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Hitman42 [59]
3 years ago
14

Multiplying or dividing vectors by scalars results in a. vectors. b. scalars. c. vectors if multiplied or scalars if divided. d.

scalars if multiplied or vectors if divided.
Physics
2 answers:
klemol [59]3 years ago
8 0
Let A = i+j+k be a vector and B = 3 be any scalar, 
Multiplying A and B, 
AB = (i+j+k)3 = 3i+3j+3k 

Which is a new vector whose direction is same as the old but it's 3 times greater in length  than the old vector(i+j+k).

Now, dividing A and B,
A/B = (i+j+k)/3 = \frac{1}{3}i +  \frac{1}{3}j + \frac{1}{3}k

Which is again a new vector whose direction is same as the old but now it's 1/3 times small in length than the old vector. 

Direction is same because we multiplied by positive scalar. If we multiply A by suppose -1, -4, -1000000 or any negative number, it's direction will reverse. 

Thus, if we multiply a vector with scalar, it's length increases. If we divide, it shrinks. 
Darina [25.2K]3 years ago
4 0

Always vectors. Remeber that multiplication and division are the same thing. When you want to divide by 2, you are really just multiplying by (1/2). Also remember that multiplying a vector by a scalar does not change the vector into a scalar. It keeps it a vector.

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In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a
katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

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3 0
2 years ago
Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
Inessa05 [86]

Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

6 0
3 years ago
I meed help with these 2 questions plz
ludmilkaskok [199]

-- 30N

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4 0
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Juliette [100K]
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3 years ago
What do you mean by acceleration due to gravity? ​
likoan [24]

It is the acceleration of an object in free fall

Explanation:

When an object is in free fall, it is subjected only to one force: the force of gravity, which pulls the object downward, with a magnitude (near the Earth's surface) which is given by

F=mg

where

m is the mass of the object

g=9.8 m/s^2 is the acceleration due to gravity

We can apply Newton's second law to the object in free fall:

F=ma

where

F is the net force on the object

a is the acceleration of the object

m is the mass

However, since there is only the force of gravity acting on the object, the net force is equal to the force of gravity: so we can equate the two equations, obtaining that

mg = ma\\\rightarrow a = g

Which means that the acceleration of an object in free fall (acted upon the force of gravity only) is equal to the acceleration due to gravity, g=9.8 m/s^2.

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4 0
3 years ago
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