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maxonik [38]
3 years ago
15

Consider a ball rolling down the decreasing slope inside a semicircular bowl (the slope is steep at the top rim, gets less steep

toward the bottom, and is zero (no slope) at the bottom). As the ball rolls from the rim downward toward the bottom, its rate of gaining speed
Physics
1 answer:
satela [25.4K]3 years ago
8 0
The answer would be:
<span>It's rate of gaining speed decreases.
The rate at which speed changes is called acceleration, 
You can think of this problem as an inclined plane. But the angle of an inclined plane is constantly decreasing.
We know that on a frictionless inclined plane acceleration of an object is:
</span>a=gsin(\theta)
<span>Where g is the gravitational acceleration of the Earth and \theta is the angle of an inclined plane. 
Using our analogy, the ball would start on an inclined plane with a 90-degree angle and that angle would continue to decrease to zero. 
The sine function is 1 at 90 degrees and is equal to zero at 0 degrees. Since our acceleration is proportional to the sine, and sine function is decreasing with the angle, our acceleration is also decreasing.

</span>
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A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle
Yuri [45]

Answer:

(a). Index of refraction are n_{red} = 1.344 & n_{violet} = 1.406

(b). The velocity of red light in the glass v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

The velocity of violet light in the glass v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

Explanation:

We know that

Law of reflection is

n_1 \sin\theta_{1} = n_2 \sin\theta_{2}

Here

\theta_1 = angle of incidence

\theta_2 = angle of refraction

(a). For red light

1 × \sin 56.6 = n_{red} × \sin 38.4

n_{red} = 1.344

For violet light

1 × \sin 56.6 = n_{violet} × \sin 36.4

n_{violet} = 1.406

(b). Index of refraction is given by

n = \frac{c}{v}

n_{red} = 1.344

v_{red} = \frac{c}{n_{red} }

v_{red} = \frac{3(10^{8} )}{1.344}

v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

v_{violet} = \frac{3(10^{8} )}{1.406}

v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

This is the velocity of violet light in the glass.

8 0
3 years ago
A small immersion heater is rated at 255 W . The specific heat of water is 4186 J/kg⋅C°.
tangare [24]

Answer:

0.2093 s

Explanation:

First, we the energy needed to heat the cup of soup

Q = cm(t₂-t₁).................. Equation 1

Where c = specific heat capacity of water, m = mass of water, t₂ = Final temperature, t₁ = initial temperature.

Given: c = 4186 J/kg.°C, m = density of water×volume of water,

Where density of water = 1 kg/m³, volume of water = 250/1000000 = 0.00025 m³,

Therefore, m = 1×0.00025  = 0.00025 kg, t₂ = 62°C, t₁ = 17 °C

Substitute into equation 1

Q = 4186(0.00025)(62-17)

Q = 47.0925 J.

Secondly we look for the time using

W = Q/t................. Equation 2

Where W = power rating of the heater.

make t the subject of the equation

t = Q/W.............. Equation 3

Given: W = 225 W

Substitute into equation 4

t = 47.0925/225

t = 0.2093 s

3 0
4 years ago
In a pith ball experiment, the two pith balls are at rest. The magnitude of the tension in each string is |T|=0.55N, and the ang
Len [333]

Answer:

Explanation: Two pith ball will repel each other . they will remain balaced due to tension in the spring whose one component balances the weight and the other balances the repulsive force on each.

The gravitational force will be balanced by T cos 27.33 and the electrostatic repulsive force will be balanced by T sin27.33

So

Fg =T cos 27.33

= .55 X .888

= .49 N

Fq = T sin27.33

=.55 x .459

= .25 N.

5 0
3 years ago
A projectile is fired vertically from earth's surface with an initial speed of 7.3 km/s. neglecting air drag, how far above the
Dominik [7]
We can solve the problem by using conservation of energy. 

In fact, initially the projectile has only kinetic energy, which is given by
K= \frac{1}{2}mv^2
where m is the projectile's mass while v=7.3 km/s=7300 m/s is its initial velocity.

At the point of maximum height, the speed of the projectile is zero, so it only has gravitational potential energy which is equal to
U=mgh
where g is the gravitational acceleration and h is the maximum height of the projectile.

Since the energy must be conserved, we can equalize K and U to find the value of h:
\frac{1}{2}mv^2=mgh
h= \frac{v^2}{2g}= \frac{(7300 m/s)^2}{2 \cdot 9.81 m/s^2}=2.72 \cdot 10^6 m = 2720 km
5 0
4 years ago
PLEASE ANSWER QUICK!!!!!!!!!!!!!
Leno4ka [110]
The answer to your question is True.
5 0
3 years ago
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