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3 years ago

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Our attitudes influence

1 answer:

Ad libitum [116K]3 years ago

4 0

Our attitudes influence "A. our beliefs and understanding and how we respond to them." In general, attitude influences all these things, and the more positive the attitude the more likeliness there is for positive change.

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A bowling ball has a mass of 7.2 kg and a weight of 70.6 N. It moves down the bowling alley at 1 m/s and strikes a pin with a fo

Inga [223]

The answer is 15.0N its explained in newtons third law hope this helps:)

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3 years ago

Read 2 more answers

devlian [24]

Answer:

c

Explanation:

6 0

3 years ago

Solve each problem. Be sure to show your work and give a final answer with units rounded to the given number of significant figu

Alex Ar [27]

Answer:

1. the voltage will be 2.35×12.5 = 29.4V

2. the resistance would be 9.0/6.2= 1.45ohms

3. in series they will add up thus 4+8+12= 24ohms

4. in parallel it will be 2.18ohms

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3 years ago

A car traveling on a straight road at 15.0 meters per second accelerates uniformly to a speed of 21.0 meters per second in 12.0

oksano4ka [1.4K]

<h2>Option 3, 216 m is the correct answer.</h2>

Explanation:

We have initial velocity, u = 15 m/s

Time, t = 12 seconds

Final velocity, v = 21 m/s

We have equation of motion v = u + at

Substituting

21 = 15 + a x 12

a = 0.5 m/s²

Now we have equation of motion v² = u² + 2as

21² = 15² + 2 x 0.5 x s

s = 216 m

Displacement = 216 m

Option 3, 216 m is the correct answer.

8 0

3 years ago

A sky diver with a mass of 70kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kV^

xeze [42]

Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

$v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}$

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

$v_{100}=33.81$

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

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3 years ago