Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
It traveled 4 centimeters.
Explanation:
In a speed versus time graph, the distance travelled is given by the area under the graph.
In this graph we have the following:
- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s
- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s
Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:
Answer:
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Explanation:
The distance travelled on the rough ice is equal to the width of the rough ice.
distance d = 5.0 m
Initial speed u = 9.2 m/s
Final speed v = 5.8 m/s
The time taken to move through the rough ice can be calculated using the equation of motion;
d = 0.5(u+v)t
time t = 2d/(u+v)
Substituting the given values;
t = 2(5)/(9.2+5.8)
t = 2/3 = 0.66667 second
The acceleration is the change in velocity per unit time;
acceleration a = ∆v/t
a = (v-u)/t
Substituting the values;
a = (5.8-9.2)/0.66667
a = -5.099974500127
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Thank you for posting your question here at brainly. Below is Yoland's study:
<span>Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m.
</span>
I think the answer is "She can use constructive interference to generate a wave with an amplitude of 1.5 m."
