Answer:
Final speed of the car, v = 24.49 m/s
Explanation:
It is given that,
Initial velocity of the car, u = 0
Acceleration,
Time taken, t = 7.9 s
We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :
v = 24.49 m/s
So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.
Answer:
Explanation:
From the Question we are told that
Mass of A and B is 60kg
Speed of A=2m/s
Speed of B=1m/s
Mass of bag =5kg
Generally the momentum of the astronaut A and bag is mathematically given as
Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag
Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1
Therefore
Answer:
A) F = - 8.5 10² N, B) I = 21 N s
Explanation:
A) We can solve this problem using the relationship of momentum and momentum
I = Δp
in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction
v₀ = 8.50 m / s
v_f = -8.50 m / s
F t = m v_f -m v₀
F =
let's calculate
F =
F = - 8.5 10² N
B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations
v² = v₀² - 2g (y- y₀)
as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0
v =
calculate
v =
v = 14 m / s
to calculate the momentum we use
I = Δp
I = m v_f - mv₀
when it hits the ground its speed drops to zero
we substitute
I = 1.50 (0-14)
I = -21 N s
the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is
I = 21 N s
Answer:
Position xf is farther away from the sensor than x0, and ax is negative
Explanation:
Area of trapezoidal are=
=
=
=0.6875 m
As the area is positive therefore displacement from xo is positive
ax=(change in velocity)/(Time)
ax=
Answer:
a)
b)
Explanation:
Given that
v(t) = 5 t i + t² j - 2 t³ k
We know that acceleration a is given as
Therefore the acceleration function a will be
The acceleration at t = 2 s
a= 5 i + 2 x 2 j - 6 x 2² k m/s²
a=5 i + 4 j -24 k m/s²
The magnitude of the acceleration will be
a= 24.83 m/s²
The direction of the acceleration a is given as
a)
b)