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Natasha_Volkova [10]

3 years ago

10

Who’s good at algebra?

2 answers:

lutik1710 [3]3 years ago

5 0

Answer:

I'm alright................

k0ka [10]3 years ago

3 0

What’s the question

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Consider a small child sliding down two differently shaped slides that have the same coefficient of kinetic friction. How must t

timama [110]

Answer:

Having a bigger angle above the horizontal

Explanation:

Applying the energy conservation theorem:

$K_1+U_1+W_{ext}=K_2+U_2$

The kinetic energy is reduced because of the work done by the friction force.

The friction force is given by:

$F_f=F_N*\µ$

so the friction force depends on the Normal force, because the slide has an angle the normal force is given by:

$F_N=m.g*cos(\theta)$

So when the angle of the slide is bigger, the friction force decreases, for example:

for 45 degrees:

$F_N=m.g*cos(45)\\F_N=0.70(m.g)\\$

for 75 degrees:

$F_N=m.g*cos(75)\\F_N=0.26(m.g)\\$

as you can see if the angle is bigger above the horizontal, the friction force is reduced and so the work done by that force. We didn't have to change the height of the slide, so the potential gravitational energy remains the same.

6 0

3 years ago

Which of the following is the correct relationship among the acceleration, velocity, and position?

Damm [24]

Answer:1

Explanation:

If

s=displacement

v=velocity of particle

a=acceleration of particle

acceleration can be written as rate of change of velocity

so $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

multiply and divide by $ds$

$a=\frac{\mathrm{d} v}{\mathrm{d} t}\times \frac{ds}{ds}$

$a=\frac{\mathrm{d} v}{\mathrm{d} s}\times \frac{\mathrm{d} s}{\mathrm{d} t}$

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ads=vdv$

option 1 is correct

5 0

3 years ago

Read 2 more answers

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

6 0

3 years ago

In a ray diagram showing refraction, the incident ray and the refracted ray...

Amanda [17]

C because of the positioning of the initial ray of refraction

4 0

2 years ago

Which energy source is formed when organic matter is trapped underground without exposure to air or moisture?

REY [17]

C. Coal Is The Answer

8 0

3 years ago

Read 2 more answers