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sesenic [268]
3 years ago
8

Olympic skier Tina Maze skis down a steep slope that descends at an angle of 30∘ below the horizontal. The coefficient of slidin

g friction between her skis and the snow is 0.10.
I found Maze's acceleration which is 4.1 m/s^2. But how do you determine Maze's speed 4.0 s after starting?
Physics
1 answer:
LekaFEV [45]3 years ago
5 0

net force on Maze during his sliding downwards motion

F = mgsin30 - \mu mg cos30

now we will have

F = mg(sin30 - 0.10cos30)

now to find the acceleration we will have

a = \frac{F}{m}

a = g(sin30 - 0.10cos30)

a = 9.81(0.5 - 0.10(0.866))

a = 4.1 m/s^2

now to find the speed after 4 s is given by kinematics

v_f - v_i = at

here we know that

v_i = 0

v_f - 0 = 4.1(4)

v_f = 16.22 m/s

so Maze speed after 4 s will be 16.22 m/s

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The heat of fusion for water is the amount of energy needed for water to
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The potential difference between the plates of an air-filled parallel-plate capacitor with a plate separation of 4 cm is 51 V. W
Salsk061 [2.6K]

Answer:

Explanation:

Electric field between plates of a parallel plate capacitor is uniform .

In a uniform electric field , relation between electric field and potential gradient is as follows

electric field = potential gradient               [ E = - dV / dl ]

in the given case ,

dV = 51 V  ,

dl = 4 cm

=  4 x 10⁻² m

E = 51 /  4 x 10⁻²

= 12.75 x 10² V / m

= 1275 V / m

6 0
3 years ago
Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and m
Anika [276]
6x2=12m
6x18=108
12m+108
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8 0
3 years ago
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the
kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0
3 years ago
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