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Andre45 [30]
3 years ago
9

A skateboarder, starting from rest, rolls down a 10.9-m ramp. When she arrives at the bottom of the ramp her speed is 6.74 m/s.

(a) Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 26.6 ° with respect to the ground, what is the component of her acceleration that is parallel to the ground
Physics
1 answer:
JulijaS [17]3 years ago
4 0

Answer:b

Explanation:b

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Numerical Problems
Evgesh-ka [11]

Answer:

a = 5 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 10 [m/s]

t = time = 2 [s]

a = acceleration [m/s²]

Now replacing:

20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ]

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Carmen simplifies the expression (4y+8x+6)+25+(4x+6y+7). The coefficient of the variable y in her simplified answer is
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The answer of the problem is 12x+10y+38
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3 years ago
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A 68 kg runner exerts a force of 59N. What is the acceleration of the runner?
Korvikt [17]

Explanation:

<em><u>Newton's First Law ,</u></em>

F = ma

a = m/F

a = 68 / 59

a = 1.15 m/s2

4 0
4 years ago
What would be the speed of a car if it covers a distance of 400 km in 5 hours?​
iragen [17]

Answer:

80km/hr

Explanation:

Hope this helps! :>

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8 0
3 years ago
A boy walks a distance of 100 meters to the right at a steady speed of 1.00 m/s. Then he stops for 30 seconds. Then returns back
Mamont248 [21]

Explanation:

speed=\frac{Distance}{time}

In the figure attached:

A boy travels AB distance of 100 m, with speed of 1.00 m/s

AB = 100 m

1.00 m/s=\frac{100 m}{t_1}

t_1=100 s

After reaching at point B he stops there at for 30 seconds.

t_2= 30 s

After , 30 seconds he he comes back to his initial position that is A with steady speed of 1.00 m/s.

Distance covered from B to A= 100 m

Time taken by him during coming back=t_3

1.00=\frac{100 m}{t_3}

t_3=100 s

After returning to to point A he turns left and moves towards point C with speed of 1.5 m/s for 2 minutes.

Distance of AC = ?

t_4=2 min= 120 s

1.5 m/s=\frac{AC}{120 s}

AC = 180 m

The total time of his round trip is:T

T=t_1+t_2+t_3+t_4=100 s+30 s(stop)+ 100 s(returning)+120 s=350 s

The total distance: D

D = AB + BA (returning) + AC=100 m + 100 m + 180 m = 380 m[/tex]

The total displacement of boy:

Displacement is the shortest distance between the initial point and final point.

He first walked to point B from A and then came back to A . And after that walking to point C from A.So, the final displacement (d) is from A to C.

d = AC = 180 m

The total displacement of boy is 180 m.

The average speed of the boy is given by:

=\frac{AB+BA+AC}{t_1+t_2+t_3+t_4}=\frac{D}{T}

\frac{D}{T}=\frac{380 m}{350 s}=1.085 m/s

The average velocity of the boy is given by:

=\frac{AC}{t_1+t_2+t_3+t_4}=\frac{d}{T}

\frac{d}{T}=\frac{180 m}{350 s}=0.5142 m/s

5 0
4 years ago
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