Answer:
θ = 90º
Explanation:
The velocity is given by
v =
calculate
v = 3 i ^ + √2 j ^ + 2t k ^
acceleration is defined by
a = dv / dt
a = 2 k ^
one way to find the angle is with the dot product
v. a = | v | | a | cos θ
cos θ= v.a / | v | | a |
Let's look for the value of each term
v. a = 4 t
| v | = =
| a | = 2
they ask us for the angle for time t = 0
v. a = 0
| v | = √11 = 3.317
we substitute
cos θ = 0 /√11
cos θ = 0
therefore the angles must be θ = 90º
Answer:
a. 1/1000 sec
Explanation:
Shutter speed is the length of time that the film you’re photographing is being exposed to the scene in film photography. However, in digital photography, shutter speed is the length of time that the image sensor sees the scene the photographer is trying to capture.
For shutter speeds, the greater the denominator the higher the speed and the lower the denominator, the lower the speed.
Thus, the fastest one is option A.
Answer:
The speed of this particle is constantly .
Explanation:
Position vector of this particle at time :
.
Write as a column vector to distinguish between the components:
.
Both and are constants. Therefore, and would also be constants with respect to . Hence, and .
Differentiate (component-wise) with respect to time to find the velocity vector of this particle at time :
.
The speed (a scalar) of a particle is the magnitude of its velocity :
.
Therefore, the speed of this particle is constantly (a constant.)
Average speed is defined as total distance moved in total interval of time
so it is given as
now here is we show distance by "d" and time by"t"
then we will have mathematical expression as follows
Answer:
It takes <u>2.93 s</u> for the object to reach ground from the height of 42 m.
Explanation:
Given:
Displacement of the object is,
As it is dropped, initial velocity of the object is,
Acceleration of the object is equal to acceleration due to gravity and is equal to,
Now, using Newton's equation of motion and plugging in the values, we get:
So, it takes 2.93 s for the object to reach ground from the height of 42 m.