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3241004551 [841]
3 years ago
14

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.460 m/s. The to

tal mass of the sled, man, and rock is 95.0 kg. The mass of the rock is 0.310 kg and the man can throw it with a speed of 15.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled (in m/s) if the man throws the rock forward (i.e., in the direction the sled is moving).
Physics
2 answers:
xz_007 [3.2K]3 years ago
6 0

Answer:

= 0.417 m/s

Explanation:

Momentum before throwing the rock: m*V = 95.0 kg * 0.460 m/s

= 44.27 N*s  

A) man throws the rock forward

mass of rock m1 = 0.310 kg

V1 = 15.5 m/s, in the same direction of the sled with the man

sled and man:

m2 = 95 kg - 0.310 kg = 94.69 kg

v2 = ?

Conservation of momentum:

momentum before throw = momentum after throw

44.27N*s = 0.310kg * 15.5m/s + 94.69kg*v2

⇒ v2 = [44.27 N*s - 0.310 * 15.5N*s ] / 94.69 kg

= 0.417 m/s

AleksandrR [38]3 years ago
6 0

Answer:

0.41 m/s.

Explanation:

Mt = 95 kg

Mr = 0.31 kg

Vr = 15.5m/s

Ut = 0.46 m/s

Mass of the man and sled = (95 - 0.31) kg

= 94.69 kg

Using conservation of momentum equation,

Momentum before the throw = momentum after the throw

95 × 0.46 = 0.31 × 15.5 + 94.69 × V2

43.7 = 4.805 + 94.69 V2

V2 = 0.41 m/s.

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13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
3 years ago
A particle moves in a straight line and has acceleration given by a(t) = 12t + 10. Its initial velocity is v(0) = −5 cm/s and it
soldi70 [24.7K]

Answer:

The position function is s_{t}=2t^3+5t^2-5t+9.

Explanation:

Given that,

Acceleration a =12t+10

Initial velocity v_{0} = -5\ cm/s

Initial displacement s_{0}=9\ cm

We know that,

The acceleration is the rate of change of velocity of the particle.

a = \dfrac{dv}{dt}

The velocity is the rate of change of position of the particle

v=\dfrac{dx}{dt}

We need to calculate the the position

The acceleration is

a_{t} = 12t+10

\dfrac{dv}{dt} = 12t+10

a_{t}=dv=(12t+10)dt

On integration both side

\int{dv}=\int{(12t+10)}dt

v_{t}=6t^2+10t+C

At t = 0

v_{0}=0+0+C

C=-5

Now, On integration again both side

v_{t}=\int{ds_{t}}=\int{(6t^2+10t-5)}dt

s_{t}=2t^{3}+5t^2-5t+C

At t = 0

s_{0}=0+0+0+C

C=9

s_{t}=2t^3+5t^2-5t+9

Hence, The position function is s_{t}=2t^3+5t^2-5t+9.

7 0
3 years ago
A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a
asambeis [7]

Answer:

The recoil speed of the man and rifle is v_{man}=0.016 ms^{-1}.

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

m=\frac{F}{g}

Calculate the combined mass of the man and rifle.

m_{man,rifle}=\frac{650+25}{g}

Put g=9.8 ms^{-2}.

m_{man,rifle}=\frac{650+25}{9.8}

m_{man,rifle}=68.88 kg

The expression for the conservation of momentum is as follows as;

m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}

Here, m_{man,rifle} is the mass of the man and rifle,  m_{rifle} is the mass of the rifle,u_{man},u{bullet}  are the initial velocities of the man and bullet and v_{man},v{man,rifle} are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

m_{bullet}=4.5 g

m_{bullet}=.0045 kg

Put m_{bullet}=.0045 kg,m_{man,rifle}=68.88 kg , u_{man,rifle}=0, v_{bullet}= 240 ms^{-1} and u_{bullet}=0.

m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)

0=(68.88)v_{man,rifle}+(.0045)(240)

0=(68.88)v_{man,rifle}+1.08

(68.88)v_{man,rifle}=\frac{-1.08}{68.88}

v_{man,rifle}=-0.016 ms^{-1}  

Therefore, the recoil speed of the man and rifle is v_{man}=0.016 ms^{-1}.

3 0
3 years ago
Definition of definite​
KIM [24]

Google definition: a statement of the exact meaning of a word, especially in a dictionary.

the degree of distinctness in outline of an object, image, or sound, especially of an image in a photograph or on a screen.

My definition: The meaning or deintion of any word.

5 0
3 years ago
Using that transfers and third laws explain what happened when you dribble a basketball
Leno4ka [110]
The ball is using an reaction and opposite reaction, so when you dribble a basketball you push the ball with downward force and the ground pushes the ball back up thus making the opposite reaction.
5 0
3 years ago
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