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3 years ago

You indicate that a symbol

Physics

1 answer:

Goryan [66]3 years ago

8 0

Answer:

B. over the symbol.

Explanation:

vectors are represented with a symbol carrying an arrow head with also indicates direction

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Determine the change in electric potential energy of a system of two charged objects when a -2.1-C charged object and a -5.0-C c

elena55 [62]

Answer:

Change in electric potential energy ∆E = 365.72 kJ

Explanation:

Electric potential energy can be defined mathematically as:

E = kq1q2/r ....1

k = coulomb's constant = 9.0×10^9 N m^2/C^2

q1 = charge 1 = -2.1C

q2 = charge 2 = -5.0C

∆r = change in distance between the charges

r1 = 420km = 420000m

r2 = 160km = 160000m

From equation 1

∆E = kq1q2 (1/r2 -1/r1) ......2

Substituting the given values

∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)

∆E = 94.5 × 10^9 (3.87 × 10^-6) J

∆E = 365.72 × 10^3 J

∆E = 365.72 kJ

6 0

3 years ago

By counting the number of waves that pass a certain point each second, we will know the _______ of the wave.

Travka [436]

The answer is A. Frequency

6 0

3 years ago

Read 2 more answers

What happens when a force is applied to an object in stable equilibrium

Katarina [22]

Answer:

the object is no longer in equilibrium .

7 0

3 years ago

A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e

alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

$K.E=\frac{1mv^2}{2}$

K.E= kinetic energy = $0.991\times 10^{-19}J$

m= mass of an electron = $9.109\times 10^{-31}kg$

v= velocity of object = ?

Putting in the values in the equation:

$0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}$

$v=466462m/s$

The speed necessary for the electron to have this energy is 466462 m/s

4 0

3 years ago

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The barrel of a rifle has a length of 0.855 m. A bullet leaves the muzzle of a rifle with a speed of 553 m/s. What is the accele

Novay_Z [31]

Answer:

Acceleration of the bullet will be 1778835.6 $m/sec^2$

Explanation:

We have given length of the barrel refile s= 0.855 m

When the bullet leaves the muzzle its velocity is 553 m/sec

So final velocity v = 553 m/sec

Initial velocity will be 0 that is u = 0 m/sec

According to third equation of motion $v^2=u^2+2as$

$553^2=0^2+2\times a\times 0.855$

$a=178835.6m/sec^2$

5 0

4 years ago