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ollegr [7]
3 years ago
7

Choose the +x-direction to point to the right. • Object 1 has a mass 1.66 kg and is moving to the right at 11.2 m/s. • Object 2

has a mass 6.59 kg and is moving to the left at 10.4 m/s. • What is x-component of the total momentum of the system in kg m/s? • Assume 3 significant figures for your final answer.
Physics
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

M = 49.4kgm/s (towards the left)

Explanation:

Momentum is the product of mass and velocity of an object

Momentum = mass * velocity

Momentum of Object 1 with mass 1.66 kg moving to the right at 11.2 m/s, is expressed as:

M1 = 1.66 * 11.2

M1 = 18.592kgm/s

Momentum of Object 2 with mass 6.59 kg moving to the left at 10.4 m/s is expressed as:

M2 = 6.59 * -10.4

M2 = -68.536kgm/s (negative since it is moving towards the left)

The total momentum will be the sum of momentum along the x-component as shown:

M = M1+M2

M = 18.592kgm/s--68.536kgm/s

M = -49.944kgm/s

M = 49.4kgm/s (towards the left)

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ikadub [295]

Answer:

This is one of the hottest summers America has ever seen and Jeremiah Warren is taking advantage of the heat wave and doing a little cooking…in his car!

It was about 98 or 99 degrees in Texas, so Jeremiah thought he’d head outside to see if it was hot enough to actually cook food.  He buttered up a frying pan on his dashboard and filled it with an egg, cookie dough, a hot dog and bacon (mmm…bacon!).  He set up another frying pan on the sidewalk and a few hours later he came back for a mid-day snack.

Explanation:

4 0
3 years ago
A wheel is rotating at a rate of 2.0 rev every 3.0s. through what angle, in radian , does the wheel rotate in 1.0s?​
igomit [66]

Answer:

4 \pi

Explanation:

First of all, let's convert 2.0 rev into radians:

1.0 rev = 2\pi rad\\2.0 rev = 4 \pi rad

This means that the angular speed of the wheel is

\omega = 2.0 rev/s = 4 \pi rad/s

The angle through which the wheel rotates in a time t is given by

\theta=\omega t

And substituting t=1.0 s, we find

\theta=(4\pi rad/s)(1.0 s)=4 \pi rad

7 0
3 years ago
In an elastic collision momentum is conserved as is
zzz [600]

In an elastic collision momentum is conserved as well as the kinetic energy

Explanation:

In physics, there are two types of collisions:

  • Elastic collision: in an elastic collision, the total momentum of the system is  conserved, and the total kinetic energy of the system is conserved as well. This is because there are no internal frictions acting on the system, so the energy is conserved. An example of elastic collision is (approximately) that occurring between two billiard balls.
  • Inelastic collision: in an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not. In fact, due to the presence of internal frictions, part of the total energy is converted into thermal energy and sound during the collision, and therefore "wasted", so the final total kinetic energy is less than the initial one. An example of inelastic collision is the collision between two cars. The maximum amount of kinetic energy is lost when the two objects stick together after the collision; in this case, we talk about perfectly inelastic collision.

Therefore, the complete sentence is

In an elastic collision momentum is conserved as well as the kinetic energy

Learn more about collisions:

brainly.com/question/13966693#

brainly.com/question/6439920

#LearnwithBrainly

7 0
3 years ago
Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on
leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

q_1=50\ nC at y=6\ m

q_2=-80\ nC at x=-4\ m

q_3=70\ nC at y=-6\ m

Electric Potential is given by

V=\frac{kQ}{r}

Distance of q_1 from x=8\ m

d_1=\sqrt{6^2+8^2}

d_1=\sqrt{36+64}

d_1=10\ m

similarly d_2=8-(-4)

d_2=12\ m

d_3=\sqrt{(-6)^2+8^2}

d_3=\sqrt{36+64}

d_3=10\ m

Potential at x=8\ m is

V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}

V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]

V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}

V_{net}=9\times 5.33

V_{net}=47.97\approx 48\ V

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Answer:

Please give options

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