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Stolb23 [73]
2 years ago
5

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20 g bullets at 965 m/s. The mass of the hunter (

including his gun) is 72.5 kg, and the hunter holds tight to the gun after firing it. Find the recoil velocity of the hunter if he fires the rifle (a) horizontally and (b) at 56.0∘ above the horizontal.
Physics
1 answer:
ohaa [14]2 years ago
8 0

Answer:

a) V_H=0.056 (-i)m/s

b) V_H=0.031 (-i)m/s

Explanation:

By conservation of the linear momentum:

P_o = P_f

0 = M_H*V_H+m_b*V_b

V_H = m_b/M_H*V_b

When the bullet is shot horizontally V_b = 965m/s

V_H = -0.056m/s

When the bullet is shot at 56.0° above the horizontal,

V_b = 965*[cos(56),sin(56)]m/s=[539.62;800.02]m/s

Since on the y-axis ground will absorb the shock, we only make the calculations on the x-axis:

V_H = -0.031m/s

V_H = -0.056m/s

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You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
2 years ago
1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup
azamat

Answer:

\dot{W} = 339.84 W

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

\dot{W} =\dot{m}\frac{V^{2}}{2}

here \dot{m}is mass flow rate and given as

\dot{m} = \rho*Q

\dot{W} =\rho*Q\frac{V^{2}}{2}

Putting all value to get minimum power

\dot{W} =1.18*9*\frac{8^{2}}{2}

\dot{W} = 339.84 W

7 0
2 years ago
Two students are playing paddle ball with a 5 kg spongy ball. If the ball is thrown at the batter with a speed of 5 m/s and boun
fenix001 [56]

Answer:

75 kgm/s

Explanation:

Impulse: This can be defined as the product of mass and change in velocity. The S.I unit is kgm/s.

From the question,

I = m(v-u)................... Equation 1

Where I = impulse, m = mass, v = final velocity, u = initial velocity.

Let the direction of the initial velocity be the positive direction.

Given: m = 5 kg, v = -10 m/s (bounce off), u = 5 m/s.

Substitute into equation 1

I = 5(-10-5)

I = 5(-15)

I = -75 kgm/s.

The negative sign tells that the impulse act in the same direction as the final velocity of the ball

Hence,

I = 75 kgm/s

3 0
2 years ago
As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t
Andreyy89

Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other  .So both the shells lose their charges .The positive half shell also loses all its charges

When we separate the half shells , there will be no deflection  in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.

4 0
2 years ago
Read 2 more answers
A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concer
777dan777 [17]

Answer:

E. The ocean gains more entropy than the iron loses.

Explanation:

When there is a spontaneous process , entropy of the system increases . Here hot iron is losing entropy and ocean is gaining entropy . Net effect will be gain of entropy . That means entropy gained by ocean is more than entropy lost by iron .

Hence option E is correct .

8 0
2 years ago
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