Which is equivalent to (^3squareroot 125)^x ?
2 answers:
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Let's find the perimeter first by adding all the side lengths together. There are two missing side lengths, which we can find by taking the side opposite them and subtract the side behind the missing side from them.
I've attached a diagram showing how to find the side lengths.
Now let's add all the side lengths together: 4 cm + 8 cm + 8 cm + 2 cm + 4 cm + 6 cm = 32 cm
The perimeter of this figure is 32 cm.
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To find the area we can divide this figure into 2 rectangles. On the second attached diagram, I've split the figure into two and labeled the length and width of each rectangle.
Area of a rectangle = length * width
Area of red labeled rectangle = 4 cm * 2 cm = 8 cm²
Area of yellow labeled rectangle = 8 cm * 4 cm = 32 cm²
Add the two areas together: 8 + 32 = 40
The area of the figure is 40 cm².
Answer:
(a) The velocity after 2 second is 5.9 m/s
The velocity after 4 second is -13.7 m/s.
(b) The projectile reaches its maximum height after 2.60 s of projection.
(c)The maximum height that is attained by the projectile is 37.18 m.
(d)Therefore the projectile hits the ground after 5.36 seconds of projection.
(e)The velocity of the projectile when it hits the ground is 27.03 m/s
Step-by-step explanation:
Given that, a projectile shot vertically upward from a point 4 m above the ground with a initial velocity of 25.5 m/s.
The height of the projectile after t seconds is
where h is in meter.
(a)
We use the formula
v=u+at
V= final velocity
u = initial velocity = 25.5 m/s
a = acceleration= acceleration due to gravity= 9.8 m/s²
Since the object moves upward direction and acceleration due to gravity is downward direction. So here a= -9.8 m/s.
v(2)= 25.5+(-9.8)×2
=25.5-19.6
=5.9 m/s
And when t= 4
v(4)= 25.5+(-9.8)×4
=25.5-39.2
= -13.7 m/s
The velocity after 2 second is 5.9 m/s
The velocity after 4 second is -13.7 m/s.
(b)
At its maximum height,the velocity of the projectile is zero. i.e v=0
∴0=25.5+(-9.8)t
⇒9.8t=25.5
⇒t = 2.60 s
The projectile reaches its maximum height 2.60 s after projection.
(c)
To find the maximum height, we are putting t= 2.60 in this equation .
=37.18 m
The maximum height that is attained by the projectile is 37.18 m.
(d)
When the projectile hits the ground the height will be zero i.e h=0
From the equation of height we get
⇒t= -0.15 ,5.36
Therefore it hits the ground after 5.36 seconds of projection.
(e)
To find the velocity we use the formula v=u+at
Here v = final velocity=?
u=25.5 m/s,
t = 5.36 s
a= -9.8m/s²
v=25.5+(-9.8)5.36
= -27.03 m/s
Negative sign denoted that the motion of the projectile is downward direction.
The velocity of the projectile when it hits the ground is 27.03 m/s.