What's the name of a star that does 'not move'?

An electron is accelerated through 2400 V from rest and then enters a region in which there is a uniform 1.70 T magnetic field.

aalyn [17]

Answer:

Explanation:

Let v be the velocity acquired by electron in electric field

V q = 1/2 m v²

V is potential difference applied on charge q , m is mass of charge , v is velocity acquired

2400 x 1.6 x 10⁻¹⁹ = .5 x 9.1 x 10⁻³¹ x v²

v² = 844 x 10¹²

v = 29.05 x 10⁶ m /s

Maximum force will be exerted on moving electron when it moves perpendicular to magnetic field .

Maximum force = Bqv , where B is magnetic field , q is charge on electron and v is velocity of electron

= 1.7 x 1.6 x 10⁻¹⁹ x 29.05 x 10⁶

= 79.02 x 10⁻¹³ N .

Minimum force will be zero when electron moves along the direction of magnetic field .

5 0

3 years ago

18. Compared to its weight on Earth, a 5 kg object on the moon will weigh A. the same amount. B. less. C. more.

s2008m [1.1K]

Answer:

B. less

Explanation:

acceleration due to gravity on Earth, g = 9.8 m/s²

acceleration due to gravity on Moon, g = 1.6 m/s²

Given mass of the object as, m = 5 kg

Weight of an object is given as, W = mg

Weight of the object on Earth, W = 5 x 9.8 = 49 N

Weight of the object on Moon, W = 5 x 1.6 = 8 N

Therefore, the object weighs less on the moon compared to its weight on Earth.

The correct option is "B. less"

8 0

3 years ago

Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

8 0

3 years ago

Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th

bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

the mass of planet B to be, Mₓ = M

the radius of planet A to be, R₁

the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁² - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂² [where, R₂ = 2R₁]

= GM/4R₁² -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

brainly.com/question/13299152

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6 0

2 years ago

What percent of the world lumber harvest goes to paper production 10% 20% 25% 50%

sergejj [24]

The answer is 50% of the paper production

8 0

3 years ago