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Genrish500 [490]
3 years ago
7

An object that’s charged has more electrons than protons. An object that’s charged has fewer electrons than protons. An object t

hat’s charged has the same number of electrons and protons.
Physics
1 answer:
Levart [38]3 years ago
3 0
An element is neutral when the charged is equal to zero. This means that the number of electrons and protons are equal. When an object has a NEGATIVE charge, its electrons are more than its protons. Also, the object has a POSITIVE charge when the number of its electrons are fewer compared to the number of the protons. 
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John added 150 grams of salt to 2.5 liters of water. what is the concentration of the salt solutions​
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2 years ago
Two blocks a and b ($m_a>m_b$) are pushed for a certain distance along a frictionless surface. how does the magnitude of the
Yuki888 [10]

Answer:

the magnitude of the work done by the two blocks is the same.

Explanation:

The work done by block a on block b is given by:

W_a = F_a d

where Fa is the force exerted by block a on block b, and d is the distance they cover.

The work done by block b on block a is given by:

W_b = F_b d

where Fb is the force exerted by block b on block a, and d is still the distance they cover.

For Newton's third law, the force exerted by block a on block b is equal to the force exerted by block b on block a, therefore

F_a = F_b

and so

W_a=W_b

3 0
3 years ago
A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski
Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0
3 years ago
A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci
Simora [160]

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

4 0
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