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3 years ago

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A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t

hat speed, how fast will he be traveling?

1 answer:

gregori [183]3 years ago

5 0

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

$p_i = (M+m)u$

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

$p_f = MV+mv$

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

$p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s$

So, the skater is moving at 2.34 m/s (forward) after the shot.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

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Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

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2 years ago

musickatia [10]

Answer:

C

Explanation:

4 0

2 years ago

A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops,

ZanzabumX [31]

Answer:

5) 13 revolutions (approximately)

Explanation:

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2α*θ Formula (1)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Data:

ω₀ = 18 rad/s

ωf = 0

α = -2 rad/s² ; (-) indicates that the wheel is slowing

Revolutions calculation that turns the wheel until it stops

We apply the formula (1)

ωf²= ω₀² + 2α*θ

0 = (18)² + 2( -2)*θ

4*θ = (18)²

θ = (18)²/4 = 81 rad

1 revolution = 2π rad

θ = 81 rad * 1 revolution / 2πrad

θ = 13 revolutions approximately

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3 years ago

To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1

dedylja [7]

Answer:

$2.2\times 10^{-5} s$

Explanation:

We are given that

The wavelength of sound wave= $\lambda=3.3 cm=3.3\times 10^{-2}m/s$

1 cm/s= $10^{-2}m/s$

Speed of sound wave,v= $1522 m/s$

We have to find the period of the wave.

We know that

Frequency= $\nu=\frac{v}{\lambda}$

Using the formula

Frequency = $\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4}$ Hz

Time period= $\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s$

Using identity: $\frac{a^x}{a^y}=a^{x-y}$

Hence, the time period of the wave= $2.2\times 10^{-5} s$

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A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an

Vanyuwa [196]

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .

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3 years ago