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3 years ago

14

While riding a skateboard, Nina ran into a rough patch of pavement, but she thought she could ride right over it. Instead, the s

kateboard
stopped but Nina didn't. Her _______ kept her moving forward, so she fell off the skateboard.

2 answers:

meriva3 years ago

5 0

Answer:

ineria

Explanation:

Jet001 [13]3 years ago

3 0

Answer:

inertia

Explanation:

While riding a skateboard, Nina ran into a rough patch of pavement, but she thought she could ride right over it. Instead, the skateboard stopped but Nina didn't. Her (Blank) kept her moving forward, so she fell off the skateboard.inertia

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What is the force needed to move an 225 kg car a distance of 150 meters

Kobotan [32]

Physics

mass = 225 kg

weight = m × a = 2250 N

distance = 150 m

force : ______?

Answers :

W = F × s

W = 2250 × 150

W = 337500 ✅

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3 years ago

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What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

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3 years ago

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If

Salsk061 [2.6K]

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R = 60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀ = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

$I(t) = I_o(1-e^{-\frac{t}{T}})$

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

$I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A$

Therefore, the current in the circuit 7 ms later is 0.2499 A

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3 years ago

Describe how the design of a vacuum flask keeps the liquid inside hot ?

JulijaS [17]

The gap between the two flasks is partially evacuated of air creating a near vacuum which significantly reduces heat transfer by conduction or convection

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3 years ago

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A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically dr

yan [13]

Answer:

49.79 m/s

Explanation:

Given:

Initial velocity of the roller coaster is, $u=0\ m/s$

Vertical drop or the displacement of the roller coaster is, $y=-126.5\ m$

The displacement is negative as the motion is in downward direction.

Now, as the motion is in vertical direction only, the acceleration of the roller coaster will be due to gravity acting in the downward direction.

So, the acceleration of the roller coaster is, $a=g=-9.8\ m/s^2$

Now, using the following equation of motion:

$v^2=u^2+2ay$

Where, 'v' is the velocity of the roller coaster at the bottom.

Plug in all the given values and solve for 'v'. This gives,

$v^2=0^2+2(-9.8)(-126.5)\\\\v^2=2479.4\\\\v=\sqrt{2479.4}\\\\v=49.79\ m/s$

Therefore, the speed of the roller coaster at the bottom of the drop is 49.79 m/s.

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3 years ago