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Anastasy [175]
3 years ago
11

What is used to measure motion

Physics
1 answer:
dem82 [27]3 years ago
5 0
Speed is the measure<span> of </span>motion<span>.</span>
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A leaky 10-kg bucket is lifted from the ground to a height of 14 m at a constant speed with a rope that weighs 0.5 kg/m. Initial
Lina20 [59]

solution:

Weight of bucket = 10kg

Length or distance =14m

Weight of rope=0.5kg/m

At any point x of the rope,

=(0.5)(14-x)

=(7-0.5x)

Since the water finishes draining at 14m level and total weight of water is 42kg

Total mass=(7-0.5x)+(42-3x)+10=(59-3.5x)kg

Force=(9.8)(59-3.5x)

work w =\lim_{n \to \infty }\sum_{i \to 1}^{n}(9.8)(59-3.5x)\Delta x\\=\int_{0}^{14}(9.8)(59-3.5x)dx\\=9.8\int_{0}^{14}(59-3.5x)dx\\9.8((59x-\frac{3.5x^2}{2})){_{0}}^{14}\\9.8(59(14)-\frac{3.5(14)^2}{2})\\=4733.4\\therefore,\\W=4733.4J\approx 4733J

5 0
3 years ago
Refer to the figure for this question: Four particles form a square. the charges are q1=q4=Q and q2=q3=q. (Part A) What is Q/q i
nikitadnepr [17]
Draw a vector diagram. The net force on particle 1 = F12 + F13 + F14 These forces have to be added as vectors. 
We will resolve our forces along the direction 1-4 F12 (tot) = -kQq / a^2 in the direction of particle 4 F12 = -kQq *sin (45) / a^2 F12 = -kQq /( a^2 * sqrt(2) ) 
By symetry this is the same as F13 F13 = -kQq /( a^2 * sqrt(2) ) 
F14 = -kQQ / (Sqrt(2)*a) ^ 2 
For net force on particle 1 : 
F12+F13+F14 = 0 -2kQq /( a^2 * sqrt(2) ) + -kQQ / (Sqrt(2)*a) ^ 2 = 0 
Some simple manipulation should give you : 
Q/q = -2 sqrt(2) 
4 0
3 years ago
Read 2 more answers
Someone please help me will give BRAILIEST
Daniel [21]
It causes the greenhouse effect
7 0
3 years ago
To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
A loop with a break in it that prevents current from flowing is called a(n)_______<br> circuit.
Annette [7]

Answer:

an open circuit

Explanation:

7 0
4 years ago
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