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taurus [48]
3 years ago
12

Which option is part of designing a set of experimental procedures?

Physics
2 answers:
zloy xaker [14]3 years ago
4 0
B identifying the types of data to be gathered
vesna_86 [32]3 years ago
4 0
The answer is B. Identifying the types of data to be gathered
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If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
4 years ago
A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on t
Phoenix [80]

Answer:

a.I=981.34 N*s

b.v_f=3.96 m/s

c.v_{f1}=3.63m/s

d.y_f=0.673m

Explanation:

Given: m=67kg, h=0.720m, 0

a.

I=\int\limits^{t_1}_{t_2} {F(t)} \, dt

F(t)=9200*t-11500t^2

I=\int\limits^{0.8s}_{0s}{9200*t-11500*t^2} \, dt

I=4600*t^2-3833.3*t^3|(0.80,0)

I=2944-1962.66=981.35

I=981.34 N*s

b.

v_f^2=v_i^2+a*y'

Starting from the rest

v_f^2=0+2*9.8m/s^2*0.80s

v_f^2=15.68

v_f=\sqrt{15.68m^2/s^2}=3.96 m/s

c.

I_{total}=p_f

I_1-m*g*d=m*v_{f1}-m*v_f

981.34-67kg*9.8m/s^2*0.720=67.0kg*v_{f1}-67.0kg*(-3.96m/s)

Solve to vf

v_{f1}=3.63m/s

d.

v_f^2=v_i^2+2*a*y_f'

y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2

y_f=0.673m

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3 years ago
When light travels through a small hole, it appears to an observer that the light spreads out, blurring the outline of the hole.
Scilla [17]

wave theory

this observation supports the theory of light as a wave

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3 years ago
Speed in any given direction is called <br> A. Trust<br> B.velocity<br> C.mass<br> D.acceleration
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B. velocity is the speed in any given direction.
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