Answer:
6m/s
Explanation:
the original momentum = mass x velocity = 8x (60+10) = 560
momentum after = mass x velocity of the school bag + mass x velocity of the boy = 10x20 + 60x A
200+60A = 560
A=6
The appropriate response is Gallium. It is a concoction component with image Ga and nuclear number 31. It is in gathering 13 of the occasional table and subsequently has similitudes to alternate metals of the gathering, aluminum, indium, and thallium.
Explanation:
Given parameters:
Distance = 15miles north = 24140.2m
Initial velocity = 0m/s
Final velocity = 4m/s
Unknown:
Speed, velocity and acceleration = ?
Solution:
The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.
Speed =
The speed of the student is 4m/s
Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;
Velocity =
The velocity of the student is 4m/s due north
Acceleration is the change in velocity with time;
To find the acceleration, we use
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
4² = 0² + 2x a x 24140.2
a =
= 0.00033m/s²
Explanation:
The given data is as follows.
Length of beam, (L) = 5.50 m
Weight of the beam, (
) = 332 N
Weight of the Suki, (
) = 505 N
After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

= 0
x = 
= 
= 0.986 m
Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.
Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.
Answer:
The total frictional force is 358.0 newtons
Explanation:
Power is the amount of average work (W) an object does on a period of time (Δt):

Remember average work is average force (F) times displacement (Δs):

but displacement over time is average speed
, then:
(1)
That is, the power of the car is the force the engine does times the speed of the car. As the question states, if the car is at constant velocity then the power developed is used to overcome the frictional forces exerted by the air and the road, that is by Newton's first law, the force the motor of the car does is equal the force of frictional forces. So, to find the frictional forces we only have to solve (1) for F:

Knowing that 1hp is 746W then 30hp=22380W and 1 mile = 1609m then 140 mph = 225308
=
, then:
