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Nostrana [21]
3 years ago
11

Which measurement is equivalent to 350 milliliters?

Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0
350 milliliters equals to .35 liters
Therefore, 3.5 times 10^-1 would be the only correct answer since that is the only one that matches .35 liters
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If a machine increases force it must apply the force over a long distance? True or false
Eduardwww [97]

Answer: The machine must apply the force over a shorter distance. That's because a machine doesn't change the amount of work and work equals force times distance. Therefore, if force increases, distance must decrease

FALSE

HOPE THIS HELPS

4 0
3 years ago
Material that slows down or resists the flow of current. these are used in electric circuits to help control the flow of current
Nataly_w [17]
It's called an insulator

3 0
3 years ago
On a hot day, the temperature of a 65,000-L swimming pool increases by 1.20°C. What is the net heat transfer during this heating
vichka [17]

Answer:

326149.2 KJ

Explanation:

The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:

Q = m*cv*ΔT

Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.

A 65000 L swimming pool will have a mass of:

65000L *\frac{1m^3}{1000L} * \frac{1000kg}{1m^3} = 65000 kg

The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.

We replace the data and get:

Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ

3 0
3 years ago
Two people on ice skates push against each other. The person on the left is 50kg, the person on the right is 60kg. If the person
Zarrin [17]

I got -3.6 m/s but I had to do conservation of momentum for this question. Which involves Newtons third law but with simply that law I do not know how to complete this question. If you would like me to post my work I will though! Sorry

6 0
3 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
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