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3 years ago

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A 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.8 m/s left. Afterwards, the 282 kg car m

oves right at 0.800 m/s. What is the momentum of the 155 kg car afterwards? (Unit=kg*m/s)

1 answer:

svetoff [14.1K]3 years ago

7 0

Answer:

482.4kgm/s

Explanation:

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Two guitars are playing together producing a beat frequency. The first guitar is playing at a frequency of 867 Hz and the beat i

vlabodo [156]

Answer:

The possible frequency of guitar are 842 HZ and 892 HZ.

Explanation:

given the beat frequency = $\frac{1}{0.04}$ = 25 HZ

we know that beat frequency = difference of frequencies of two instruments

therfore 25 = x-867 or 25 = 867-x

therefore x = 842 or 892

6 0

3 years ago

A rope can withstand a maximum tension of 75 N before snapping . What maximum speed can you spin a 6.0 kg rock in a vertical cir

butalik [34]

Answer:

The maximum speed of the stone without breaking the rope is 4.33 m/s.

Explanation:

Given;

maximum tension the rope can withstand, T = 75 N

mass of the rock, m = 6 kg

radius of the circle, r = 1.5 m

As the stone spin in a vertical circle, it performs circular motion with centripetal acceleration directed inwards. The maximum centripetal force during this motion will be equal to maximum tension on the rope.

$F_c,_m_a_x = T = \frac{MV_m_a_x^{2} }{r}\\\\V_m_a_x^{2} = \frac{Tr}{M}\\\\V_m_a_x = \sqrt{\frac{75*1.5}{6}}\\\\V_m_a_x = 4.33 \ m/s$

Therefore, the maximum speed of the stone without breaking the rope is 4.33 m/s.

7 0

4 years ago

A chair of weight 125 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 3

iris [78.8K]

Answer:

N = 148.10N

Explanation:

GIVEN

Weight =125 N

force = 35

angle =42°

since there is no vertical acceleration $a_{y} = 0$ from

free body diagram

$\Sigma fa_{y} = ma_{y} = 0$

$N-W-Fsin42\degree = 0\\\\N= W+Fsin42\\N = 125+35\times 0.66\\N = 148.10 N$

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4 years ago

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

I am Lyosha [343]

Answer:

$2.09\ \text{m/s}$

$22298.4\ \text{J}$

Explanation:

m = Mass of each the cars = $1.6\times 10^4\ \text{kg}$

$u_1$ = Initial velocity of first car = 3.46 m/s

$u_2$ = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

$mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}$

Speed of the three coupled cars after the collision is $2.09\ \text{m/s}$ .

As energy in the system is conserved we have

$K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}$

The kinetic energy lost during the collision is $22298.4\ \text{J}$ .

6 0

3 years ago

Determine the initial velocity of the ball if it reaches a height of 15 meters.

jasenka [17]

Answer:

the initial velocity of the ball is 17.14 m/s

Explanation:

Given;

maximum height reached by the ball, h = 15 m

let the initial velocity of the ball = u

at maximum height, the final velocity of the ball is zero, v = 0

The initial velocity of the ball is calculated by using the following upward motion kinematic equation;

v² = u² - 2gh

0 = u² - 2(9.8 x 15)

u² = 294

u = √294

u = 17.14 m/s

Therefore, the initial velocity of the ball is 17.14 m/s

4 0

3 years ago