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3 years ago

13

when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv

ing the charge?

1 answer:

sveta [45]3 years ago

5 0

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

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Why is it good that adolescents prune their brains

Aneli [31]

Answer:

Synaptic pruning is an essential part of brain development. By getting rid of the synapses that are no longer used, the brain becomes more efficient as you age.

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3 years ago

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

4 0

3 years ago

What would be the magnitude of the electric field 0.75 m from a 0.63 C master charge and what would be the force on a 0.50 C tes

bagirrra123 [75]

The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

<h3>Electric field on the master charge</h3>

E = kq/r²

where;

q is magnitude of master charge
r is distance of separation
k is Coulomb's constant

E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

where;

E is electric field
q is the test charge

F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

Learn more about electric field here: brainly.com/question/14372859

#SPJ1

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2 years ago

3. An object of mass 90 kg travels down a slide.

Gwar [14]

Answer:

3) Ep = 13243.5[J]

4) v = 17.15 [m/s]

Explanation:

3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.

m = mass = 90 [kg]

h = elevation = 15 [m]

Potential energy is defined as the product of mass by gravity by height.

$E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]$

This energy will be transformed into kinetic energy.

Ek = 13243.5 [J]

4) The velocity can be determined by defining the kinetic energy, as shown below.

$E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]$

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3 years ago

A colloidal liquid in a has is called

S_A_V [24]

Aerosol is the answer

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3 years ago