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vitfil [10]
3 years ago
6

You are on a building with a snowball. You see your favorite physics teacher walking by about 10 m away from you. How fast must

you throw the snow ball to hit him (assume you hit him on the foot and you throw the snow ball horizontally from a height of 5.5m)
Physics
1 answer:
tankabanditka [31]3 years ago
5 0

You must throw the snow ball by velocity 9.43 m/s horizontally to

hit him

Explanation:

You throw the snow ball horizontally from a height 5.5 meters

Your favorite physics teacher is 10 m away from you

You hit him on the foot

1. The initial velocity of the ball is v

2. The vertical component of v is zero because you throw the ball

     horizontally

3. The vertical distance h is 5.5 m

4. The horizontal distance x is 10 m

We need to find how fast you must throw the snow ball to hit him

→ h = v_{y} t + \frac{1}{2} g t²

→ h = -5.5 m ⇒ height below the point of thrown

→ g = -9.8 m/s²

→  v_{y} = 0 ⇒ vertical component of v

→ -5.5 = (0) t + \frac{1}{2} (-9.8) t²

→ - 5.5 = -4.9 t²

Divide both sides by -4.9

→ t² = 1.12245

Take √ for both sides

→ t = 1.06 seconds

→ x = v_{x} t

→ v_{x} = v

→ x = 10 m , t = 1.06

→ 10 = v(1.06)

Divide both sides by 1.06

→ v = 9.43 m/s

You must throw the snow ball by velocity 9.43 m/s horizontally to

hit him

Learn more:

You can learn more about velocity in brainly.com/question/13103128

#LearnwithBrainly

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A 1000-kg car is moving at 30 m/s around a horizontal unbanked curve whose diameter is 0.20 km. What is the magnitude of the fri
omeli [17]

Answer:

4500 N

Explanation:

When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:

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Therefore, a body with mass m, will feel a force f:

f = m v^2/r

Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:

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The car will not slide if     f = fs,   i.e.

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That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration

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7 0
4 years ago
Read 2 more answers
Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

5 0
4 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
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Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

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3. If I run 150m in 30 seconds, what speed will I have been running at?
Radda [10]

Answer:

speed = distance/time

Explanation:

speed = 150/30

speed =5m/s

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7 0
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