speed is 0.5 v.
g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :
So, the height of the projectile above the ground is 
. Hence, this is the required solution.
Answer:
Explanation:
mass m = 3 kg
spring constant be k
k x .8 = 40 N
k = 40 / .8 = 50 N /m
angular frequency ω = √ ( k / m )
= √ ( 50 / 3 )
= 4.08 rad /s
Let amplitude of oscillation be A .
1/2 k A² = 1/2 m v²
50 A² = 3 x 1²
A = .245 m = 24.5 cm
For displacement , the equation of SHM is
x = A sinωt
= 24.5 sin4.08 t
x = 24.5 sin4.08 t
Here, angle 4.08 t is in radians .
Here in crash test the two forces are acting on the dummy in two different directions
As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.
So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors
so we can say
here given that
now we will plug in all data in the above equation
so it will have net force 4501.9 N which will be reported by sensor
The question is whether the statement is true or false.
The answer if false.
Explanation:
It is exactly the opposite. The soccer ball will hit the ground with greater velocity.
Since the soccer ball is thrown upward, when it returns to the same heigth from which it was throwm it will have a velocity downward, which will make that the soocer ball reaches the ground at the bottom of the clif with greater velocity than the volleball.
The greater the velocity with which the soccer ball is thrown upward, the greater its velocity when reaches the same point from which it was thrown, and the greater the velocity with which it will hit the ground at the bottom of the clif.
