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2 years ago

Force usually has the symbol F. It is a vector. What is the correct way to write the symbol to show that it is a vector?

Physics

1 answer:

const2013 [10]2 years ago

4 0

Answer:

(B) F with a line on top

Explanation:

There is no one single 'correct' way to denote a vector. Most typical is a line over the symbol: $\overline{F}$ , or an arrow over the symbol: $\vec F$ . (However, if you want to use your own vector notation, for example, by printing the symbol in bold, then that is fine, too, as long as you clearly define this upfront),

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An object moves from the position +16m to the position +43m in 12s. What us the total displacement

NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

=43m -16m

=27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V = (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12 ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0

3 years ago

Which of the following is accurate in describing converging and diverging lenses?

kenny6666 [7]

B. Both of these types of lenses have the ability to produce real images.

3 0

2 years ago

Read 2 more answers

Please help me answer this

ivann1987 [24]

Answer:

A is the correct answer

Hope it helps

4 0

2 years ago

The position vector of a particle of mass 1.70 kg as a function of time is given by r with arrow = (6.00 î + 5.70 t ĵ), where r

ollegr [7]

Answer:

The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

Explanation:

Given that,

Mass = 1.70 kg

Position vector $r=(6.00\hat{i}+5.70 t \hat{j})$

We need to calculate the angular velocity

The velocity is the rate of change of the position of the particle.

$v = \dfrac{dr}{dt}$

$v=\dfrac{d}{dt}(6.00\hat{i}+5.70 t \hat{j})$

$v=5.70\hat{j}$

We need to calculate the angular momentum of the particle

Using formula of angular momentum

$L=r\cdot p$

Where, p = mv

Put the value of p into the formula

$L=m(r\times v)$

Substitute the value into the formula

$L=1.70(6.00\hat{i}+5.70 t \hat{j}\times5.70\hat{j})$

$L=1.70\times34.2$

$L=58.14\ kgm^2/s$

Hence, The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

7 0

2 years ago

An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at

Deffense [45]

Answer:

$v_{o}=-14.60m/s$

Explanation:

<u>Kinematics equation for first Object:</u>

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

but:

$v_{o}=0m/s$ The initial velocity is zero

$y_{o}=20m$

it reach the water at in instant, t1, y(t)=0:

$0=y_{o}-1/2*g*t_{1}^{2}$

$t_{1}=\sqrt{2y_{o}/g}=\sqrt{2*20/9.81}=2.02s$

<u>Kinematics equation for the second Object:</u>

The initial velocity is zero

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

but:

$y_{o}=20m$

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s

$0=y_{o}+v_{o}t_{2}-1/2*g*t_{2}^{2}$

$v_{o}=1/(t_{2})*(1/2*g*t_{2}^{2}-y_{o})=(1/1.02)*(1/2*9.81*1.02^{2}-20)=-14.60m/s$

The velocity is negative, because the object is thrown downwards.

6 0

2 years ago