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natali 33 [55]
3 years ago
15

Which quantities are vectors

Physics
2 answers:
Zarrin [17]3 years ago
8 0

Answer:

For example, displacement, velocity, and acceleration are vector quantities, while speed (the magnitude of velocity), time, and mass are scalars. To qualify as a vector, a quantity having magnitude and direction must also obey certain rules of combination.

Explanation:

gayaneshka [121]3 years ago
6 0
Velocity, and acceleration are vector quantities, while speed the magnitude of velocity time, and mass are scalars.
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Una muestra de agua (líquida) de 1220 kg se encuentra a 0° C y baja se temperatura hasta -29°C mientras se congela en el proceso
S_A_V [24]

Answer:

-148,6 MJ

Explanation:

Dado que la liberación de calor al medio ambiente es;

H = mcθ

Dónde;

m = masa de agua

c = capacidad calorífica específica del agua

θ = aumento de temperatura

H = 1220 kg × 4200 × [-29-0]

H = -148,6 MJ

5 0
3 years ago
If the force of a golf club on a golf ball is 200 N forward, what will the force of the ball on the club be? A. 200 N forward B.
Korolek [52]

The ball should put 200 N of force towards the golfer.

Newton's Third Law is every action has an equal and opposite reaction.

It's the ball exerting 200 N of force towards the club as well, but the opposite reaction is that it flies away.

8 0
4 years ago
Read 2 more answers
As the Earth orbits the sun, the average distance between their centers at any given time is 1.5x10^11 meters. If the mass of th
natta225 [31]

Answer:

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Explanation:

8 0
3 years ago
A block slides to a stop as it goes 47 m across a level floor in a time of 6.35 s. a) What was the initial velocity? b) What is
UNO [17]

Answer :

(a) The initial velocity is, 14.8 m/s

(b) The acceleration is, -2.33m/s^2

Explanation :

By the 1st equation of motion,

v=u+at     ...........(1)

where,

v = final velocity = 0 s

u = initial velocity

t = time = 6.35 s

a = acceleration

The equation 1 will be:

0=u+a(6.35}

u=-6.35a       ..........(2)

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2     ...........(3)

where,

s = distance = 47 m

Now substitute equation 2 in 3, we get:

47=(-6.35a)\times (6.35)+\frac{1}{2}\times a\times (6.35)^2

By solving the term, we get:

a=-2.33m/s^2

The acceleration is, -2.33m/s^2

Now we have to calculate the initial velocity.

Using equation 2, we gte:

u=-6.35a

u=-6.35s\times (-2.33m/s^2)

u=14.8m/s

The initial velocity is, 14.8 m/s

5 0
3 years ago
Find equation tangent to a circle at given point
vlabodo [156]

the equation of the tangent line must be passed on a point A (a,b) and perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle, its center is C(0,0). And we assume that the tangent line passes to the point A(2.3).

</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle. 

<span>Let's find the equation of the line parallel to the radius.</span>

<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>

det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.

let's find the equation of the line perpendicular to this previous line.

let M a point which lies on the line. so MA.AC=0 (scalar product), 

it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent



7 0
4 years ago
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