The classification and collection of data that are in the form of numbers is called

A stationary speed gun emits a microwave beam at 2.10x10^10 Hz. It reflects off a car and returns 638 Hz lower. What is the spee

stiv31 [10]

Answer:

The answer is 4.55 m/s!

Explanation:

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6 0

3 years ago

How many significant digits are in the number 204.0920<br>

monitta

Answer:

Seven

Explanation:

The rules for significant digits are:

Non-zero digits are always significant.
Zeros between significant digits are also significant.
Trailing zeros are significant only after a decimal point.

Here, the 2, 4, 9, and 2 are significant because they are non-zero digits.

The first two 0s are significant because they are between significant digits.

The last 0 is significant because it is a trailing zero after a decimal point.

Therefore, all seven digits are significant.

3 0

3 years ago

Read 2 more answers

A bowling ball with a mass of 8 kg is moving at a speed of 5 m/s. What is its<br> kinetic energy?

natali 33 [55]

Answer:

100J

Explanation:

Kinetic energy=1/2mv^2

Kinetic energy=(1/2 x 8)x5^2

Kinetic energy=4x25

Kinetic energy=100

100J

5 0

3 years ago

A 0.0780 kg lemming runs off a

kotegsom [21]

Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

velocity (v) = 4.84 m/s

gravity (g) = 9.81m/s^2

1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

so KE = 0.91 J

2. Next, solve for Potential energy (PE)

PE = mgh

(0.0780kg)(9.81m/s^2)(5.36m) = 4.10 J

so PE = 4.10 J

3. Mechanical Energy , E = KE + PE

Plug in values for KE and PE

KE + PE = 0.91J + 4.10 J = 5.01 J

4 0

3 years ago

A mass spectrometer was used in the discovery of the electron. In the velocity selector, the electric and magnetic fields are se

Mama L [17]

Answer:

Explanation:

Radius of dee, r = 8 mm = 0.008 m

Electric field, e = 400 V/m

Magnetic field, B = 4.7 x 10^-4 T

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

(a) Let v is the speed of electrons.

$v = \frac{Bqr}{m}$

$v = \frac{4.7\times 10^{-4}\times 1.6\times 10^{-19}\times 0.008}{9.1 \times 10^{-31}}$

v = 661098.9 = 661099 m/s

(b)

$\frac{e}{m}=\frac{1.6 \times 10^{-19}}{9.1\times 10^{-31}}$

e / m = 1.76 x 10^14 C / kg

(c) Let K be the kinetic energy

K = 0.5 x mv²

K = 0.5 x 9.1 x 10^-31 x 661099 x 661099

K = 1.99 x 10^-19 J

K = 1.24 eV

So, the potential difference is

V = 1.24 V

(d) if the acceleration voltage is doubled

V = 2 x 1.24 = 2.48 V

So, Kinetic energy

K = 2.48 eV

K = 2.48 x 1.6 x 10^-19 = 3.968 x 10^-19 J

Let v is the speed

K = 0.5 x mv²

3.968 x 10^-19 = 0.5 x 9.1 x 10^-31 x v²

v = 933856.5 m/s

Let the new radius is r.

$r=\frac{mv}{Bq}$

$r=\frac{9.1\times 10^{-31}\times 933856.5}{4.7\times 10^{-4}\times 1.6\times 10^{-19}}$

r = 0.0113 m = 1.13 cm

7 0

3 years ago