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Anarel [89]
2 years ago
11

The 0.01 kg marble is dropped from rest at A through the smooth glass tube and accumulate in the basket at C as shown in Figure

Q2(b). Determine: i) the velocity of the marble at B ii) the horizontal distance R of the basket from the end of the tube, and iii) the speed at which the marble falls into the basket.​
Physics
1 answer:
Anuta_ua [19.1K]2 years ago
4 0

Crazy Wally Ok Ok ok hhahahaha

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andrew-mc [135]
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5 0
3 years ago
What is the current in a series circuit that has two resistors (4.0 ohms and 7.5 ohms) and a power source of 9.0 volts?
Olin [163]

-- First, we have to decide how to handle the two resistors.

The effective resistance of resistors in series is the sum
of their individual resistances.  That is, they act like a single
resistor, whose resistance is the sum of all of them.

So in this question, the 4.0 ohms and the 7.5 ohms act like a
single resistor of 11.5 ohms.

-- The current in the circuit is

                       (the supply voltage) / (the total resistance)

                   =                (9.0 volts)  /  (11.5 ohms)

                   =                      0.783... Ampere        (rounded)

3 0
3 years ago
Read 2 more answers
N which order did the events forming our solar system occur?
brilliants [131]

Answer:

The solar nebula became hot and dense because of that it pulling in more gas. This flattened into a rotating disk. It  spun  faster and faster, forming the Sun.

Explanation:

hope this helps

4 0
2 years ago
Read 2 more answers
A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit
Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

3 0
3 years ago
Ian’s school is exactly 6 blocks north of his house. What is his average velocity on a walk from home to school that takes him 1
choli [55]

Answer:

.5 units north

<em><u>or</u></em>

55 ft/minutes(<em>squared</em>) north

Explanation:

.5 is what your info gives me but if i take that the average distance a block is it is 660ft than the answer is 55.

4 0
3 years ago
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