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3 years ago

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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3 years ago

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>

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

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3 years ago