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Lyrx [107]
1 year ago
5

A block of ice is sliding down a ramp of slope 40 to the horizontal. What is the rate of acceleration of the block? Assume the f

orce of friction is not significant.

Physics
1 answer:
uysha [10]1 year ago
8 0

The given problem can be exemplified in the following diagram:

Since there is no friction or any other external force, the only force acting in the direction of the movement is the component of the weight of the block, therefore, applying Newton's second law:

\Sigma F=ma

Replacing the values:

mg\sin 40=ma

We may cancel out the mass:

g\sin 40=a

Using the gravity constant as 9.8 meters per square second:

9.8\frac{m}{s^2}\sin 40=a

Solving the operations:

6.3\frac{m}{s^2}=a

Therefore, the acceleration is 6.3 meters per square second.

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Five different forces act on an object. Is it possible for the net force on the object to be zero?
bazaltina [42]
No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
4 0
3 years ago
To measure the strength of an earthquake, you can use either a _____ scale or _____ scale.
allsm [11]
To measure the strength of an earthquake, you can use either a Richter scale or Mercalli scale. Richter scale uses the amplitued of the wave and the distance from the source. Mercalli scale uses observations of people and is not considered to be scientific as Richter scale.
8 0
3 years ago
Read 2 more answers
f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected
siniylev [52]

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

Weight=\frac{E_M-r_f}{A\sigma_M^2}

For the equity, here as

  • Weight is percentage of the equity which is to be calculated
  • {E_M-r_f} is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
  • A is the risk aversion factor which is given as 4.
  • \sigma_M is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59

By substituting values.

Weight=\frac{E_M-r_f}{A\sigma_M^2}\\Weight=\frac{8.30\%}{4(20.59\%)^2}\\Weight=0.4894 =48.94\%

So the weight of equity is 48.94%.

Now the weight of T bills is given as

Weight_{T-Bills}=1-Weight_{equity}\\Weight_{T-Bills}=1-0.4894\\Weight_{T-Bills}=0.5105=51.05\%\\

So  the weight of T-bills is 51.05%.

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

7 0
3 years ago
Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an
castortr0y [4]

Answer:

  • Fx = -9.15 N
  • Fy = 1.72 N
  • F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

  F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

  f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

  = -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

  ≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

  ≈ (-9.15309, 1.71982)

The resultant component forces are ...

  • Fx = -9.15 N
  • Fy = 1.72 N

Then the magnitude and direction of the resultant are

  F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

  F∠γ ≈ 9.31∠-10.6°

4 0
3 years ago
A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in-
Leto [7]

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

            F = m a₀          

           a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

           x = v₀ t + ½ a t²

           x = ½ a₀ t²

           t² = 2x / a₀

           28² = 2x /a₀          (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

           

in the direction of axis a, there is a motor so its force is F/2

               

the acceleration on this axis is

          a = F/2m

          a = a₀ / 2

so if we use the distance equation

             x = v₀ t + ½ a t²

as part of rest v₀ = 0

             x = ½ (a₀ / 2) t²

             

let's clear the time

             t² = (2x / a₀)  2

we substitute the let of equation 1

             t² = 28² 2

             t = 28 √2

             t = 39.60 s

4 0
3 years ago
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