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Annette [7]
3 years ago
13

What is the momentum of a 10kg ball moving at a velocity of 4m/s east?​

Physics
1 answer:
Katyanochek1 [597]3 years ago
6 0

Answer:

40 kg.m/s

Explanation:

Momentum, p is defined as the product of mass and velocity of an object. Numerically, it is represented as, p=mv where m is mass of the object and v is the velocity in which the object moves, with keen observation on the direction before and after collision. Substituting 10 kg for m and 4 m/s for v then momentum, P=10*4=40 kg.m/s

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give an example of situation in which an automobile driver can have a centripetal acceleration but no tangential speed
Mars2501 [29]

There is no need for tangential acceleration when moving in a circle at a constant speed.

<h3>What is centripetal acceleration?</h3>

centripetal acceleration refers to the speed at which a body moves through a circle. Due to the fact that velocity is a vector quantity (i.e., it has both a magnitude, the speed, and a direction), when a body travels in a circle, its direction is constantly changing, which causes a change in velocity, which results in an acceleration.

<h3>Which is an example of centripetal acceleration?</h3>

Centripetal acceleration occurs when you spin a ball on a string above your head. A car experiences centripetal acceleration when it is being driven in a circle. Additionally, a satellite in orbit around the Earth experiences centripetal acceleration.

To know more about tangential acceleration :

brainly.com/question/14993737

#SPJ9

6 0
1 year ago
The electric flux through a spherical surface is 1.4 ✕ 105 N · m2/C. What is the net charge (in C) enclosed by the surface?
Anit [1.1K]

Answer:

The  value is   Q_{net} =  1.239 *10^{-6} \  C

Explanation:

From the question we are told that

   The electric flux is \Phi =  1.4*10^{5} \  N\cdot m^2/C

     

Generally the net charge is mathematically represented as

    Q_{net} =  \Phi *  \epsilon_o

Here \epsilon_o is the permetivity of free space with value  

       \epsilon_o =  8.85*10^{-12}  \  \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

So

   Q_{net} =  1.4*10^5 *  8.85*10^{-12}

=>   Q_{net} =  1.239 *10^{-6} \  C

8 0
3 years ago
Which property of table salt is also a property of other ionic compounds
Jet001 [13]

s alluded to in the other answers, salt refers to any ionic compound that doesn't have “oxides” in it. Table salt is sodium chloride. Going down the periodic table, the first column contains lithium, sodium, potassium, rubidium, cesium, and francium. This group (alkali metals) of atoms (and their corresponding positive ions) gets larger in the order shown above. Therefore, their ionic bonds with chloride (or any nonmetal) gets smaller. The trend of their corresponding compounds is a decreasing hardness, decreasing melting point, decreasing boiling point, and decreasing thermal stability. These are the major periodic trends of these corresponding compounds. Other metal ions generally have higher positive charges on them. This makes the ionic bonds considerably larger and you can probably surmise most of their corresponding properties listed above. However, the details of their lattice structures may cause the overall trend to vary.

3 0
3 years ago
Read 2 more answers
What are the two factors that affect the frictional force between two surfaces?
pishuonlain [190]

Answer:

The amount of force and the angle between them.

Explanation:

6 0
3 years ago
A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge
Troyanec [42]

Answer:

I_{l} =44.84 \mu A

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire =  1.30×10⁻² Ω  

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))

\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))

I_{l} = \dfrac{V}{R}

I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}

I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))

I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))

I_{l} =44.84 \times 10^{-6}\A

I_{l} =44.84 \mu A

3 0
3 years ago
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