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3 years ago

Which statements describe how a machine can help make work easier? Check all that apply.

Physics

1 answer:

nikdorinn [45]3 years ago

6 0

Answer:

It can exert a greater force through a greater distance than the force used on the machine.

Explanation:

A machine is a device that makes work easier and so it should exert greater force than the force applied on it.

You might be interested in

Choose a substance you're familiar with. what are its physical and chemical properties? How would you measure its density? What

Gekata [30.6K]

you can look at magnesium, it can react with oxygen to form oxides. (chemical) it is malleable and a solid at room temperature. (physical)

to measure its density, the mass and volume can be worked out and from this density too. look up the equation, it is quite easy :)
physical changes -- it can be melted, and oxidized

the chemical changes of oxidation magnesium looses electrons to form oxides, this is a chemical reaction- chemical change..--- use to get the density use (rho) or density D = M/V

6 0

2 years ago

A small metal object is tied to the end of a string and whirled round a circular path of radius 30cm. The object makes 20 oscill

LUCKY_DIMON [66]

Answer:

(i) The angular speed of the small metal object is 25.133 rad/s

(ii) The linear speed of the small metal object is 7.54 m/s.

Explanation:

Given;

radius of the circular path, r = 30 cm = 0.3 m

number of revolutions, n = 20

time of motion, t = 5 s

(i) The angular speed of the small metal object is calculated as;

$\omega = \frac{20 \ rev}{5 \ s} \times \frac{2 \pi \ rad}{1 \ rev} = \frac{40\pi \ rad}{5 \ s} = 8\pi \ rad/s = 25.133 \ rad/s$

(ii) The linear speed of the small metal object is calculated as;

$v = \omega r\\\\v = 25.133 \ rad/s \ \times \ 0.3 \ m\\\\v = 7.54 \ m/s$

6 0

3 years ago

Why does a light bike have more kinetic energy

FrozenT [24]

Because it doesn't use energy it uses mechanical and kinetic

3 0

2 years ago

Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat

STALIN [3.7K]

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

$B = \frac{\mu_0}{2}\frac{Ia^2}{(z^2+a^2)^{3/2}}$

Where,

a = Radius

z = Distance to the magnetic field

I = Current

$\mu_0 =$ Permeability constant in free space

Our values are given as

$D=2a = 16cm \rightarrow$ diameter of the sphere then,

$a = 0.08m$

Thus z = a

$B = \frac{\mu_0}{2}\frac{Ia^2}{(a^2+a^2)^{3/2}}$

$B = \frac{\mu_0I}{2(2^{3/2})a}$

$B = \frac{\mu_0 I}{2^{5/2}a}$

Re-arrange to find I,

$I = \frac{2^{5/2}Ba}{\mu_0}$

$I = \frac{2^{5/2}(3*10^{-12})(8*10^{-2})}{4\pi*10^{-7}}$

$I = 1.08*10^{-6}A$

Therefore the current at the pole of this sphere is $1.08*10^{-6}A$

5 0

2 years ago

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: =E−Ryn2 In this equation Ry stands for

skelet666 [1.2K]

Answer:

λ = 162 10⁻⁷ m

Explanation:

Bohr's model for the hydrogen atom gives energy by the equation

$E_{n}$ = - k²e² / 2m (1 / n²)

Where k is the Coulomb constant, e and m the charge and mass of the electron respectively and n is an integer

The Planck equation

E = h f

The speed of light is

c = λ f

E = h c /λ

For a transition between two states we have

$E_{n}$ - $E_{m}$ = - k²e² / 2m (1 / $n_{f}$ ² -1 / $n_{i}$ ²)

h c / λ = -k² e² / 2m (1 / $n_{f}$ ² - 1/ $n_{i}$ ²)

1 / λ = (- k² e² / 2m h c) (1 / $n_{f}$ ² - 1/ $n_{i}$ ²)

The Rydberg constant with a value of 1,097 107 m-1 is the result of the constant in parentheses

Let's calculate the emission of the transition

1 /λ = 1.097 10⁷ (1/10² - 1/8²)

1 / λ = 1.097 10⁷ (0.01 - 0.015625)

1 /λ = 0.006170625 10⁷

λ = 162 10⁻⁷ m

3 0

3 years ago