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2 years ago

2. The amount of charge and the distance from the charge determine the...

Physics

2 answers:

olasank [31]2 years ago

8 0

Answer:

Strength of the electric field

Explanation:

The amount of charge and the distance from the charge determine the strength of the electric field.

Mathematically, it can be written as :

$E=\dfrac{1}{4\pi \epsilon_0}\dfrac{q}{r^2}$

where,

q is the electric charge

r is the distance between the charges

It can be also defined as the electric force per unit electric charge. Hence, the correct option is (c) " Strength of the electric field ".

cricket20 [7]2 years ago

7 0

Answer:

the answer is c Strength of the electric field

Explanation:

it is that bcause

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Which instrument is launched into the atmosphere to collect pressure, temperature, humidity, wind speed, and other data?

koban [17]

C. Radiosonde is the answer

the above mentioned is not correct

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3 years ago

A long point-object, mass = 1.0 kg, moves in a circular path at a radial distance = 0.5 m from the axis of rotation. What is the

Vinvika [58]

Answer: $0.25\ kg-m^2$

Explanation:

Given

mass of Point object $m=1 kg$

Distance $r=0.5 m$

Since mass is moving in circular path therefore every time mass is at distance of r from center .

Also Moment of Inertia tells about the distribution of mass over the given region with respect to center of mass.

Therefore $I=mr^2$

$I=1\times 0.5^2$

$I=0.25\ kg-m^2$

5 0

3 years ago

A girl rolls a ball up an incline and allows it to re- turn to her. For the angle and ball involved, the acceleration of the bal

zalisa [80]

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

7 0

2 years ago

A soccer player icks a rock horizontally off a 40m high cliff into a pool f water if the player hears the sound of the splash s

Semenov [28]

Answer:

v = 9.936 m/s

Explanation:

given,

height of cliff = 40 m

speed of sound = 343 m/s

assuming that time to reach the sound to the player = 3 s

now,

time taken to fall of ball

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 40}{9.8}}$

t = 2.857 s

distance

d = v x t

d = v x 2.875

time traveled by the sound before reaching the player

$t_0 = t - t_{fall}$

$t_0 = 3 - 2.875$

$t_0 = 0.143 s$

distance traveled by the wave in this time'

r = 0.143 x 343

r= 49.05 m

now,

we know.

d² + h² = r²

d² + 40² = 49.05²

d =28.387 m

v x 2.875=28.387 m

v = 9.936 m/s

7 0

2 years ago

Topic Gravitational force amd firld strength.. help me please

I am Lyosha [343]

The gravitational force between m₁ and m₂ has magnitude

$F_{1,2} = \dfrac{Gm_1m_2}{x^2}$

while the gravitational force between m₁ and m₃ has magnitude

$F_{1,3} = \dfrac{Gm_1m_3}{(15-x)^2}$

where x is measured in m.

The mass m₁ is attracted to m₂ in one direction, and attracted to m₃ in the opposite direction such that m₁ in equilibrium. So by Newton's second law, we have

$F_{1,2} - F_{1,3} = 0$

Solve for x :

$\dfrac{Gm_1m_2}{x^2} = \dfrac{Gm_1m_3}{(15-x)^2} \\\\ \dfrac{m_2}{x^2} = \dfrac{m_3}{(15-x)^2} \\\\ \dfrac{(15-x)^2}{x^2} = \dfrac{m_3}{m_2} = \dfrac{60\,\rm kg}{40\,\rm kg} = \dfrac32 \\\\ \left(\dfrac{15-x}x\right)^2 = \dfrac32 \\\\ \left(\dfrac{15}x-1\right)^2 = \dfrac32 \\\\ \dfrac{15}x - 1 = \pm \sqrt{\dfrac32} \\\\ \dfrac{15}x = 1 \pm \sqrt{\dfrac32} \\\\ x = \dfrac{15}{1\pm\sqrt{\dfrac32}}$

The solution with the negative square root is negative, so we throw it out. The other is the one we want,

$x \approx 6.74\,\mathrm m$

5 0

2 years ago