PLEASE HELP WILL GIVE BRAINIEST!!!!!!

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:

It tells the speed of car. At this time the speed of car is 80km/h , means if the car runs constantly at this speed then it will cover 80 kilometres in 1 hour.

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

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

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