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marshall27 [118]
3 years ago
14

A metal ion (X) with a charge of 4+ is attracted to a nonmetal ion (Z) with a

Physics
1 answer:
Keith_Richards [23]3 years ago
8 0

Answer:

A) X3Z4

Explanation:

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Calculate the magnitude and direction of the electric field 2.0 m from a long wire that is charged uniformly at λ = 4.0 × 10-6 C
Len [333]

Answer: 71.93 *10^3 N/C

Explanation: In order to calculate the electric field from long wire we have to use the Gaussian law, this is:

∫E*dr=Q inside/εo  Q inside is given by: λ*L then,

E*2*π*r*L=λ*L/εo

E= λ/(2*π*εo*r)= 4* 10^-6/(2*3.1415*8.85*10^-12*2 )= 71.93 * 10^3 N/C

6 0
3 years ago
Compare and contrast potential and kinetic energy
inysia [295]
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7 0
3 years ago
Read 2 more answers
two horses are pulling a 325 kg wagon, initially at rest. The horses exert 250 N and 178 N forward forces, respectively. Ignorin
Dovator [93]

Answer:

AFter 3.5 s, the wagon is moving at:   4.62\,\,\frac{m}{s}

Explanation:

Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):

Net force = 250 N + 178 N = 428 N

Therefore, the acceleration from Newton's 2nd Law is:

F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:

v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}

7 0
3 years ago
A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a co
Flura [38]

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = -F_Dd

Where:

d = Distance traveled = 450 m

Therefore,

\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd and

v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh  -F_Dd}{ \frac{1}{2}m}  = v_1^2 + 2gh -\frac{   2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}

= 658.714 m²/s²

v₂ = 25.665 m/s

Her speed at the bottom of the slope = 25.665 m/s.

6 0
3 years ago
What is the weight of a 7.0 kilogram bowling ball on the surface of the moon
Artist 52 [7]
The acceleration on surface of moon =1.67m/s^2

Weight =mass ×acceleration
=7×1.67
=11.69N
8 0
3 years ago
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