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poizon [28]
3 years ago
11

Another name for a pivot point is the:

Physics
1 answer:
Makovka662 [10]3 years ago
7 0

Answer:

I think it's c .... fulcrum

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What is the current in a 120V circuit if the resistance is 20Ω?
Kaylis [27]

We have: I=\frac{U}{R}=\frac{120}{20}=6A

ok done. Thank to me :>

6 0
2 years ago
A bus is moving at a speed of 150km/hr. Begins to slow at a constant rate of 3.0m/s each second. Find how far it goes before sto
ladessa [460]

Answer:

Distance = 13.9 meters

Explanation:

Given the following data;

Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s

Decelerating speed = 3m/s

To find the distance travelled with this speed;

Distance = maximum speed/decelerating speed

Distance = 41.67/3

Distance = 13.9 meters

Therefore, the bus would travel a distance of 13.9 meters before stopping.

4 0
2 years ago
A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0
3 years ago
Calculate the mass (in SI units) of (a) a 160 lb human being; (b) a 1.9 lb cockatoo. Calculate the weight (in English units) of
kondaur [170]

Explanation:

1. Force applied on an object is given by :

F = W = mg

(a) A 160 lb human being, F = 160 lb

g = acceleration due to gravity, g = 32 ft/s²

m=\dfrac{F}{g}

m=\dfrac{160\ lb}{32\ ft/s^2}

m = 5 kg

(b) A 1.9 lb cockatoo, F = 1.9 lb

m=\dfrac{F}{g}

m=\dfrac{1.9\ lb}{32\ ft/s^2}

m = 0.059 kg

2. (a) A 2300 kg rhinoceros, m = 2300 kg

W=2300\ kg\times 32\ ft/s^2=73600\ lb

(b) A 22 g song sparrow, m = 22 g = 0.022 kg

W=0.022\ kg\times 32\ ft/s^2=0.704\ lb

Hence, this is the required solution.

5 0
3 years ago
Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values
Vlad1618 [11]

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and  4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and  1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and  2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and  2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and  8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and  8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge   Right Charge Resulting force(N)

1μC                     4μC                  90 N

4μC                   1μC                    90 N

2μC                  2μC                    90 N

1μC                    2μC                   45 N

1μC                  8μC                    180 N

2μC                  8μC                  360 N

5 0
3 years ago
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