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AleksAgata
1 month ago
14

Stella is driving down a steep hill. She should keep her car __________ to help _________. a. light, it speed up in a higher b.

gear, slow her vehicle in a lower c. gear, slow her vehicle in a lower d. gear, speed up her vehicle
Physics
2 answers:
inna [3.1K]1 month ago
6 0
b. gear, slow her vehicle in a lower. c. The appropriate choice remains: When driving downhill, utilizing a lower gear enhances engine braking effect and mitigates the risk of brake overheating. Operating in a lower gear causes the engine to rev higher than typical, which is part of the engine braking system. This is crucial for handling steep inclines because excessive brake use can result in failures.
Ostrovityanka [3.2K]1 month ago
3 0
Both B and C are correct since they present identical options proposed in the question. Utilizing a lower gear in a vehicle aids in managing speed while descending a hill. It also helps preserve the brakes, which can overheat and lead to failures if overused on slopes. Shifting to a lower gear allows the engine to help reduce speed, although brakes may still be applied with less force. In this scenario, Stella should downshift to maintain control of her speed while descending the hill for safety.
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Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
Yuliya22 [3333]

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵ kgm² s⁻¹

Explanation:

a)Given

m = 5500 kg

l = 44 m

The moment of inertia for one blade

I= 1/3 x m l²

where m denotes the mass of the blade

l represents the length of each blade.

Substituting the necessary values, the moment of inertia for one blade is

I= 1/3 x 5500 x 44²

I= 35.49 x 10⁵ kg.m²

Total moment of inertia for 3 blades

I= 3 x 35.49 x 10⁵ kg.m²

I= 106.48 x 10⁵ kg.m²

b) The angular momentum 'L' is calculated using

L =I x ω

where,

I= the moment of inertia of the turbine i.e 106.48 x 10⁵ kg.m²

ω= angular velocity =2π f

f represents the frequency of rotation of the blade i.e 13 rpm

f = 13 rpm=>= 13 / 60 revolutions per second

ω = 2π f => 2π  x  13 / 60 rad / s

L=I x ω =>106.48 x 10⁵ x   2π  x  13 / 60

  = 144.97 x 10⁵  kgm² s⁻¹    

7 0
1 month ago
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Yuliya22 [3333]

Details that are not provided: the problem figure is included.

We can address the exercise by applying Poiseuille's law. This law indicates that for a fluid flowing in a laminar manner within a confined pipe,

\Delta P = \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P represents the pressure difference across the two ends

\mu denotes the viscosity of the fluid

L signifies the length of the pipe

Q=Av indicates the volumetric flow rate, where A=\pi r^2 is the cross-sectional area of the tube and v refers to the fluid's velocity

r stands for the pipe's radius.

This law can be utilized for the needle, allowing us to compute the pressure difference between point P and the needle's end. In this scenario, we have:

\mu=0.001 Pa/s is the dynamic viscosity of water at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Substituting these values into the formula yields:

\Delta P = 3200 Pa

This pressure difference specifies the value between point P and the needle's termination. As the end of the needle is under atmospheric pressure, the gauge pressure at point P, relative to atmospheric pressure, is exactly 3200 Pa.

8 0
3 months ago
In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc
kicyunya [3294]

Answer:

Explanation:

Let T represent the tension in the swing.

At the peak mg-T=\frac{mv^2}{r}

where v denotes the velocity needed to maintain the circular motion.

r equals the distance from the rotation point to the center of the ball, which is L+\frac{d}{2} (with d being the ball's diameter).

The threshold velocity can be expressed as mg-0=\frac{mv^2}{r}

To determine the velocity at the bottom, we can use energy conservation principles at both the top and bottom positions.

At the top E_T=mg\times 2L+\frac{mv^2}{2}

Energy at the bottom E_b=\frac{mv_0^2}{2}

By comparing the two states using conservation of energy, we find v_0^2=4gr+gr

v_0^2=5gr

v_0=\sqrt{5gr}

v_0=\sqrt{5g\left ( \frac{d}{2}+L\right )}

6 0
2 months ago
A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete
Sav [3153]

Answer:

407 steps

Explanation:

Based on the question,

P = mgh/t........... Equation 1

Where P stands for power, m is mass, g denotes gravity, h is height, and t represents time.

Rearranging the equation to solve for h, we have:

h = Pt/mg............. Equation 2

Providing values: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Given constant: g = 9.8 m/s²

By substituting into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

Calculating number of steps: 6525/16

The resulting number of steps = 407 steps

6 0
2 months ago
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