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2 years ago

Calculate the speed of a car if it covers the distance of 100m in 5seconds.

Physics

2 answers:

NeTakaya2 years ago

8 0

Answer:

200m/s

Explanation:

$v = \frac{displacement}{time}$

Oksi-84 [34.3K]2 years ago

4 0

Answer:

the answer is 20

Explanation:

distance =100m

time=5seconds

speed=?

so distance/time

=100/5

=20

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Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is 180.00 km/h and

snow_lady [41]

Answer:

Explanation:

From the given information:

The coordinate axis is situated in the east and north direction.

So, the north will be the y-axis and the east will be the x-axis

Similarly, the velocity of the plane in regard to the air in the coordinate system will be $v_{P/A} = v( cos \theta \ i + sin \theta \ j)$

where:

$v_{P/A}$ = velocity of the plane in regard to the air

v = velocity

θ = angle of inclination of the plane with respect to the horizontal

replacing v = 180 km/ and θ = 20° in above equation, then:

The velocity of the airplane in the coordinate system as:

$v_{P} = v_o( cos \phi \ i + sin \phi \ j)$

where;

$v_p$ = velocity of the airplane

$v_o$ = velocity

∅ = angle of inclination with regard to the base axis;

Then; replacing $v_o$ = 150 km/h and ∅ = 30°

Therefore, the velocity of the plane in the system is :

$v_p = v_A + v_{P/A}$

$v_A= v_P -v_{P/A}$ --- (1)

$v_A=$ ( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j

$v_A=$ (-39.24 km/h)i + (13.44 km/h) j

The magnitude is:

$v_A= (-39.24 km/h)i + (13.44 km/h) j$

$|v_A|^2 = \sqrt{ (-39.24 km/h)^2+ (13.44 km/h)^2}$

$v_A$ = 41.48 km/h

The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.

The angle of motion is:

tan θ = 39.24/13.44

tan θ = 2.9

θ = $tan ^{-1} (2.9)$

θ = 70.97°

The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.

5 0

3 years ago

what if l was standing and all of continent are coming together and becomes pangea.What will happen???

azamat

Answer:

not really

Explanation:

pangea took million of years to separate. Years by Years the continents move. for example The two continents are moving away from each other at the rate of about 2.5 centimeters (1 inch) per year. And there movent is due to the activity beneath the earths crust. The plates are moved by currents in the magma at the very surface of the earths mantle , which then in turn cause things like earthquakes and volcanic eruptions.

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3 years ago

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What were some benefits to the american television from analog broadcast to digital broadcast?

Elanso [62]

Answer:

nothing no no no

Explanation:

nothing cause no

8 0

2 years ago

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1. El puente mas largo del mundo es el puente Akashi Kaikyo, en Japón. El puente mide 3910 m de largo y está construido con acer

bearhunter [10]

Answer:

It's 1.0000042 times longer in summer than in winter. It represents a 1.6 centimeters difference between seasons.

Explanation:

The linear coefficient of thermal expansion for steel is about $1.2*10^{-7}\°C^{-1}$ . From the equation of linear thermal expansion, we have:

$L_f=L_0(1+\alpha\Delta T)$

Taking the winter day as the initial, and the summer day as the final, we can take the relationship between them:

$L_{summer}=L_{winter}[1+(1.2*10^{-7}\°C^{-1})(30\°C+5\°C)]\\\\L_{summer}=(1.0000042)L_{winter}$

It means that the bridge is 1.0000042 times longer in summer than in winter. If we multiply it by the length of the bridge, we obtain that the difference is of about 1.6 centimeters between the two seasons.

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3 years ago

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

5 0

1 year ago

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Calculate the speed of a car if it covers the distance of 100m in 5seconds​.

Calculate the speed of a car if it covers the distance of 100m in 5seconds.