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3 years ago

14

This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters: what is the correspon

ding time interval?
A. 2 seconds
B. 4 seconds
C. 5 seconds
D. 10 seconds
E. 15 seconds

1 answer:

shtirl [24]3 years ago

6 0

Answer: The correct answer is option (D).

Explanation:

The area under the line is 10 meters that is displacement which means that area enclosed by the right angled triangle is 10 unit squares.

So, the it can be written as:

$\text{area of triangle}=\frac{1}{2}\times velocity\times time$

$10=\frac{1}{2}\times 2.0 m/s\times time$

$time =10 seconds$

Hence the, the correct answer is option (D).

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The concept required to solve this problem is associated with potential energy. Recall that potential energy is defined as the product between mass, gravity, and change in height. Mathematically it can be described as

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$\Delta h = -16.1m-0m$

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3 years ago

The maximum wavelength For photoelectric emissions in tungsten is 230 nm. What wavelength of light must be use in order for elec

notka56 [123]

Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

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h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

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λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

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Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

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3 years ago

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

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$F_1=\frac{Kq_1q_2}{r^2}$

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

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levacccp [35]

0.77 m/s2 directed 35° south of west

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3 years ago