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Vladimir79 [104]

2 years ago

9

Which of the following will exert a force on a magnet? another magnet a nearby piece of glass a nearby piece of plastic a nearby

electric current

2 answers:

spin [16.1K]2 years ago

8 0

The magnet will attract or repell the other magnet so magnet is one answer

the glass obviously doesn't exert a force

the ellectric current makes the ellectrons and protons, in the current, allign so that they attract or repell the other magnet

gladu [14]2 years ago

4 0

The nearby magnet and the electric current both will.

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A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

2 years ago

Suppose you need 7.0m of Grade 70 tow chain, which has a diameter of /38" and weighs 2.16/kgm, to tow a car. How would you calcu

Zina [86]

Answer:

<em>M = l × m</em>

Explanation:

M = total mass

l = total length

m = mass per unit length

Note: The unit of weight in the question i.e. /kgm is wrong. The correct unit is kg/m.

4 0

2 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a very light rope that goes over an ideal pulley. If the

AleksAgata [21]

Answer:

acceleration = 2.4525‬ m/s²

Explanation:

Data: Let m1 = 3.0 Kg, m2 = 5.0 Kg, g = 9.81 m/s²

Tension in the rope = T

Sol: m2 > m1

i) for downward motion of m2:

m2 a = m2 g - T

5 a = 5 × 9.81 m/s² - T

⇒ T = 49.05‬ m/s² - 5 a Eqn (a)‬

ii) for upward motion of m1

m a = T - m1 g

3 a = T - 3 × 9.8 m/s²

⇒ T = 3 a + 29.43‬ m/s² Eqn (b)

Equating Eqn (a) and(b)

49.05‬ m/s² - 5 a = T = 3 a + 29.43‬ m/s²

49.05‬ m/s² - 29.43‬ m/s² = 3 a + 5 a

19.62 m/s² = 8 a

⇒ a = 2.4525‬ m/s²

5 0

2 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

kobusy [5.1K]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Q₁ = 572.8 $* 10^{-9}$ C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

8 0

2 years ago

A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus

dexar [7]

Answer:

$a=5m/s^2$

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

$v_{1}=150m/s$

and the final velocity is :

$v_{2}=0 m/s$

so, the change in velocity is:

$\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) = 150 m/s$

and the change in time , according to the problem:

$\Delta t=30s$

So, the aceleration is:

$a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2$

6 0

2 years ago