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belka [17]
4 years ago
11

When an object loses electrons, it acquires a(n) ____________________ electric charge?

Physics
1 answer:
madam [21]4 years ago
4 0
It acquires a positive electric charge.
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When an object threw to the free space to make an angle of 25 degree at an initial speed of 15 m/sec, the ball takes time to rea
alekssr [168]

Answer:

The horizontal distance traveled by the projectile is 15.23 m.

Explanation:

Given;

angle of projection, θ = 25⁰

initial velocity of the projectile, u = 15 m/s

time of flight, t = 1.12 s

The the travelling path of the object is calculated as the range of the projectile

R = u_x t\\\\R = (15\ Cos \ 25^0) \times 1.12\\\\R = 13.595 \times 1.12\\\\R = 15.23 \ m

Therefore, the horizontal distance traveled by the projectile is 15.23 m.

8 0
3 years ago
What is the average acceleration of a tennis ball that has an initial velocity of 6.0 m/s [E] and a final velocity of 7.3 m/s [W
Marizza181 [45]

Given :

The average acceleration of a tennis ball that has an initial velocity of 6.0 m/s.

and a final velocity of 7.3 m/s.

It is in contact with a tennis racket for 0.094 s

To Find :

The average acceleration of the tennis ball.

Solution :

We know, average acceleration is given by :

a_{avg}=\dfrac{Final \ velocity-Initial\ velocity}{Time\ Taken}\\\\a_{avg}=\dfrac{7.3-6.0}{0.094}\ m/s^2\\\\a_{avg}=13.83\ m/s^2

Therefore, average velocity is given by 13.83 m/s².

Hence, this is the required solution.

7 0
3 years ago
Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second
Sophie [7]

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light   λ  = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

=  2.09 /1.65  

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

3 0
3 years ago
Cart a having a mass of 150 kg initially moving to the right at a speed of 8 m/s collides with cart b with a mass of 150 kg, ini
kakasveta [241]

The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s

<h3>Conservation of Linear Momentum</h3>

Given Data

  • Mass of cart one M1  = 150kg
  • Initial Velocity U1 = 8m/s
  • Final VelocityV1 = 5 m/s

Mass of cart two M2  = 150kg

Velocity U2 = 6m/s

Applying the principle of conservation of linear momentum we have

M1U1+M2U2 = M1V1+ M2V2

a. what is the speed of cart b after collision

substituting our given data we have

150*8+ 150*6 = 150*5+150*V2

1200 + 900 = 1200+ 150V2

2100 - 1200 = 150V2

900 = 150V2

Divide both sides by 150

V2 = 900/150

V2 = 6m/s

b. what is the total momentum of the system before and after collision

Total Momentum in the system is

Total momentum = Momentum before Impact+ Momentum after Impact

Total momentum = M1U1+M2U2 + M1V1+ M2V2

Total momentum = 1200 + 900 + 1200+ 900

Total momentum = 4200 kg m/s

Learn more about Conservation of Linear Momentum here:

brainly.com/question/7538238

6 0
2 years ago
What happens when an object with a lower density is placed in a container with an
maksim [4K]
The only thing that definitely happens in every such case is:
The container becomes heavier.
5 0
3 years ago
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