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3 years ago

heres free wallpapers and pnts! :)) (phone and PC and chromebook wallpapers) (credits to those who made them !)

Physics

1 answer:

stiks02 [169]3 years ago

8 0

Answer: :DD

Explanation:

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How do I solve this

lesantik [10]

multiply grav pull by mass of astro maybe with a calculator

7 0

4 years ago

.3.3: Populating a vector with a for loop. Write a for loop to populate vector userGuesses with NUM_GUESSES integers. Read integ

Reptile [31]

Answer:

#include <iostream>

#include <vector>

using namespace std;

int main() {

const int NUM_GUESSES = 3;

vector<int> userGuesses(NUM_GUESSES);

int i = 0;

int uGuess = 0;

for(i = 0; i <= userGuesses.size() - 1; i++){

cin >> uGuess;

userGuesses.at(i) = uGuess;

}

cout << endl;

return 0;

}

Explanation:

First inbuilt library were imported. Then inside the main( ) function, 3 was assigned to NUM_GUESSES meaning the user is to guess 3 numbers. Next, a vector was defined with a size of NUM_GUESSES.

Then a for-loop is use to receive user guess via cin and each guess is assigned to the vector.

8 0

3 years ago

Jasmine and her partner want to test to see if there is a relationship between how much of her solar panel is exposed to sunligh

katen-ka-za [31]

Any one trial might have been done incorrectly.

7 0

3 years ago

two teams are playing tug of war. team a pulls to the right with a force of 450n .team b pulls to the left with a force of 415 n

Alex_Xolod [135]

Explanation:

It is given that, two teams are playing tug of war.

Force applied by Team A, $F_A=450\ N$

Force applied by Team B, $F_B=415\ N$

We need to find the net force acting on the rope. It is equal to :

$F_{net}=F_A-F_B$

$F_{net}=450-415$

$F_{net}=35\ N$

So, the net force acting on the rope is 35 N and it is acting toward right. Hence, this is the required solution.

4 0

3 years ago

Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi

yanalaym [24]

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron= $v=6.1\times 10^7m/s$

Charge on electron, $q=e=1.6\times 10^{-19} C$

Mass of electron, $m_e=9.1\times 10^{-31} kg$

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

$V=\frac{v^2m_e}{2e}$

Where V=Potential difference

$m_e=$ Mass of electron

v=Velocity of electron

Using the formula

$V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}$

$V=10581.59 V$

Hence, the potential difference=10581.59 V

8 0

3 years ago