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3 years ago

What are all the moon phases

2 answers:

8 0

New Moon
Waxing Crescent
First Quarter
<span>Waxing Gibbous
</span>Full Moon
<span>Waning Gibbous
</span>Last Quarter
Waning Crescent

;)

photoshop1234 [79]3 years ago

5 0

New moon, waxing crescent, first quarter, waxing gibbous, full moon, waning gibbous, third quarter (or last quarter) , waning crescent .

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A guitar vibrates in frequency with a tuning fork when the fork is held against its body. This is a case of

Butoxors [25]

Energy transfer the energy from the tuning fork is being transferred to the guitar<span />

4 0

3 years ago

Muốn đun sôi 200g nước từ 30 độ cần cung cấp nhiệt lượng bao nhiêu :

Aleks [24]

Answer:

cần cung cấp 70 độ vì nước sôi ở 100°C

Explanation:

7 0

3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$

The frictional force on the block W₂, $F_{f2}$ ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ $F_{w2}$ = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + $\mathbf{F_{w2}}$

The frictional force from the ground, $\mathbf{F_{f1}}$ = N×μ + $\mathbf{F_{f2}}$ = P

Where;

P = The horizontal force applied to the block

P = (W₁ + $\mathbf{F_{w2}}$ ) × μ + $\mathbf{F_{f2}}$

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0

3 years ago

Parents have a major role in observing how young children grow at home. Create and discuss simple developmental monitoring tips

Rudik [331]

Answer:

by observation of their children

4 0

2 years ago

for the primitive yo-yo in fig, to find the downward acceleration of the cylinder and the tension in the string. you can take th

mixas84 [53]

The downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.

<h3>Downward acceleration of the cylinder</h3>

The downward acceleration of the solid cylinder is determined from the principle of conservation of angular momentum as shown below;

Iα = Tr

where;

I is moment of inertia of the solid cylinder
α is angular acceleration of the cylinder
T is tension in the string
r is length of the string

α = Tr/I

$\frac{a}{R} = \frac{Tr}{I} \\\\\frac{a}{R} = \frac{Tr}{\frac{1}{2} MR^2}\\\\\frac{a}{R} =\frac{2Tr}{MR^2} \\\\a = \frac{2Tr}{MR}$

where;

a is the downward acceleration of the solid cylinder
R is radius of the cylinder

Thus, the downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.

Learn more about acceleration here: brainly.com/question/605631

#SPJ1

7 0

2 years ago