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ycow [4]
3 years ago
9

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\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.

7 0
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.
5 0
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}}

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3 0
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}.

8 0
3 years ago
What is a mass spectrometer? How does it work?
ZanzabumX [31]

Answer:

Mass spectrometry is an analytical technique that measures the mass-to-charge ratio of ions. The results are typically presented as a mass spectrum, a plot of intensity as a function of the mass-to-charge ratio.

Explanation:

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7 0
2 years ago
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