What’s the question here?
0.77 m/s2 directed 35° south of west
net force = (-17,-12)
net force = mass * acceleration
(-17,-12) = 27 * (x-acceleration,y-acceleration)
(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)
angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.
magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)
The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
<h3>What is concave mirror?</h3>
A concave mirror has a reflective surface that is curved inward and away from the light source.
Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.
<h3>
Object distance of the concave mirror</h3>
Apply mirrors formula as shown below;
1/f = 1/v + 1/u
where;
- f is the focal length of the mirror
- v is the object distance
- u is the image distance
when image height = object height, magnification = 1
u/v = 1
v = u
Substitute the given parameters and solve for the distance of the object from the mirror's vertex
1/f = 1/v + 1/v
1/f = 2/v
v = 2f
v = 2(19.5 cm)
v = 39 cm
Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
Learn more about concave mirror here: brainly.com/question/27841226
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Below are the 5 main indicators of chemical change.
Chemical change indicators:<span>
Color change
</span>Temperature change
Precipitate formation<span>
Odor
Bubble formation
I hope this helps!</span>
Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''
Vy in this problem will follow this equation =
where g is the gravity acceleration
This is equation (1)
For Y(t) :
We suppose yi = 0
This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)
So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)
2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s
The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m
