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hjlf
3 years ago
12

Ivan finished 4/5 of his math homework problems and all 3 questions for his science homework before dinner. If he finished a tot

al of 43 problems before dinner, what is the total of math homework problems he was assigned?
Physics
2 answers:
goldenfox [79]3 years ago
7 0

Answer:

50 Is the correct answer

Explanation:

Because <u>4/5 = 0.8</u> an if you divide 50 by 0.8. 50/0.8 = 40.

And he <u>finished</u> 40 questions in math, and 3 in science.

So he has a total of 50 questions in math.

devlian [24]3 years ago
6 0

he made 40 I hope it helps you

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Two spheres are cut from a certain uniform rock. One has radius 4.50 cm. The mass of the other is five times greater. Find its r
Sonja [21]
Vol of sphere = 4/3 pi r^2.density of sphere = mass/volume.mass = densityxvolumesphere 1. mass = density x 4/3 pi 4.5^2sphere 2 5mass = density x 4/3 pi r^25=4/3 pi r^2 divided by 4/3 pi 4.5^25=r^2 divided by  4.5^25x4.5^2=r^2root(5x4.5^2)=r4.5 root 5 = r
4 0
2 years ago
Read 2 more answers
A friend insists that electric current is the same as electrical energy. Work with a partner to develop a rebuttal
kotykmax [81]

Answer: The electrons flowing through the wire are referred to as a quantity of electricity, and the flow of electricity is referred to as “an electric current.”

Explanation: Hope it Helps have a blessed day

4 0
3 years ago
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

6 0
2 years ago
Can someone please help me
guajiro [1.7K]
ANSWER:
C. Small, minimize

Hope it helps u!
5 0
3 years ago
As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -
postnew [5]

Answer:

-2.3 × 10^-9 Coulombs(C).

Explanation:

So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;

=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "

=> " Assuming the shuttle is a conducting sphere of radius 15 m".

So, in order to estimate the value for the charge we will be making use of the equation below:

Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.

Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.

So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.

= -2.3 × 10^-9 Coulombs(C).

4 0
3 years ago
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