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Reptile [31]
3 years ago
12

A child sleds down a hill with an acceleration of 2.94 m/s2. If her initial speed is 0.0 m/s and her final speed is 17.5 m/s, ho

w long does it take her to travel from the top of the hill to the bottom?
Physics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

Explanation:

Given that the child accelerates uniformly and that both initial (v_{o}) and final speeds (v_{f}), measured in meters per second, and acceleration (a), measured in meters per square second, are known, we proceed to use the following kinematic equation to determine the time taken to travel from the top of the hill to the bottom (t), measured in seconds, is:

t = \frac{v_{f}-v_{o}}{a} (1)

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 17.5\,\frac{m}{s} and a = 2.94\,\frac{m}{s^{2}}, then the time taken is:

t = \frac{17.5\,\frac{m}{s}-0\,\frac{m}{s} }{2.94\,\frac{m}{s^{2}} }

t = 5.952\,s

The child will take 5.952 seconds to travel from the top of the hill to the bottom.

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luda_lava [24]
A ) 
T = mB g + mB a
T + mA a - mA g sin 35° = (Mi) mA g cos 35°
------------------------------------------------------------
T = 2.7 · 9.81  + 2.7 a
T = 26.487 + 2.7 a
26.487 + 2.7 a + 2.7 a - 2.7 · 9.81 · 0.574 = 0.15 · 2.7 · 9.81 · 0.819
5.4 a + 26.487 - 15.2023 = 3.2539
5.4 a = 8.0296
a = 1.487 ≈ 1.5 m/s²
B )
T = 2,7 · 9.81 = 26.487 
26.487 - 15.2035 = (Mi) · 2.7 · 9.81 · 0.819
11.2835 = (Mi) · 21.69
(Mi) = 11.2835 : 21.69 = 0.52
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4 years ago
When light reflects from a surface, there is a change in its
alexandr402 [8]
The answer is: none of the above.

Explanation:

When light reflects from a surface, the frequency, wavelength, and speed do not change. They remain the same.
6 0
3 years ago
Which vector should be negative?
pentagon [3]

Answer:

 The velocity of a falling object

Explanation:

  The positive X axis is towards right and positive Y axis is towards up, so North direction is positive

  A vector with less than 1 magnitude is not negative, because its magnitude may be in between 0 and 1 which is positive vector.

  Any vector whose magnitude is greater than 1 is never be a negative vector.

  The velocity of a falling object is towards bottom, that is towards negative Y axis. So that vector is negative.

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At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m2. A fan is 30 m from the loudspeak
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Explanation:

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3 years ago
In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

Learn more about centripetal and angular acceleration here:

brainly.com/question/2562955

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#LearnwithBrainly

8 0
3 years ago
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