Answer:
Yes. The fact that an object moves at constant velocity implies that its speed is also constant. Note that the converse statement isn't necessarily true.
Explanation:
Velocity is a vector. For two vectors to be equal to each other,
- their magnitudes (sizes) need be the same, and
- they need to point in the same direction.
In motions, the magnitude of an object's velocity is the same as its speed.
If the car moves with a constant velocity, that means that
- the magnitude of its velocity, the speed of the car, is constant;
- also, the direction of the car's motion is also constant.
In other words,
.
Note that the arrow here points only from the velocity side to the speed side. It doesn't point backward because knowing that the speed of an object is constant won't be sufficient to prove that the velocity of the object is also constant. For example, for an object in a uniform circular motion, the speed is constant but the direction keeps changing. Hence the velocity isn't constant.
Answer:
The answer is below
Explanation:
The amplitude decreases by 2% during each oscillation. Hence the decrease in amplitude can be represented by an exponential decay in the form:
y = abˣ; where x ad y are variables, a is the initial value and b is the factor.
Let y represent the amplitude after x oscillations. Since the initial amplitude is 10 cm, hence:
a = 10 cm, b = 2% = 0.02.
Therefore:
y = 10(0.02)ˣ
The amplitude after 25 oscillations is gotten by substituting x = 25 into the equation. Hence:
y = 10(0.02)²⁵
y= 3.355 * 10⁻⁴² cm
The amplitude after 25 oscillations is 3.355 * 10⁻⁴² cm
Pronouns in the nominative case may function as Subjects.
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We went to the game.
You played the game.
He/she/they watched the game.
I’m guessing b because hydrogen is in your room and maybe eliminate to o but it also can be D
my final is: B!
Answer:
117,30m
Explanation:
I think the situation here is a horizontal projection to the ground. So in order to find the distance the formula = Ut, where U is the initial speed and t is the Time of flight. To get the time of flight in this case =√2h/g where h is the height and g is gravity. so to get the time = √2×300÷9.81 =7.821 .so range =ut which is equal to the time multiplied by 15m/s =117.30m
