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antoniya [11.8K]
3 years ago
9

Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver springs upward with an initial speed o

f 2.22 m/s from a 3.0-m board. (a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.] (b) What is the highest point he reaches above the water?
Physics
1 answer:
adell [148]3 years ago
7 0

Answer:

Part a)

v_f = 7.99 m/s

Part b)

y = 3.25 m

Explanation:

Part a)

Since the diver is moving under gravity

so here its acceleration due to gravity will be uniform throughout the motion

so here we will have

v_f^2 - v_i^2 = 2 a y

here we have

v_i = 2.22 m/s

y = -3 m

v_f^2 - (2.22)^2 = 2(-9.81)(-3)

v_f = 7.99 m/s

Part b)

at highest point of his motion the final speed will be zero

so we will have

v_f^2 - v_i^2 = 2 a (\Delta y)

0 - 2.22^2 = 2(-9.81)(y - 3)

y = 3.25 m

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Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s

The interaction time to avoid that the water balloon breaks is 0.029s

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3 years ago
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Which of the following elements is commonly found in the Earth's crust, living matter, oceans, and atmosphere? A. carbon B. argo
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g Initially, the motorcycle travels along a straight road with a speed of 35 m/s (this is almost 80 mph). The maximum decelerati
astraxan [27]

Given:

Initial speed of the motorcycle (u) = 35 m/s

Final speed of the motorcycle (v) = 0 m/s (Complete Stop)

Maximum deceleration of the motorcycle (a) = -1.2 m/s²

Required Equation:

\boxed{\bf{ v = u + at}}

Answer:

By substituting values in the equation, we get:

\rm \longrightarrow 0 = 35 + ( - 1.2)t \\  \\  \rm \longrightarrow  0 = 35 - 1.2t \\  \\  \rm  \longrightarrow 35 - 1.2t = 0 \\  \\  \rm  \longrightarrow 35- 35 - 1.2t = 0 - 35 \\  \\  \rm  \longrightarrow  - 1.2t =  - 35 \\  \\  \rm \longrightarrow  \dfrac{ - 1.2t}{ - 1.2}  =  \dfrac{ - 35}{ - 1.2}  \\  \\  \rm \longrightarrow  t = 29.167 \: s

\therefore Time taken by motorcycle to come to a complete stop (t) = 29.167 s

4 0
3 years ago
Consider a system of two particles: ball A with a mass m is moving to the right a speed 2v and ball B with a mass 3m is moving t
arlik [135]

Answer:

Explanation:

Answer:

Explanation:

Given that,

System of two particle

Ball A has mass

Ma = m

Ball A is moving to the right (positive x axis) with velocity of

Va = 2v •i

Ball B has a mass

Mb = 3m

Ball B is moving to left (negative x axis) with a velocity of

Vb = -v •i

Velocity of centre of mass Vcm?

Velocity of centre of mass can be calculated using

Vcm = 1/M ΣMi•Vi

Where M is sum of mass

M = M1 + M2 + M3 +...

Therefore,

Vcm=[1/(Ma + Mb)] × (Ma•Va +Mb•Vb

Rearranging for better understanding

Vcm = (Ma•Va + Mb•Vb) / ( Ma + Mb)

Vcm = (m•2v + 3m•-v) / (m + 3m)

Vcm = (2mv — 3mv) / 4m

Vcm = —mv / 4m

Vcm = —v / 4

Vcm = —¼V •i

3 0
3 years ago
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