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givi [52]
3 years ago
7

El vector a tiene componented ax= 2.70 cm, ay=2.25 cm; y el vector b tiene componentes bx= 0.30 cm, by= 1.75 cm. Calcular las co

mponentes de la resultante a+b
Physics
1 answer:
Talja [164]3 years ago
3 0

Answer:

Explanation:

Given that,

A vector A has x component to be 2.7cm and y component to be 2.25cm

Then,

A = 2.7•i + 2.25•j

A vector B has x component of 0.30cm and y component of 1.75cm

B = 0.3•i + 1.75•j

So, we want to find A+B

Addition of vectors

Generally

(a•i + b•j) + (c•i + d•j) = (a+c)•i +(b+d)•j

Vectors are added component wise.

So,

A + B = (2.7•i + 2.25•j) + (0.3•i + 1.75•j)

A + B = (2.7 + 0.3)•i + (2.25 + 1.75)•j

A + B = 3•i + 4•j

We can also find it magnitude and direction

Generally,

A = a•i + b•j

|A| = √(a²+b²)

<A = arctan(b/a)

So,

|A+B| = √(3²+4²) = √9+16 = √25

|A+B| = 5

And it's direction

< = arctan(y/x)

< = arctan(4/3)

< = 53.13°

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Alexxandr [17]
Hope it cleared your doubt.

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3 years ago
Plz help ASAP I'll mark as brainliest ​
gogolik [260]

Hi there!

1.

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

2.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

3.

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

4.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

8 0
2 years ago
Describe the advantages and disadvantages of keeping the road clear of ice
harina [27]
A advantage is less car accidents and a disadvantage is, in order to keep the roads clear of ice is a chemical. Like salt. Which is bad for the animals.
4 0
3 years ago
The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?
postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

Initial\ Distance = 2cm ---- r_1

New\ Distance = 4cm --- r_2

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

F_1 = \frac{kq_1q_2}{r_1^2} --- Coulomb's law

Substitute 2 for r1

F_1 = \frac{kq_1q_2}{2^2}

F_1 = \frac{kq_1q_2}{4}

The new force (F2) is

F_2 = \frac{kq_1q_2}{r_2^2}

Substitute 4 for r2

F_2 = \frac{kq_1q_2}{4^2}

F_2 = \frac{kq_1q_2}{4*4}

F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}

Substitute F_1 = \frac{kq_1q_2}{4}

F_2 = \frac{1}{4}*F_1

F_2 = \frac{F_1}{4}

The new force is 1/4 of the previous force.

3 0
2 years ago
Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The
tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q  x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q  /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0
3 years ago
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