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3 years ago

Define an element and give 5 examples of elements that are important to life .

Physics

1 answer:

fenix001 [56]3 years ago

8 0

Answer:

A pure substance consisting only of atoms with the same number of protons in their nuclei-these appear on the periodic table

Oxygen

Hydrogen

Carbon

Sulfur

Phosphate

Nitrogen

Magnesium

Calcium

Potassium

Chlorine

(I know that these are more examples than needed, but you can use any)

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What type of Psychologist was B.F. Skinner?

Sladkaya [172]

Answer:

Behaviorist

Explanation:

BF skinner was big on behaviorism and produced massive amounts of support for operant conditionining. He literally had a Skinner box where he did experiments with animals regarding conditionining .

4 0

2 years ago

An object with a mass of 70 kilograms is supported at a height of 8 meters above the ground, what's the potential energy of the

Olegator [25]

Answer:

5488 Joules

Explanation:

Use the formula for the potential energy:

$E_p = mgh = 70kg\cdot 9.8\frac{m}{s^2}\cdot 8m = 5488 J$

6 0

3 years ago

A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Negle

Artist 52 [7]

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

θ₀ = 90° - 5.22°

7 0

3 years ago

What is the potential energy of a 5kg rock that is 7m high on a hill

Goryan [66]

Explanation:

Gravitational potential energy = mgh = (5)(9.81)(7) = 343.35J.

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

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