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Paraphin [41]
3 years ago
5

EXPERTS/ACE and people that wanna help 4 sure only!

Physics
1 answer:
mario62 [17]3 years ago
3 0
That first one you have selected (3,-3) works in both equations so it's correct.
good job.

you can do this guess and test method with multiple choice answers. If it works in both equations it is the solution. Otherwise use substitution or elimination to combine the two into one equation in only one variable. Then you can solve for the one variable first and use it to solve for the other.

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What is the energy in space left over from the big bang called
Lilit [14]
It is called the CMBR, which stands for cosmic microwave background radiation. It was discovered by Arno Penzias and Robert Wilson in 1964.
5 0
3 years ago
What is the meaning of the word "force"?
WARRIOR [948]

Answer: Strength or energy as an attribute of physical action or movement.

Explanation:

5 0
3 years ago
a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if
klio [65]

Answer:

Velocity of the two balls after collision: 60\; \rm m \cdot s^{-1}.

100\; \rm J of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

  • Mass of the first ball: 100\; \rm g = 0.1\; \rm kg.
  • Mass of the second ball: 400\; \rm g = 0.4 \; \rm kg.

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass m and velocity v is: p = m \cdot v.

Momentum of the two balls before collision:

  • First ball: p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}.
  • Second ball: p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}.
  • Sum: 10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1} given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be 30\; \rm kg \cdot m \cdot s^{-1}. The mass of the two balls, combined, is 0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg. Let the velocity of the two balls after the collision v\; \rm m \cdot s^{-1}. (There's only one velocity because the collision had sticked the two balls to each other.)

  • Momentum after the collision from p = m \cdot v: (0.5\, v)\; \rm kg \cdot m \cdot s^{-1.
  • Momentum after the collision from the conservation of momentum: 30\; \rm kg \cdot m \cdot s^{-1}.

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about v:

0.5\, v = 30.

v = 60.

In other words, the velocity of the two balls right after the collision should be 60\; \rm m \cdot s^{-1}.

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass m and velocity v is \displaystyle \frac{1}{2}\, m \cdot v^{2}.

Kinetic energy before the collision:

  • First ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Second ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Sum: 500\; \rm J + 500\; \rm J = 1000\; \rm J.

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

  • Mass of the two balls, combined: 0.5\; \rm kg.
  • Velocity of the two balls right after the collision: 60\; \rm m\cdot s^{-1}.

Sum of the kinetic energies of the two balls right after the collision:

\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J.

Therefore, 1000\; \rm J - 900\; \rm J = 100\; \rm J of kinetic energy would be lost during this collision.

7 0
3 years ago
As a roller coaster car crosses the top of a 50-m-diameter loop-the-loop, its apparent weight is the same as its true weight.
scZoUnD [109]

Answer:22.36 m/s

Explanation:

Given

the diameter of loop d=50 m

the radius of loop r=25 m

At the top position, we can write,

weight and Normal reaction combination will provide the centripetal force i.e.

R+W=\frac{mv^2}{r}

R=W\quad \quad [\text{apparent weight =Actual weight}]

2W=2mg=\frac{mv^2}{r}

v=\sqrt{2gr}

v=\sqrt{2\times 10\times 25}

v=22.36\ m/s

3 0
3 years ago
A rubber ball of mass 18.5 g is dropped from a height of 1.90 m onto a floor. The velocity of the ball is reversed by the collis
kati45 [8]

Answer:

J=0.211kg.m/s

Explanation:

The impulse-momentum theorem states:

J=\Delta p\\J=m(v_a-v-b)\\where:\\m=mass\\v_a=velocity\_after\\v_b=velocity\_before

The velocity before the impact is given by:

(v_b)^2=2.a.\Delta y\\v_b=\sqrt{2*(9.8m/s^2)1.90m}=6.10m/s(-\hat{j})

For the velocity after the impact:

(v_f)^2=(v_a)^2+2.a.\Delta y\\(0)^2=(v_a)^2+2.(-9.8m/s^s).(1.45m)\\\\v_a=\sqrt{2*9.8m/s^2*1.45m}\\v_a=5.33m/s(\hat{j})

so:

J=18.5*10^{-3}kg(5.33m/s(\hat{j})-6.10m/s(-\hat{j}))\\\\J=18.5*10^{-3}kg(5.33m/s(\hat{j})+6.10m/s(\hat{j}))\\\\J=0.211kg.m/s

3 0
3 years ago
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