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liberstina [14]
3 years ago
12

I need answer fast please

Physics
1 answer:
Hatshy [7]3 years ago
4 0

gold density = 38.6 *10^-3 / 2 *10^-6 = 19300 Kg/m3 Iron density = 46.8 *10^-3 / 6 *10^-6 = 7800 Kg/m3 water mass = density*volume = 1 * 10 = 10 grams = 0.01 Kg

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When air is blown across the top of an open <span>water bottle, air molecules in the bottle vibrate at a particular frequency and sound is produced in a process called "refraction". 
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3 years ago
Someone is trying to balance a (110cm) plank with certain forces.
Mashutka [201]

Answer:

a. i. 30 Nm ii. This moment is a clockwise positive moment.

b. i. 15 Nm ii, This moment is a counter-clockwise negative moment.

c. i. The plank will not balance. ii. The plank would tip up.

d. 150 N

Explanation:

a) Calculate the moment of the 60N force (about O), then name its type.

i. Calculate the moment of the 60N force (about O)

Since moment = Force × perpendicular distance from point of moment ,

M = Fd

Since F = 60 N and d = 50 cm = 0.5 m

M = 60 N × 0.5 m = 30 Nm

ii. Then name its type.

This moment is a clockwise positive moment.

b) Calculate the moment of the 30N force (about O), then name its type.

i. Calculate the moment of the 30N force (about O),

Since moment = Force × perpendicular distance from point of moment ,

M' = F'd'

Since F' = 30 N and d' = 50 cm = 0.5 m

M' = 30 N × 0.5 m = 15 Nm

ii. Then name its type.

This moment is a counter-clockwise negative moment.

c) Will the plank balance? If not, which way will it tip?

i. Will the plank balance?

The plank will balance if the net moment on it is zero

So net moment, M' = positive moment - negative moment = M - M' = 30 Nm   - 15 Nm = 15 Nm

Since the net moment on the plank is M" = 15 Nm ≠ 0,<u>the plank will not balance.</u>

ii Which way will it tip?

The plank would tip in the direction of the greater moment since the net moment is positive. <u>This moment tilts the plank in a clockwise direction, so the plank would tip up.</u>

d) What extra force would be needed at (B) to balance the plank?

The extra force must balance the net moment,

So M" = F"d" where F" = force and d" = distance of force from O = 10 cm = 0.10 m

F" = M"/d"

= 15 Nm/0.10 m

= 150 N

4 0
3 years ago
Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

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3 years ago
WILL GIVE BRAINLIEST! QUICK PLEASE HELP!!!
LenKa [72]

Answer:Weight is the result of gravity . The gravitational field strength of Earth is 10 N/kg (ten newtons per kilogram). This means an object with a mass of 1 kg would be attracted towards the centre of Earth by a force of 10 N. We feel forces like this as weight.

Explanation:

3 0
3 years ago
Read 2 more answers
Please help me this is timed . <br> Find x if a = 3.0 m/s^2
storchak [24]

Answer:

x = 50 N

Explanation:

Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.

Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.

_______

F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a

m (mass which we are given) = 150 kg

a (acceleration which we are given) = 3.0m/s

________

So F = m × a → F2 - F1 = m × a →

500 - F1 = 150 × 3.0 → 500 - F1 = 450 →

-F1 = -50 → F1 = 50

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3 years ago
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