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spin [16.1K]
3 years ago
9

According to newton���s second law of motion, the acceleration of an object equals the net force acting on the object divided by

the object���s
a. weight.
b. mass.
c. velocity.
d. momentum.
Physics
1 answer:
Leto [7]3 years ago
7 0
Newton's 2nd law: F = m * a ⇒ a = F / m
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Scientific Inquiry incorporates many scientific methods.<br> A) true<br> B) false
Shtirlitz [24]
True fhfhhj ufg ffgh fdh fhgg f jc fg gt
7 0
4 years ago
A 17 kg box experiences an applied force of +175 N and a force of friction of -125 N. While experiencing these unbalanced forces
Flura [38]

Answer: (b)

Explanation:

Given

Mass of box m=17\ kg

Applied force F=175\ N

Friction force f=-125\ N

Box travels a distance of s=15\ m

time taken t=5.25\ s

Net unbalanced force F_{net}=175-125=50\ N

Work done by the unbalanced force

\Rightarrow W=F_{un}\cdot s\\\Rightarrow W=50\times 15\\\Rightarrow W=750\ N

Power developed by unbalanced force

\Rightarrow P=\dfrac{W}{t}\\\\\Rightarrow P=\dfrac{750}{5.25}\\\\\Rightarrow P=142.85\approx 1423\ W

Thus, option (b) is correct.

7 0
3 years ago
A Pitot-static probe is used to measure the speed of an aircraft flying at 3000 m. If the differential pressure reading is 3300
adell [148]

Answer:

   v = 306.76 Km/h

Explanation:

given,

height of the aircraft = 3000 m

differential pressure reading = 3300 N/m²

density of air = 0.909 Kg/m³

speed of aircraft = ?

Assuming the air flowing above air craft is in-compressible, irrotational and steady so, we can use Bernoulli's equation to solve the problem.

using Bernoulli's equation

          \dfrac{v^2}{2} = \dfrac{\Delta P}{\rho}

where ρ is the density of the air at 3000 m

          v= \sqrt{\dfrac{2 \times \Delta P}{\rho}}

          v= \sqrt{\dfrac{2 \times 3300}{0.909}}

          v = \sqrt{7260.726}

                 v = 85.21 m/s

          v= 85.21 \times \dfrac{3600}{1000}

                 v = 306.76 Km/h

4 0
3 years ago
when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its c
andrey2020 [161]

Answer:

ΔD = 2.29 10⁻⁵ m

Explanation:

This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

          ΔA = 2α A ΔT

the area is

         A = π r² = π D² / 4

we substitute

         ΔA = 2α π D² ΔT/4

as they do not indicate the initial temperature, we assume that ΔT = 75ºC

    α = 1.7 10⁻⁵ ºC⁻¹

we calculate

          ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

          ΔA = 6.49 10⁻⁷ m²

by definition

           ΔA = A_f- A₀

           A_f = ΔA + A₀

           A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

           A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

           A_f = 2,551 10⁻⁴ m²

the area is

           A_f = π D_f² / 4

           A_f = \sqrt{4  A_f /\pi }

           D_f = \sqrt{4 \ 2.551 10^{-4} /\pi }

           D_f = 1.80229 10⁻² m

the change in diameter is

           ΔD = D_f - D₀

           ΔD = (1.80229 - 1.8) 10⁻² m

           ΔD = 0.00229 10⁻² m

           ΔD = 2.29 10⁻⁵ m

5 0
3 years ago
WRONG ANSWERS WILL BE REPORTED
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The answer to number 9 is C, decreased muscle tension. I believe number 10 is D, prepares an organism to respond to stress.
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