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3 years ago

7

How can we avoid injuries in school or community when engaging in physical activities?

1 answer:

Mazyrski [523]3 years ago

3 0

Answer:

we should not fight with each other

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How does charging by conduction occur?

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Hello!
Charging by conduction involves the contact of a charged object to a neutral object. Suppose that a positively charged aluminum plate is touched to a neutral metal sphere. The neutral metal sphere becomes charged as the result of being contacted by the charged aluminum plate.
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3 years ago

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When pressing the accelerator on a car, the ___________ energy of the fuel is transformed into the _____________ energy that mak

lesya [120]

B. Chemical; Mechanical

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If you could travel at the speed of light, how long would it take to travel from our solar

Illusion [34]

Answer:

4.689 years

Explanation:

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3 years ago

a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/

LiRa [457]

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

$0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\$

we isolate th (needed to reach the maximum height hmax)

$th = \frac{v_{0}*sin(\alpha)}{g}$

The formula describing vertical distance is:

$y = Vy * t-g* t^{2} / 2$

So, given y = hmax and t = th, we can join those two equations together:

$hmax = Vy* th-g*th^{2}/2$

$hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)$

if we launch a projectile from some initial height h all you need to do is add this initial elevation

$hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)$

$hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft$

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3 years ago

a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy

Dvinal [7]

Answer:

$F_{net}= 85.41\ N$

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

$F_a = mg sin \theta$

now for upward motion

$F_{net} = mg sin \theta + air\ drag$

$F_{net} = mg sin \theta + mg sin \theta$

$F_{net} = 2 mg sin \theta$

$F_{net} = 2\times 50 \times 9.8 sin 5^0$

$F_{net}= 85.41\ N$

3 0

3 years ago