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spin [16.1K]
2 years ago
6

Draw a simple circuit that lights up a bulb. ​

Physics
2 answers:
coldgirl [10]2 years ago
5 0

Hope this helps

Ps- U can pick between these two pictures

Please mark as brainliest

BlackZzzverrR [31]2 years ago
4 0

\:  \:  \:  \:  \:

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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
Dependent variable is the _______ that happens because of the UI independent variable. (No answer choices provided)
Damm [24]

Answer: One is called the dependent variable and the other the independent variable. The independent variable is the variable the experimenter changes or controls and is assumed to have a direct effect on the dependent variable.

4 0
3 years ago
A scientist is studying a sample of matter the matter has no definite shape or volume and it can conduct an electric current. Th
Anna35 [415]
D. plasma ..................
4 0
3 years ago
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Stopping distance of vehicles When brakes are applied to a moving vehicle, the distance it travels before stopping is called sto
DochEvi [55]

Answer:

Stopping distance = 40m

Explanation:

Given the following :

Initial speed of vehicle before applying brakes = 72km/hr

Converting km/hr to m/s:

72km/hr = [(72 * 1000)m] / (60 * 60)

72km/hr = 72,000m / 3600s

72km/hr = 20m/s

Deceleration after applying brakes (-a) (negative acceleration) = - 5m/s^2

From the 3rd equation of motion:

v^2 = u^2 + 2as

Where v = final Velocity ; u= Initial Velocity ; a = acceleration and s = distance

Final velocity when the car stops will be 0

Therefore ;

v^2 = u^2 + 2as

0 = 20^2 + 2(-5)(s)

0 = 400 - 10s

10s = 400

s = 400/10

s = 40m

Therefore, the stopping distance of the car = 40 meters

7 0
3 years ago
What is one reason supporting genetic modifications to food
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Health and safety are two reasons
4 0
3 years ago
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