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mars1129 [50]
3 years ago
15

The diagram shows a 15-kg box resting on a wedge which has an angle of inclination of 30° and a coefficient of friction of 0.15.

Find the value of the vector Fg.
Physics
2 answers:
telo118 [61]3 years ago
6 0
We calculate first for the normal force, Fn, which is equal to the product of the force and the cosine of the given angle,
                                         Fn = (15 kg)(9.8 m/s²)(cos 30°)
                                          Fn = 127.30 N
Then, to calculate for the Fg, we multiply the normal force by the given coefficient.
                                         Fg = (127.30 N)(0.15) = 19 N
The nearest value is letter A. 
lara31 [8.8K]3 years ago
6 0
The correct answer is B. 147.15 N
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Explanation:

Let \textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}} and \textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}

The sum of the two vectors is

\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}

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The difference between the two vectors can be written as

\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}

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The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st
irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, 31 {\rm {N}/{mm}}. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

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\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J

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Read 2 more answers
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The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

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The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

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From above we know that -

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                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

                ⇒ h=\ \frac{(24.2)^2}{2\times 9.8}

                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


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