In this case, the movement is uniformly delayed (the final rapidity is less than the initial rapidity), therefore, the value of the acceleration will be negative.

1. The following equation is used:

a = (Vf-Vo)/ t

a: acceleration (m/s2)

Vf: final rapidity (m/s)

Vo: initial rapidity (m/s)

t: time (s)

2. Substituting the values in the equation:

a = (5 m/s- 27 m/s)/6.87 s

3. The car's acceleration is:

a= -3.20 m/ s<span>^2</span>

5 0

3 years ago

A transformer has input voltage and current of 12 V and 4 A

Advocard [28]

Explanation:

It is given that,

Input voltage, $V_i=12\ V$

Input current, $I_i=4\ A$

Output current, $I_o=0.8\ A$

Number of turns in the secondary side of transformer, $N_s=1177$

We need to find the number of turns in the primary side of the transformer. The current to the number of turns in the input and output is given by :

$\dfrac{N_s}{N_p}=\dfrac{I_p}{I_s}$

Substituting all the above values

So,

$N_p=\dfrac{N_sI_s}{I_p}\\\\N_p=\dfrac{1177\times 4}{0.8}\\\\N_p=5885$

So, the number of turns in primary side of the transformer is 5885.

4 0

3 years ago

Which of the following is a common feature among the four inner planets?

denpristay [2]

<span>rocky surface

(outer plants are made of gas)</span>

6 0

3 years ago

A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in

eimsori [14]

Answer:

Electric field at radius r inside the solid sphere is

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:Given

The radius of the solid sphere is

The charge on the solid sphere is

The inner radius of the shell is

The outer radius of the shell is

The total charge on the shell is

PART(A)

The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be

The charge enclosed by the radius r inside the solid sphere is

rho=

According to gauss law

PART(B)

The electric field at radius r where

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

The charge enclosed by the radius r where

According to gauss law

Read more on Brainly.com - brainly.com/question/13242041#readmore

3 0

3 years ago