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cluponka [151]
3 years ago
6

Which will have a larger momentum when moving at the same speed: a 2,000-kg truck or a 1,000-kg sedan

Physics
1 answer:
Alina [70]3 years ago
5 0

Answer:

2000 kg

Explanation:

Given that Which will have a larger momentum when moving at the same speed: a 2,000-kg truck or a 1,000-kg sedan

According to the definition of momentum, momentum is the product of mass and velocity.

That is,

Momentum = mass × velocity

Since velocity or speed is the same, then, the one of higher mass will have a greater momentum.

Therefore, the 2000 kg truck will have the greater momentum.

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When an object is turning a corner what direction is the acceleration?
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A laser pulse of duration 25 ms has a total energy of 1.4 J. The wavelength of this radiation is
SpyIntel [72]

Answer:

n = 4 x 10¹⁸ photons

Explanation:

First, we will calculate the energy of one photon in the radiation:

E = \frac{hc}{\lambda}\\\\

where,

E = Energy of one photon = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of radiation = 567 nm = 5.67 x 10⁻⁷ m

Therefore,

E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{5.67\ x\ 10^{-7}\ m}

E = 3.505 x 10⁻¹⁹ J

Now, the number of photons to make up the total energy can be calculated as follows:

Total\ Energy = nE\\1.4\ J = n(3.505\ x\ 10^{-19}\ J)\\n = \frac{1.4\ J}{3.505\ x\ 10^{-19}\ J}\\

<u>n = 4 x 10¹⁸ photons</u>

8 0
3 years ago
Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
kirza4 [7]

Answer:

The value is U  = 0.06 \  J

Explanation:

From the question we are told that

The value of charge on each three point charge is

q_1 = q_2 = q_3 =q=  1.05 \mu C  =  1.05 *10^{-6} \  C

The length of the sides of the equilateral triangle is r  =  0.500 \

Generally the total potential energy is mathematically represented as

U  = k *  [ \frac{q_1 *  q_2}{r}  +  \frac{q_2 *  q_3}{r}   + \frac{q_3 *  q_1}{r} ]

=> U  = k * 3 * \frac{q^2}{r}

Here k is coulomb constant with value k = 9*10^{9}\  kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

=>    U  = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }

U  = 0.06 \  J

6 0
3 years ago
A long, solid rod 5.3 cm in radius carries a uniform volume charge density. Part A If the electric field strength at the surface
Blizzard [7]

Answer:

\rho = 7.35\times \muC/m^3

Explanation:

given,

Radius of the solid rod, R = 5.3 cm

Electric field strength,E = 22 kN/C

Let the volume charge density be ρ

From Gauss law

  E = \dfrac{Rl}{2\epsilon_0}

 ε₀ is the permitivity of free space

  R is the radius of the rod

 and also,

\rho=\dfrac{2E\epsilon_0}{R}

ρ is the volume charge density

\rho=\dfrac{2\times 22\times 10^{3}\times 8.854\times 10^{-12}}{0.053}

\rho = 7.35\times 10^{-6}\ C/m^3

\rho = 7.35\times \muC/m^3

Hence, the volume charge density is equal to \rho = 7.35\times \muC/m^3

6 0
3 years ago
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