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Elza [17]
3 years ago
11

Which factors will increase the speed of a sound wave in the air?

Physics
1 answer:
Dafna11 [192]3 years ago
7 0
A higher temperature, stiffer materials, and less dense materials increase the speed of sound.
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Which are most likely to be used during a routine visit to the dentist?
Paraphin [41]

Answer:

The most likely items to be used are;

Ultrasound and X-rays

Explanation:

A routine visit to a dentist consists of two areas of activities, including;

a) Dental examination and check up

b) Oral prophylaxis, and dental cleaning

The dental examination may involve the use of X-rays, which allow the detection of cavities between the teeth

The dental cleaning can be carried out with the use of an ultrasound cleaner, which allow the cleaning of sensitive teeth without hurting the patient

Therefore, the items most likely to be used during a routine dental visit are ultrasound and X-rays

8 0
3 years ago
Read 2 more answers
One terminal of a car battery is said to be connected to “ground.” since it is not really connected to the ground, what is meant
GalinKa [24]
This expression means that the negative terminal (-) is connected to the metal chassis or engine, which means that all voltages used for the electrical devices in the car are measured with respect to the car's chassis or engine. 
Today's vehicles have a negative ground system, which means that the vehicle's steel frame or chassis is directly connected to the negative side of the battery via the negative battery cable. 
3 0
3 years ago
Two double bonds means that the total number of electrons being shared in the molecule is
torisob [31]

One double bond consists 4 electrons, so 2 double bonds means 8 electrons

7 0
3 years ago
You leave your home at 1pm. At 3pm, you are 100 km east of your house. What was your average velocity in km/hr
Rasek [7]

Answer:

50km/h

Explanation:

Average velocity = change in distance (or distance travelled) divided by/ the change in time (or time taken.)

The change in distance has been given as 100km.

The change in time is 3pm-1pm = 2 hours.

Therefore the average velocity was 100/2 = 50km/h (to the east).

Hope this helped!

4 0
3 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
3 years ago
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