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3 years ago

What is the safety rules?

Physics

2 answers:

krek1111 [17]3 years ago

6 0

Answer:

helping because if a fire you need to be a good person even if you hate somone

Black_prince [1.1K]3 years ago

5 0

Answer:

Quick question do you mean what are some safety rules

Explanation:

Crosswalk, Stop sign,

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A car of mass 1000 kg is travelling at 10 m/s. What is its kinetic energy?

Marysya12 [62]

Answer:

50,000

Explanation:

kinetic energy=1\2×mv^2

=1\2×1000kg×10×10

=50000

6 0

2 years ago

Why do we feel the effects of gravity given that the gravitational force is so weak?

almond37 [142]

Any two objects in the universe attract each other. Gravity is the force exerted by earth on you (you exert the same force on earth) but due to the fact that earth has a huge mass compared to yours, you will be attracted to earth only by a small gravitational force.

6 0

3 years ago

Read 2 more answers

A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation rela

larisa86 [58]

ΔU = -Wint

Consdier the work of of interaction is W =m*g*h - equation -1

and the Potential energy U.

Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h

Now we will write the equation for a Change in Potential energy ΔU,

ΔU = Uf - Ui

= 0-m*g*h

ΔU = -m*g*h --Equation 2

Now compare the both equation

Wint = -ΔU

we can rewrite the above equation

ΔU = -W.

So our Answer is ΔU = -W. .

5 0

3 years ago

Which of the following equations is balanced correctly? A. 3 H2O → H2 + 3 O2 B. Cl2 + 2 KBr → KCl + Br2 C. 2 C2H2 + 5 O2 → 4 CO2

posledela

A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced. That leaves C. to be correctly balanced.

8 0

2 years ago

Read 2 more answers

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago