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Nikolay [14]
2 years ago
15

If you were to walk toward a Van der Graaff generator, the effect of its electrical field would get weaker as you walked towards

it.
True or False
Physics
1 answer:
gavmur [86]2 years ago
7 0

This is False.  

If you walk towards a Vander Graaff generator and touch it, your body will be electrically charged with the same sign as the generator (that’s why your hair stands up, because it is charged with static electricity, and…electrical charges of the same sign repel each other)

Note that the Vander Graaff generator is isolated from the ground, so, if the electrical charge is increased, electrical arcs will form and fly off it to any nearby grounded object or person.


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oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student
pochemuha

Answer:

m = 63 grams

Explanation:

ω = 10 cycles/s(2π radians/cycle) = 20π rad/s

ω = √(k/m)

m = k/ω² = 250/(20π)² = 0.06332... kg

6 0
3 years ago
A .422 kg coffee mug rests on a table top 0.63 m above the floor. What is the potential energy of the mug with respect to the fl
kondaur [170]

Answer:

2.61 J

Explanation:

Since potential energy U = mgy where m = mass of object, g = acceleration due to gravity = 9.8 m/s² and y = height of object above the ground.

Now for the coffee mug, m= 0.422 kg and it is 0.63 m on a table, so it is 0.63 m above the ground. Thus, y = 0.63 m.

We compute U

U = mgy

= 0.422 kg × 9.8 m/s² × 0.63 m

= 2.605 J

≅ 2.61 J

So, the potential energy of the mug with respect to the floor is 2.61 J

4 0
3 years ago
A dragster race car can accelerate from rest to incredible speeds. In one case a dragster is able to finish the 305 m run in 3.6
Dafna1 [17]

Answer:

45.89m/s²

Explanation:

Given

Distance S = 305m

Time t = 3.64s

To get the acceleration during this run, we will apply the equation of motion:

S = ut+1/2at²

Substitute the given parameters into the formula and calculate the value of a

305 = 0+1/2 a(3.64)²

304 = 1/2(13.2496)a

304 = 6.6248a

a = 304/6.6248

a = 45.89m/s²

Hence the average acceleration during this run is 45.89m/s²

4 0
2 years ago
¿cual es la velocidad de un haz de electrones que marchan sin desviarse cuando pasan a traves de un campo magnetico perpendicula
Elina [12.6K]

Answer:

La velocidad del haz de electrones es 1.78x10⁵ m/s. Este valor se obtuvo asumiendo que el campo magnético dado (3500007) estaba en tesla y que la fuerza venía dada en nN.

Explanation:

Podemos encontrar la velocidad del haz de electrones usando la Ley de Lorentz:

F = |q|vBsin(\theta)     (1)

En donde:

F: es la fuerza magnética = 100 nN

q: es el módulo de la carga del electron = 1.6x10⁻¹⁹ C

v: es la velocidad del haz de electrones =?

B: es el campo magnético = 3500007 T

θ: es el ángulo entre el vector velocidad y el campo magnético = 90°

Introduciendo los valores en la ecuación (1) y resolviendo para "v" tenemos:

v = \frac{F}{qBsin(\theta)} = \frac{100 \cdot 10^{-9} N}{1.6 \cdot 10^{-19} C*3500007 T*sin(90)} = 1.78 \cdot 10^{5} m/s            

Este valor se calculó asumiendo que el campo magnético está dado en tesla (no tiene unidades en el enunciado). De igual manera se asumió que la fuerza indicada viene dada en nN.

Entonces, la velocidad del haz de electrones es 1.78x10⁵ m/s.  

Espero que te sea de utilidad!                                        

7 0
3 years ago
A straight wire segment 5 m long makes an angle of 30° with a uniform magnetic field of 0.37 T. Find the magnitude of the force
SIZIF [17.4K]

Answer:

The magnitude of the force on the wire is 2.68 N.

Explanation:

Given that,

Length of the wire, L = 5 m

Magnetic field, B = 0.37 T

Angle between wire and the magnetic field, \theta=30^{\circ}

Current in the wire, I = 2.9 A

We need to find the magnitude of the force on the wire. The magnetic force in the wire is given by :

F=BIL\ \sin\theta\\\\F=0.37\ T\times 2.9\ A\times 5\ m\times \ \sin(30)\\\\F=2.68\ N

So, the magnitude of the force on the wire is 2.68 N. Hence, this is the required solution.

7 0
3 years ago
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