Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
Answer:
Chicken and peanut butter lol
Explanation:
I'm pretty sure the answer is B: <span>establish public doubt
Hope this helps!</span>
Answer:
Explanation:
The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:
, where 
is the mass of the basketball and 
is the change in velocity.
Since the basketball is changing direction, its total change in velocity is:
.
Therefore, the basketball's change in momentum is:
.
Thus, the impulse on the basketball is 
(two significant figures).
Answer:
Explanation:
Given that
speed u=4×10⁶ m/s
electric field E=4×10² N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates
acceleration a=qE/m=1.6×10⁻¹⁹×4×10²/9.1×10⁻³¹=0.7×10¹⁴=7×10¹³ m/s
now we find the horizantal distance travelled by electrons hit the plates
horizantal distance X=u[2y/a]^1/2
=4×10⁶[2×2×10⁻²/7×10¹³]^1/2
=9.5cm
now we find the velocity f the electron strike the plate
v²-(4×10⁶)²=2×7×10¹³×2×10⁻²
v²=16×10¹²+28×10¹¹
v²=1.88×10¹³m/s
speed after hits =>V=4.34×10⁶ m/s
