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2 years ago

When converted to a household measurement, 9 kilograms is approximately equal to a

Physics

1 answer:

Alex17521 [72]2 years ago

4 0

Answer:

D) 19.8 lbs

Explanation:

1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.

1kg is also equal to 2.205 lbs. 9*2.205=19.8416

9 kg is also equal to 9000 grams, but grams are not a part of the household measurement system

a) 9000 grams. b) 9000 ounces. c) 19.8 ounces. d) 19.8 pounds.

This leaves us with 19.8 lbs

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In and electric circuit, where do the electrons come from that flow in the circuit

Firdavs [7]

The electrons are already there. They are freely moving through the conductor.

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3 years ago

Read 2 more answers

On an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates

11Alexandr11 [23.1K]

The acceleration due to gravity (g) on this planet is 39.44 m/s²

<h3>What is solar system?</h3>

Solar system consists of all the planets and the most importantly the center of the solar system is Sun.

Given is an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates with a period of 2 seconds.

The time period of the pendulum is

T = 2π √l/g

Squaring both sides, we get

l/g = T² / 4π²

g = 4π²l/ T²

Substitute Time period T = 2s and length l = 4m, we get

g = 4π²x 4/ 2²

g =39.44 m/s²

Thus, the acceleration due to gravity on this planet is 39.44 m/s²

Learn more about solar system.

brainly.com/question/12075871

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3 0

2 years ago

Suppose a 1500 kg speed boat is traveling with a velocity of 30 m/s what is the speed boats Monument

DanielleElmas [232]

Answer: 45000kgm/s

Explanation:

Given that:

Mass of speed boat = 1500 kg

Velocity of speedboat = 30 m/s

Speed boats Momentum = ?

The Speed boat momentum is the product of its mass and the velocity by which it moves. Momentum is a vector quantity and measured in kgm/s

i.e Momentum = mass x velocity

= 1500 kg x 30 m/s

= 45000kgm/s

Thus, the speed boats momentum is 45000kgm/s

7 0

3 years ago

A rocket ship at rest in space gives a short blast of its engine, firing 50 kg of exhaust gas out the back end with an average v

balu736 [363]

Answer:

20,000 Ns

Explanation:

mass of exhaust gases, m = 50 kg

velocity of exhaust gases, v = 400 m/s

The momentum of a body is defined as the measurement of motion of body. mathematically, it is defined as the product of mass off the body an its velocity.

change in momentum of rocket = final momentum - initial momentum

= m x v - 0

= 50 x 400

= 20,000 Ns

5 0

2 years ago

Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E

Leviafan [203]

We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
$\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}$
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2$
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
$mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\$
Because both carts have the same mass we can cancel those out:
$3v=v_A+2v_B$
Now we have our system of equation that we have to solve:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B$
Part A
We need to solve our system for $v_a$ . We will solve second equation for $v_b$ and then plug that in the first equation.
$3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}$
Now we have to plug this in the first equation:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
We will multiply the first equation with 2 and divide by m:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
Now we plug in the second equation into first one:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0$
We end up with quadratic equation that we have to solve, I won't solve it by hand.
Coefficients are:
$a=1\\
b=-6v\\
c=v^2-\frac{E}{m}$
Solutions are:
$v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}$
Part B
We do the same thing here, but we must express $v_a$ from momentum equation:
$3v=v_A+2v_B\\
v_A=3v-2v_B$
Now we plug this into our energy conservation equation:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\
3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\
3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\
3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\
3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\
3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0
$
Again we end up with quadratic equation. Coefficients are:
$a=3\\
b=-6v\\
c=6v^2-\frac{3E}{2m}$
Solutions are:
$v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}$

8 0

3 years ago