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3 years ago

suppose you increase your walking speed from 5m/s to 14m/s in a period of 1 . what is your acceleration?

1 answer:

7 0

Acceleration = (change in speed) / (time for the change.

change in speed = (ending speed) - (starting speed) = 9 m/s.

Acceleration = (9 m/s) / (period of 1) .

We don't know the units of the 'period of 1'.
If it means '1 second', then the acceleration is 9 m/s² .

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Convertir: A. 3Km a m B. 250 ma Km C. 1000Cm a m D. 10000 mm a Cm

Katen [24]

Answer:

A. 3,000,000 m

B. 0.25 km

C. 10 m

D. 1,000 cm

Explanation:

no hablo español, así que solo ingrese esto en el traductor de G*ogle

A. One kilometer equals 1000 meters, so

3,000*1,000 = 3,000,000 m

B. One meter equals 0.001 kilometer, so

250*0.001 = 0.25 km

C. One centimeter equals 0.01 meter

1,000*0.01 = 10 m

D. One milimeter equals 0.1 centimer, so

10,000*0.1 = 1,000

4 0

3 years ago

You are pushing a 30-kg block on a rough floor in a direction that is parallel to the floor. The block moves with a uniform 2 m/

kykrilka [37]

Answer:dd

Explanation:

I dunno

3 0

3 years ago

Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol

klasskru [66]

The current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

Explanation:

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

J = (1.6 $\times$ 10^-19) / 0.001 (104 $\times$ 1200 + 1016 $\times$ 500)

J = 1012480 $\times$ 10^-16 A / m^2.

J = 1.01 $\times$ 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 $\times$ 10^-9 $\times$ 0.002 = 2.02 $\times$ 10^-12

So, the current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

6 0

4 years ago

Pls help What graph indicates no reference to direction of motion, and explain why.

Komok [63]

Answer:

graph A

Explanation:

the slope of the distance-time graph is speed, speed is a scalar (with magnitudes but no direction)

but the slope for the velocity time graph is acceleration, acceleration is vector quantity ( has magnitude and direction)

4 0

3 years ago

irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano

Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

= 1 / (0.042) [ 1 parsecs = 1 arcseconds ]

= 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

brainly.com/question/16471213

#SPJ4

6 0

2 years ago