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3 years ago

Acceleration is a vector quantity. O O A. True B. False

Physics

2 answers:

iragen [17]3 years ago

4 0

The answer is A. True, Acceleration is a vector quantit

PtichkaEL [24]3 years ago

3 0

True ( I think ) ............................

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What time is the eclipse happening tonight?.

andrew11 [14]

The first location to see the partial solar eclipse begin is at 3.58 a.m. EST (08:58 UTC), the greatest point of total solar eclipse occurs at 6 a.m. EST (11:00 UTC) and the last location to see the partial eclipse end is at 8:02 a.m. EST (13:02 UTC) according to Time and Date.

3 0

2 years ago

SVEN [57.7K]

Answer:

Frequency = 1,550Hz

Explanation:

To solve this we can use the equation: $f=\frac{v}{\lambda}$

(frequency = velocity/wavelength).

We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m

Now we can substitute these values into the formula and calculate to solve:

$f=\frac{340}{0.22} \\\\f=1545.454...$

Simplify to 3 significant figures:

f = 1,550Hz

(Which I believe is just below a G6 if you were interested)

Hope this helped!

4 0

3 years ago

Name the net force between objects that allows a car to travel in a circle

iren [92.7K]

I think centripetal force ☺

6 0

3 years ago

A 23.5 kg object travels a distance of 85 m in 30 s. Find the momentum. (Hint: you must find speed first)

VLD [36.1K]

Yup, I think you add all of them

7 0

2 years ago

An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo

Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =16321.0769 N/C

The electric field strength at point (x,y) = (0cm, 20 mm) is =35321.58999 N/C

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=16321.0769 N/C

Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷ 0.000000152881

=35321.58999 N/C

8 0

2 years ago