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svlad2 [7]
3 years ago
13

Some of the principal minerals commonly found in igneous rocks are

Physics
1 answer:
Verdich [7]3 years ago
3 0
I believe that the answer is C. Hope this Helps:)))
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15. When you cannot stop safely at a yellow traffic light before entering an intersection, ______________. A. stop in the inters
Elan Coil [88]

Answer:  When you cannot stop safely at a yellow traffic light before entering an intersection, enter the intersection carefully and continue across.

Explanation: To find the correct answer, we need to know more about the traffic signal rules.

<h3>What is the traffic signal rules?</h3>
  • Red light- Indicator for the motorists to stop.
  • Green-Signal for safety and word GO.
  • Yellow- This signal let you know that the red signal is about to be displayed.
  • when it's turned on, you can start slowing down to come to a stop in anticipation of red light.
  • when we cannot stop safely at a yellow traffic light before entering an intersection, enter the intersection carefully and continue across.

Thus, we can conclude that, the option C is correct.

Learn more about the traffic signal rules here:

brainly.com/question/28044804

#SPJ4

4 0
2 years ago
Explain how the zodiacal constellations are different from the other constellations.
Temka [501]

Answer:

They lie on the Ecliptic (apparent path of the Sun around Earth)

Explanation:

There are total 88 constellations in the sky out of which only 12 are considered as zodiacal constellations or Sun signs. This is because these are the constellations which lie on the Ecliptic.

Ecliptic is the apparent path of Sun around the Earth.

In other words, it can be said that the Sun appears to reside in 12 constellations throughout the year as the Earth revolves around it.

6 0
3 years ago
The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per
Natasha_Volkova [10]

Answer:

0.2631 N/C

Explanation:

Given that:

The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m

The radius of the thick wire  r' = 0.55 mm = 0.55 × 10⁻³ m

The numbers of electrons passing through B, N = 6.0  × 10¹⁸ electrons

Electron mobility  μ =  6.0 x 10-4 (m/s)/(N/C)

= 0.0006

The number of electron flow per second is calculated as follows:

I = \frac{q}{t}

I = \frac{Ne}{t}

I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}

I = 0.96 \ A

The magnitude of the electric field is:

E = \frac{I}{ \mu n eA}

E = \frac{I}{ \mu n e(\pi r^2)}

E = \frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}

E = 0.2631 N/C

8 0
3 years ago
A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The
Oksi-84 [34.3K]

Answer:

The horizontal distance d does the ball travel before landing is 1.72 m.

Explanation:

Given that,

Height of ramp h_{1}=2.30\ m

Height of bottom of ramp h_{2}=1.69\ m

Diameter = 0.17 m

Suppose we need to calculate the horizontal distance d does the ball travel before landing?

We need to calculate the time

Using equation of motion

h_{2}=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2h_{2}}{g}}

t=\sqrt{\dfrac{2\times1.69}{9.8}}

t=0.587\ sec

We need to calculate the velocity of the ball

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2

K.E=\dfrac{7}{10}mv^2

Using conservation of energy

K.E=mg(h_{1}-h_{2})

\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})

v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})

Put the value into the formula

v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}

v=2.922\ m/s

We need to calculate the horizontal distance d does the ball travel before landing

Using formula of distance

d =vt

Where. d = distance

t = time

v = velocity

Put the value into the formula

d=2.922\times 0.587

d=1.72\ m

Hence, The horizontal distance d does the ball travel before landing is 1.72 m.

8 0
3 years ago
A body of unknown mass is attached to an ideal spring with force constant 120 N/m. It is found to vibrate with a frequency of 6.
Kaylis [27]

Explanation:

Given that,

Force constant of the spring, k = 120 N/m

Frequency of vibration, f = 6 Hz

Solution,

(a) Let T is the time period of the spring. The relation between the frequency and the time period is given by :

T=\dfrac{1}{f}

T=\dfrac{1}{6}

T = 0.167 s

(b) Let \omega is the angular frequency. It is given by :

\omega=2\pi f

\omega=2\pi \times 6

\omega=37.69\ rad

(c) The relation between angular frequency and the spring constant is given by :

\omega=\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{\omega^2}

m=\dfrac{120}{(37.69)^2}

m = 0.084 kg

or m = 84 grams

Therefore, it is the required solution.

4 0
3 years ago
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