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3 years ago

How does inertia relate to coin dropping????

1 answer:

3 0

Well Inertia means something wants to stay in place, and in reality that coin wants to stay in one place, If you placed it on an index card on a cup, and SLOWLY pulled it, it wouldn't be fast enough to overcome that force, if you pulled it quickly that coin would stay in place and drop into the cup.

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A football is thrown due north across a 40 meter river and it takes 2 seconds to cover that distance.

Alexxx [7]

Answer:

I. Speed = 20m/s

II. Velocity = 20m/s due North.

Explanation:

Given the following data;

Distance = 40m

Time = 2secs

To find the speed;

Mathematically, speed is given by the formula;

$Speed = \frac{distance}{time}$

Substituting into the equation, we have;

$Speed = \frac{40}{2}$

Speed = 20m/s.

In physics, we use the same formula for calculating speed and velocity. The only difference is that speed is a scalar quantity and as such has magnitude but no direction while velocity is a vector quantity and as such it has both magnitude and direction.

$Velocity = \frac{distance}{time}$

Therefore, the velocity is 20m/s due North.

6 0

3 years ago

A 10- kilogram block is pushed across a horizontal surface with a horizontal force of 20 N against a friction force of 10 N. The

Temka [501]

Answer:

$1m/s^2$

Explanation:

Mass of block=10 kg

Applied horizontal force =F=20 N

Friction force=f=10 N

We have to find the acceleration of block.

Net force=Applied horizontal force-friction force

$ma=F-f$

Where F= Horizontal force

f=Friction force

m=Mass of object

a=Acceleration of object

$10a=20-10=10$

$a=\frac{10}{10}=1 m/s^2$

Hence, the acceleration of the block= $1m/s^2$

4 0

3 years ago

In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?

agasfer [191]

Answer:

The frequency of the photon is $3.069\times10^{14}\ Hz$ .

Explanation:

Given that,

Energy $E=4.07\times10^{-19}\ J$

We need to calculate the energy

Using relation of energy

$E_{4}-E_{2}=\Delta E$

Where, $\Delta E$ = energy spacing

$4h\nu-2h\nu=4.07\times10^{-19}$

$\nu=\dfrac{4.07\times10^{-19}}{2h}$

Put the value of h into the formula

$\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}$

$\nu=3.069\times10^{14}\ Hz$

Hence, The frequency of the photon is $3.069\times10^{14}\ Hz$ .

4 0

3 years ago

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

4 0

3 years ago

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside thi

Zanzabum

Complete question:

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.

Answer:

The electric field inside this metal resistor is 3125 V/m

Explanation:

Given;

length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m

diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m

the potential difference due to electric field between the two ends of the resistor, V = 10 V

The electric field inside this metal resistor is given by;

ΔV = EL

where;

ΔV is change in electric potential

E = ΔV / L

E = 10 / (3.2 x 10⁻³ )

E = 3125 V/m

Therefore, the electric field inside this metal resistor is 3125 V/m

7 0

3 years ago