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anygoal [31]
4 years ago
12

A kangaroo jumps up with an initial velocity of

Physics
1 answer:
ycow [4]4 years ago
4 0

Answer:

2.3

Explanation:

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3 years ago
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Two uncharged metal spheres, #1 and #2, are mounted on insulating support rods. A third metal sphere, carrying a positive charge
tatuchka [14]

Answer:

The question is incomplete as some details are missing, here is the details ;

In this final state

a) spheres #1 and #2 both carry negative charge.

b) sphere #1 carries negative charge and #2 carries positive charge.

c) spheres #1 and #2 are still uncharged.

d) sphere #1 carries positive charge and #2 carries negative charge.

e) spheres #1 and #2 both carry positive charge.

Explanation:

From the concept of electrostatics, if a positively charged sphere is brought close to #2, there will be attraction of the opposite charges(-ve) towards it.

Now connecting a copper wire between #1 and #2, opposite charges will flow from #1 towards #2. disconnecting the copper wire makes #1 to be positively charged and #2 to be negatively charge and from the laws of attraction ; Like charges repel and unlike charges attract. the correct option is d.

5 0
3 years ago
Which of the following is an ionic compound?
const2013 [10]

compound is covalent! The bond between HI is a polar covalent bond. ... It is because Hydrogen is electropositive in nature in comparison to Iodine.

7 0
3 years ago
Two small charged spheres are 7.59 cmcm apart. They are moved, and the force on each of them is found to have been tripled. How
tiny-mole [99]

Answer:

The two small charged spheres are now 4.382 cm apart

Explanation:

Given;

distance between the two small charged sphere, r = 7.59 cm

The force on each of the charged sphere can be calculated by applying Coulomb's law;

F = \frac{kq_1q_2}{r^2}

where;

F is the force on each sphere

q₁ and q₂ are the charges of the spheres

r is the distance between the spheres

F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 =  7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 =  7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm

Therefore, the two small charged spheres are now 4.382 cm apart.

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4 years ago
What is the highest energy of radiation?
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