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ella [17]
2 years ago
11

By tilting the ramp at different angles, you can change the magnitude of the component of the gravitational force acting paralle

l to the ramp. If the downhill gravitational force is larger than the maximum value of the static friction force, the block will slide downhill. What is the largest angle (called the critical angle) that the ramp can have before the block starts to slide?
Physics
1 answer:
kow [346]2 years ago
3 0

Explanation:

As the angle increases, the component of force parallel to the incline increases and the component of force perpendicular to the incline decreases. It is the parallel component of the weight vector that causes the acceleration. Thus, accelerations are greater at greater angles of incline.

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A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.610 c. A scientist on Coruscant measures th
mamaluj [8]

Answer:

the length of the now stationary spacecraft = 89.65m

Explanation:

In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.

Thus, it has a formula;

L = L_o(√(1 - (v²/c²))

Where in this question;

L = 71m and v = 0.610 c

Thus;

71 = L_o (√(1 - ((0.61c)²/c²))

c² will cancel out to give;

71 = L_o (√(1 - 0.61²)

71 = L_o (√(1 - 0.61²)

71 = 0.792L_o

L_o = 71/0.792

L_o = 89.65m

6 0
2 years ago
A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

b)

r=(0,-1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0
2 years ago
The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect
Diano4ka-milaya [45]
Rx= 3.5 km

Ry= 2.9 km
4 0
2 years ago
Read 2 more answers
Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the s
Flura [38]

Answer:

the correct answer is A, the object goes 4 times as far

Explanation:

This is a projectile launching approach. Where the parameter we are controlling is the initial speed and they ask us how far it goes from the initial one. Let's calculate the range with a speed (vo)

        R1 = v₀² sin 2θ / g

Now let's double vo, the new speed is

         v = 2 v₀

We calculate the scope

         R2 = (2v₀)² sin 2θ / g

         R2 = 4 v₀² sin 2θ / g

         R2 = 4 R1

Therefore the correct answer is A, the object goes 4 times further

5 0
2 years ago
In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to
Mandarinka [93]

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

\beta = \frac{\lambda D}{d}

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

4 0
2 years ago
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