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ycow [4]
3 years ago
15

3. A student pushed a 10.0 kg box across a level, frictionless floor with an acceleration of 5.00 m/s.

Physics
2 answers:
solniwko [45]3 years ago
8 0
Pretty sure the answer is B
riadik2000 [5.3K]3 years ago
6 0
The answer is b 2.00N
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A force of 99 N causes a box to accelerate at a rate of 11 m/s2. What is the mass of the box? (Ignore frictional effects.)
lys-0071 [83]

We know, F = m.a

m = F/a

m = 99/11 = 9

So, The mass of the box is 9 Kg

5 0
4 years ago
Water flows horizontally through a garden hose with an inner diameter of .012 m at a speed of 7.8 m/s. it exits out a small nozz
tigry1 [53]

<span>Answer: 110.12 m/s </span>

We will use the formula A1V1 = A2V2 where 7.8 m/s is divided with 0.0085 m then multiply to 0.12 m, the result will be 110.117 or 110.12 m/s. This is related to the continuity of fluid flow in which as liquid moves horizontally, the same amount of liquid goes out as it comes in or the liquid itself do not change as it moves but the speed does when the diameter changes.

 

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IP address then call the cops
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3 years ago
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A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi
EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is,\rm d_1 = 2.2 \ cm

initial radius,

r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm

The initial crossection area;

\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times  (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2

The final crossection area;

\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2

The initial flow rate is;

R = density ×velocity ×area

\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1  = 3.947 \ m/sec

The speed of the water in the wider part will be;

From the continuity equation;

\rm A_1 V_1 = A_2V_2  \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec

Hence, the speed of the water in the wider part will be 1.194 m/sec.

To learn more about the speed, refer to the link;

brainly.com/question/7359669

#SPJ1

7 0
2 years ago
Lifting a rope A mountain climber is about to haul up a 50-m length of hanging rope. How much work will it take if the rope weig
timama [110]

Answer:

780 J

Explanation:

W=\int _{\:0}^{50}0.624xdx

4 0
2 years ago
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