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3 years ago

If the toy car weighs 250g and has a kinetic energy of 2.0 J, what is its velocity

Physics

1 answer:

sergejj [24]3 years ago

3 0

Explanation:

Kinetic energy (J) = 2j

mass= 250g

velocity=?

1kg=1000g

mass= 250/1000

mass=0.25kg

Kinetic energy (J) = ½ x mass (kg) x [velocity]² (m/s)

2=1/2 × 0.25× [velocity]²

2=0.125× [velocity]²

[velocity]² = 2/0.125

[velocity]²=16

velocity= (16)^1/2

velocity= 4 m/s

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At what position in its elliptical orbit is the speed of a planet a maximum? when it is closest to the sun when it is farthest f

Phantasy [73]

Answer:

1.when it is closest to the sun

2.when it is midway between its farthest

Explanation:

According to the law of Kepler's

T ² ∝ r³

T=Time period

r=semi major axis

We also know that time period T given as

$T=\dfrac{2\pi r}{v}$

v=Speed

$v=\dfrac{2\pi r}{T}$

$v\alpha \dfrac{r}{T}$

$v^2\alpha \dfrac{r^2}{T^2}$

$v^2\alpha \dfrac{r^2}{r^3}$

$v^2\alpha \dfrac{1}{r}$

$v\alpha \dfrac{1}{\sqrt{r}}$

So we can say that ,when r is more then the speed will be minimum and when r is low then speed will be maximum.

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3 years ago

answer this question: When you can not see what is taking place, but other senses indicate occurrences. This is called _________

Phoenix [80]

This is called an indirect observation.

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3 years ago

Early cameras were little more than a box with a pinhole on the side opposite the film. (a) What angular resolution would you ex

ollegr [7]

Answer:

angular resolution = 0.07270° = 1.269 × $10^{-3}$ rad

greatest distance from the camera = 118.20 m = 0.118 km

Explanation:

given data

diameter = 0.50 mm = 0.5 × $10^{-3}$ m

distance apart = 15 cm = 15× $10^{-2}$ m

wavelength λ = 520 nm = 520 × $10^{-9}$ m

to find out

angular resolution and greatest distance from the camera

solution

first we expression here angular resolution that is

sin θ = $\frac{1.22* \lambda }{D}$ .......................1

put here value λ is wavelength and d is diameter

we get

sin θ = $\frac{1.22*520*10^{-9}}{0.5*10^{-3}}$

θ = 0.07270° = 1.269 × $10^{-3}$ rad

and

distance from camera is calculate here as

θ = $\frac{I}{r}$ .................2

I = $\frac{15*10^{-2}}{1.269*10^{-3}}$

I = 118.20 m = 0.118 km

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3 years ago

A constant magnetic field passes through a single rectangular loop whose dimensions are 0.42 m x 0.54 m. The magnetic field has

olga_2 [115]

Answer:

$0.29887\ \text{V}$

Explanation:

A = Area = $0.42\times0.54\ \text{m}^2$

B = Magnetic field = 1.7 T

$\theta$ = Angle that magnetic field makes with the plane of the loop = $71^{\circ}$

t = Time = 0.42 s

EMF is given by

$\varepsilon=A\cos\theta\dfrac{dB}{dt}\\ =\dfrac{0.42\times 0.54\times\cos 71^{\circ}\times 1.7}{0.42}\\ =0.29887\ \text{V}$

The magnitude of the average emf induced in the loop is $0.29887\ \text{V}$ .

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3 years ago

What is a mass spectrometer? How does it work?

ZanzabumX [31]

Answer:

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3 years ago