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Igoryamba
3 years ago
11

After 20 seconds, a 200-kg object increases its velocity from 15 m/s to 40 m/s. Determine the impulse applied to the object. (Al

so show your work so I can understand it better and ty)
Physics
1 answer:
sleet_krkn [62]3 years ago
7 0

Answer:

Imp_{1-2}=5000[kg*m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of momentum, which is defined as the product of mass by Velocity.

It must be defined that the impulse after the force is applied is equal to the momentum before the impulse applied on the body.

ΣPbefore = ΣPafter

P = momentum = m*v [kg*m/s]

In this way, we will construct the following equation.

(m_{1}*v_{1})+ Imp_{1-2}=(m_{1}*v_{2})

where:

m₁ = mass of the object = 200 [kg]

v₁ = velocity of the object before the impulse = 15 [m/s]

v₂ = velocity of the object after the impulse = 40 [m/s]

Now replacing:

(200*15) + Imp_{1-2} = (200*40)\\Imp_{1-2}=5000[kg*m/s]

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Water flows through a horizontal nozzle in steady flow at the rate of 10m3/s. The inlet and outlet diameters are d1 = 0.5m and d2
Dvinal [7]

Answer:

P₁ = 2.215 10⁷ Pa, F₁ = 4.3 106 N,

Explanation:

This problem of fluid mechanics let's start with the continuity equation to find the speed of water output

        Q = A v

        v = Q / A

The area of ​​a circle is

       A = π r² = π d² / 4

Let's look at the speeds at each point

       v₁ = Q / A₁ = Q 4 /π d₁²

       v₁ = 10 4 /π 0.5²

       v₁ = 50.93 m / s

       v₂ = Q / A₂

       v₂ = 10 4 /π 0.25²

       v₂ = 203.72 m / s

Now we can use Bernoulli's equation in the colon

       P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Since the tube is horizontal y₁ = y₂. The output pressure is P₂ = Patm = 1.013 10⁵ Pa, let's clear

       P₁ = P2 + ½ rho (v₂² - v₁²)

      P₁ = 1.013 10⁵ + ½ 1000 (203.72² - 50.93²)

      P₁ = 1.013 10⁵ + 2.205 10⁷

      P₁ = 2.215 10⁷ Pa

la definicion de presion es

      P₁ = F₁/A₁

     F₁ = P₁ A₁

     F₁ = 2.215 10⁷ pi d₁²/4

     F₁ = 2.215 10⁷ pi 0.5²/4

     F₁ = 4.3 106 N

     

6 0
3 years ago
A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it
Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

L = length of the meter stick = 1 m

m = mass of the meter stick

w = angular speed of the meter stick as it hits the floor

v = speed of the other end of the stick

we know that, linear speed and angular speed are related as

v = r w\\w = \frac{v}{r}

h = height of center of mass of meter stick above the floor = \frac{L}{2} = \frac{1}{2} = 0.5 m

I = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

I = \frac{mL^{2} }{3}

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}

7 0
3 years ago
In a oxidation-reduction reaction, the oxidizing agent always undergoes reduction.
Free_Kalibri [48]

In a oxidation-reduction reaction, the oxidizing agent always undergoes reduction.

TRUE.

6 0
4 years ago
Read 2 more answers
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Julli [10]

Answer:

<em>Maximum=70 m</em>

<em>Minimum=26 m</em>

Explanation:

<u>Vector Addition </u>

Since vectors have magnitude and direction, adding them takes into consideration not only the magnitudes but also their respective directions. Two vectors can be totally collaborative, i.e., point to the same direction, or be totally opposite. In the first case, the magnitude of the sum is at maximum. Otherwise, it's at a minimum.

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6 0
4 years ago
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