Why are plane mirrors and convex mirrors unable to form real images

Distinguish between force and friction. class 8

KengaRu [80]

Answer: Friction is the resistance to motion of one object moving relative to another. For example, when you try to push a book along the floor, friction makes this difficult. Force: Force is essentially a push, or a pull action, that can lead to certain outcomes.

7 0

2 years ago

for the same type of atom with 8 protons, 9 neutrons, and 10 electrons, what type of atom or ion is it

Ilia_Sergeevich [38]

Answer:

O²⁻

Explanation:

Number of protons = 8

Number of neutrons = 9

Number of electrons = 10

What type of atom or ion is it = ?

Solution:

Protons are the positively charged particle in an atom

Neutrons do not carry any charges

Electrons are negatively charged particles

For this atom, the number of protons helps to identify what specie it is; so this is an oxygen atom.

Now,

Charge = Number of protons - Number of electrons

Charge = 8 - 10 = -2

The charge on the atom is -2 and so it is an oxygen ion with -2 charge

The ion is O²⁻

8 0

2 years ago

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

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6 0

1 year ago

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

dexar [7]

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0

3 years ago

amping equipment weighing 6.0 kN is pulled across a frozen lake by means of ahorizontal rope. The coecient of kinetic friction i

Harman [31]

Answer:

Work done = 422.45 kJ

Explanation:

given,

weight of equipment = 6 kN

coefficient of kinetic friction = 0.05

distance up to which it is pulled = 1000 m

constant acceleration = 0.2 m/s²

Work done by the camper = ?

actual acceleration acting a'

m a = m a' - μ mg

a' = a + μ g

a' = 0.2 + 0.05 x 9.8

a' = 0.69 m/s²

Work done = Force x distance

F = m a'

$F = \dfrac{6000}{9.8} \times 0.69$

F = 422.44897 N

Work done = F x d

Work done = 422.44897 x 1000

Work done = 422449 J

Work done = 422.45 kJ

4 0

3 years ago