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blsea [12.9K]
2 years ago
7

PLEASE HELP WILL GIVE BRAINIEST!!!!!!

Physics
1 answer:
seraphim [82]2 years ago
8 0
It should be Rotation
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Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s
enot [183]

Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:  

p_i = p_f  

m_1u_1 + m_2v_2 = (m_1 + m_2)v

v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}

v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s  

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:  

E(i) = E(f)  

KE(i) + PE(i) = KE(f) + PE(f)  

0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)  

v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}

Here, initial velocity is the final velocity from the first stage. Therefore:  

v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s

3. Use conservation of momentum to find the combined speed of Gayle and her brother.  

Given:

Initial velocity of Gayle and sled is, u_1(i)=10.5 m/s

Initial velocity of her brother is, u_2(i)=0 m/s

Mass of Gayle and sled is, m_1=55.0 kg

Mass of her brother is, m_2=30.0 kg

Final combined velocity is given as:

v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}  

v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79 m/s  

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:  

E(i) = E(f)  

KE(i) + PE(i) = KE(f) + PE(f)  

0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)  

v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s

6 0
3 years ago
Determine the potential difference between the ends of the wire of resistance 5 Ω if 720 C passes through it per minute.
Strike441 [17]

Answer:

The potential difference between the ends of a wire is 60 volts.

Explanation:

It is given that,

Resistance, R = 5 ohms

Charge, q = 720 C

Time, t = 1 min = 60 s

We know that the charge flowing per unit charge is called current in the circuit. It is given by :

I = 12 A

Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :

V = IR

V = 60 Volts

So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.

8 0
2 years ago
At a certain instant, the speedometer of the car indicates 80 km/h.
Scorpion4ik [409]

Answer:

  • <em>It</em><em> </em><em>tells</em><em> </em><em>the</em><em> </em><em>speed</em><em> </em><em>of</em><em> </em><em>car</em><em>.</em><em> </em><em>At</em><em> </em><em>this</em><em> </em><em>time</em><em> </em><em>the</em><em> </em><em>speed</em><em> </em><em>of</em><em> </em><em>car</em><em> </em><em>is</em><em> </em><em>8</em><em>0</em><em>k</em><em>m</em><em>/</em><em>h</em><em> </em><em>,</em><em> </em><em>means</em><em> </em><em>if</em><em> </em><em>the</em><em> </em><em>car</em><em> </em><em>runs</em><em> </em><em>constantly</em><em> </em><em>at</em><em> </em><em>this</em><em> </em><em>speed</em><em> </em><em>then</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>cover</em><em> </em><em>8</em><em>0</em><em> </em><em>kilometres</em><em> </em><em>in</em><em> </em><em>1</em><em> </em><em>hour</em><em>.</em><em> </em>

<em>.</em><em> </em>

<em>.</em><em> </em>

<em>.</em>

{\bold{\red{HOPE\:IT\:HELPS!}}}

\color{yellow}\boxed{\colorbox{black}{MARK\: BRAINLIEST!❤}}

4 0
2 years ago
State the laws of vibration of a stringed Instrument​
olchik [2.2K]
If the length and linear density are constant, the frequency is directly proportional to the square root of the tension.
7 0
3 years ago
In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the
OlgaM077 [116]

Answer:

a)   m_v = m_s ((\frac{w_o}{w})² - 1) ,  b)  m_v = 1.07 10⁻¹⁴ g

Explanation:

a) The angular velocity of a simple harmonic motion is

           w² = k / m

where k is the spring constant and m is the mass of the oscillator

let's apply this expression to our case,

silicon only

         w₉² = \frac{K}{m_s}

         k = w₀² m_s

silicon with virus

         w² = \frac{k}{m_s + m_v}

          k = w² (m_v + m_s)

in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal

           w₀²  m_s = w² (m_v + m_s)

           m_v = (\frac{w_o}{w})²  m_s - m_s

           m_v = m_s ((\frac{w_o}{w})² - 1)

b) let's calculate

          m_v = 2.13 10⁻¹⁶ [(\frac{20.4}{2.85})² - 1)]

          m_v = 1.07 10⁻¹⁴ g

4 0
3 years ago
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