I need to know the answer and explanation if possible !

Gnoma [55]

Answer:

Reference frame, also called frame of reference, in dynamics, system of graduated lines symbolically attached to a body that serve to describe the position of points relative to the body.

7 0

3 years ago

Read 2 more answers

The distance xm travelled by a particle in time t seconds is described by the equation x = 10 + 12tsquare, Find the average spee

Paraphin [41]

After 2 seconds the particle will be in position

$x(2)=10+12\cdot 2^2=10+12\cdot 4=10+48=58$

After 5 seconds the particle will be in position

$x(5)=10+12\cdot 5^2=10+12\cdot 25=10+300=310$

So, the particle will travel

$310-58=252$

meters in 3 seconds, for an average speed of

$\dfrac{252}{3}=84$

5 0

4 years ago

The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

Alexandra [31]

Answer:

$-1.144\ \mu C$

Explanation:

<u>Given:</u>

$\vec{E}$ = uniform electric field in the space = $(148.0\ \hat{i}-110.0\ \hat{j})\ N/C$

Q = Charge placed in the region = $10.4 nC\ = 1.04\times 10^{-8}\ nC$

<u>Assume:</u>

$\vec{F}$ = Electric force on the charge due to electric field

We know that the electric field is the electric force applied on a unit positive charge i.e.,

$\vec{E}=\dfrac{\vec{F}}{Q}$

This means the electric force applied on this additional charge placed in the field is given by:

$\vec{F}=Q\vec{E}\\\Rightarrow \vec{F} = 1.04\times 10^{-8}\ n C\times (148.0\ \hat{i}-110.0\ \hat{j})\ N/C\\\Rightarrow \vec{F} = (1.539\ \hat{i}-1.144\ \hat{j})\ \mu N\\$

From the above expression of force, we have the following y-component of force on this additional charge.

$F_y = -1.144\ \mu N$

Hence, the y-component of the electric force on the this charge is $-1.144\ \mu N$ .

7 0

4 years ago

If the resultant of two velocity vectors of equal magnitude is also of the same magnitude, then which statement must be correct?

Tamiku [17]

The correct option is C) The angle between the vectors is 120°.

Why?

We can solve the problem and find the correct option using the Law of Cosine.

Let A and B, the given two sides and R the resultant (sum),

Then,

$R=A=B$

So, using the law of cosines, we have:

$R^{2}=A^{2}+B^{2}+2ABCos(\alpha)\\ \\A^{2}=A^{2}+A^{2}+2*A*A*Cos(\alpha)\\\\0=A^{2}+2*A^{2}*Cos(\alpha)\\\\Cos(\alpha)=-\frac{A^{2}}{2*A^{2}}=-\frac{1}{2}\\\\\alpha =Cos(-\frac{1}{2})^{-1}=120\°$

Hence, we have that the angle between the vectors is 120°. The correct option is C) The angle between the vectors is 120°

Have a nice day!

4 0

3 years ago

A marble is dropped off the top of a building. Find it's velocity and how far below to its dropping point for the first 10 secon

Yuri [45]

consider the motion in y-direction

v₀ = initial velocity = 0 m/s

a = acceleration = g = - 9.8 m/s²

t = time

v = final velocity at any time "t"

velocity at any time is given as

v = v₀ + at

v = 0 + (- 9.8) t

v = (- 9.8) t

so at t = 1 , v = (- 9.8) (1) = - 9.8 m/s

at t = 2 , v = (- 9.8) (2) = - 19.6 m/s

at t = 3 , v = (- 9.8) (3) = - 29.4 m/s and so on

Y₀ = initial position where the marble was dropped from = 0 m

Y = final position at any time "t"

final position at any time "t" is given as

Y = Y₀ + v₀ t + (0.5) a t²

Y = 0 + (0) t + (0.5) (-9.8) t²

Y = - 4.9 t²

at t = 1 , Y = - 4.9 (1)² = - 4.9 m

at t = 2 , Y = - 4.9 (2)² = - 19.6 m

at t = 3 , Y = - 4.9 (3)² = - 44.1 m

and So on

6 0

3 years ago