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Kipish [7]
2 years ago
13

A positive charge moving up enters a magnetic field pointing out of the screen. What is the direction of the magnetic force on t

he charge?
A) Up
B) Down
C) Right
D) Left

Physics
2 answers:
erma4kov [3.2K]2 years ago
6 0
According to the right-hand thumb rule, the forefinger gives the velocity of charge, the thumb gives the magnetic force and the center finger gives the direction of magnetic field.

then, as shown in the picture, the <span>direction of the magnetic force on the charge is in the right direction.</span>

dexar [7]2 years ago
5 0

Answer:

right

Explanation:

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3. The expression 0.62 x10^3 is equivalent to...
Korolek [52]

\\ \sf\longmapsto 0.62\times 10^3

\\ \sf\longmapsto 62\times 10^{-2}\times 10^3

\\ \sf\longmapsto 62\times 10^{-2+3}

\\ \sf\longmapsto 62\times 10^1

\\ \sf\longmapsto 62\times 10

\\ \sf\longmapsto 620

5 0
1 year ago
Read 2 more answers
A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa
gavmur [86]

Answer : The correct option is, (d) 535^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of copper = 0.10cal/g^oC

c_2 = specific heat of water = 1.00cal/g^oC

m_1 = mass of copper = 120 g

m_2 = mass of water = 300 g

T_f = final temperature of mixture = 35^oC

T_1 = initial temperature of copper = ?

T_2 = initial temperature of water = 15^oC  

Now put all the given values in the above formula, we get:

120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC

T_1=535^oC

Therefore, the temperature of the kiln was, 535^oC

7 0
2 years ago
Read 2 more answers
Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo
lisabon 2012 [21]

Answer:

El peso del cartel es 397,97 N

Explanation:

La tensión dada en cada segmento del cable = 2000 N

El desplazamiento vertical del cable = 50 cm = 0,5 m

La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N

El peso del signo = 397,97 N.

8 0
2 years ago
NEED ANSWER PLEASE!!!!
olganol [36]

Answer:A, Concave

Explanation:

3 0
2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

          Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

         Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

         Em₀ = Em_f

         ½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

        v_f ² = v₀² + 2g (y₁ -y₂)

         

we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
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