Answer:
the speed of a jet plane = v = 12 m/s
Explanation:
jet plane that travels 48 meters East in 4 seconds?
<u>use the velocity as distance over time formula = v = d / t</u>
where:
t = 4 sec
d = 48 m
plugin values into the formula
v = <u> 48 m </u>
4 sec
v = 12 m/s
therefore,
the speed of a jet plane = v = 12 m/s
B......................... <span>Yes they do!</span>
The answer to your question is metaphase
Answer:
9) This is a case of deceleration
10)-0.8 ms-2
b) acceleration is the change in velocity with time
11)
a) 100 ms-1
b) 100 seconds
12) 10ms-1
13) more information is needed to answer the question
14) - 0.4 ms^-2
15) 0.8 ms^-2
Explanation:
The deceleration is;
v-u/t
v= final velocity
u= initial velocity
t= time taken
20-60/50 =- 40/50= -0.8 ms-2
11)
Since it starts from rest, u=0 hence
v= u + at
v= 10 ×10
v= 100 ms-1
b)
v= u + at but u=0
1000 = 10 t
t= 1000/10
t= 100 seconds
12) since the sprinter must have started from rest, u= 0
v= u + at
v= 5 × 2
v= 10ms-1
14)
v- u/t
10 - 20/ 25
10/25
=- 0.4 ms^-2
15)
a=v-u/t
From rest, u=0
8 - 0/10
a= 8/10
a= 0.8 ms^-2
Vertically (y-axis), horizontally(x-axis)
dy/dt = -0.125 m/s (-ve since y decreasing )
dx/dt = +0.3 m/s (+ve, x increases)
and, x = 5 m
length of the ladder = k (a constant)
k^2 = x^2 + y^2
differentiating it wrt t,
0 = 2x dx/dt + 2y dy/dt
0 = 2(5)(0.3) + 2(y)(-0.125)
y = 12
which means, when the bottom of the ladder is 5m from the wall, the top of the ladder is 12m from the bottom.
thus, k^2 = 5^2 + 12^2
length of the ladder, k = 13 m
