Answer:
a=1.024m/s
t=15.62s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X (2)
X=Xo+ VoT+0.5at^{2} (3)
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve
for this problem
Vf=16m/s
Vo=0m/s, the cart starts from the rest
X=125m
we can use the ecuation number tow to calculate the acceleration
{Vf^{2}-Vo^2}/{2.a} =X
{Vf^{2}-Vo^2}/{2.x} =a
{16^{2}-0^2}/{2(125)} =a
a=1.024m/s
to calculate the time we can use the ecuation number 1
Vf=Vo+a.t
t=(Vf-Vo)/a
t=(16-0)/1.024
t=15.62s
A projectile is any object that is thrown, dropped or otherwise laughed into the air.
1) If it is dropped - it's motion is caused by just gravitation force
2) if it is thrown - it's motion is caused by gravitation force + force you applied to though it upwards
Answer:
C and D
Explanation:
The guy above me got it wrong
Answer:
0.5A
Explanation:
V=IR where V is 1.5V, R is 3 ohms
Hence 1. 5= I x 3
1.5/3 =I
I =0.5 A
Current is 0. 5 amperes
Answer:
m = 0.25
Explanation:
Given that,
Object distance, u = -15cm
Height of the object, h = 48
Focal length, f = cm
We need to find the magnification of the image.
Let v is the image distance. Using mirror's equation.
Magnification,
Hence, the magnification of the image is 0.25.
