All categories

3 years ago

PLLLLLLZ HELP 15 POINTS

Physics

1 answer:

amid [387]3 years ago

8 0

<h2>Option A is the correct answer.</h2>

Explanation:

A 10-ω resistor and a 30-ω resistor are connected in series across a 100-V battery

Total resistance = 10 + 30 = 40 ω

We have

Voltage = Current x resistance

100 = I x 40

I = 2.5 A

In series current in all the resistors are same, that is 2.5 A

Voltage in 10ω resistor, V = I x 10 = 2.5 x 10 = 25 V

In parallel connection potential in all the resistors are same.

Voltage in 10ω resistor, V = 100 V

The ratio of the potential difference across the 10-ω resistor in the series combination to that of the 10-ω resistance connected in parallel = 25/100 = 1/4

Option A is the correct answer.

You might be interested in

Unit 5 lesson 7 physical science 12 question

Kryger [21]

I just took it 100% 11/11
1.D
2.A
3.A
4.A
5.B
6.C
7.D
8C
9A
10B
11C

8 0

3 years ago

A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c

satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

Given the data in the question;

First course : $a_{x}$ = 0.75 m/s², $d_{x}$ = 20 m, $u_{x}$ = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

$v_{x}$ ² = (10)² + (2 × 0.75 × 20)

$v_{x}$ ² = 100 + 30

$v_{x}$ ² = 130

$v_{x}$ = √130

$v_{x}$ = 11.4 m/s

for the Second Course:

$u_{y}$ = 11.4 m/s, $a_{y}$ = -1.15 m/s², $v_{y}$ = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) × $d_{y}$ )

0 = 129.96 - 2.3 $d_{y}$

2.3 $d_{y}$ = 129.96

$d_{y}$ = 129.96 / 2.3

$d_{y}$ = 56.5 m

so;

|d| = √( $d_{x}$ ² + $d_{y}$ ² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0

2 years ago

S Problem Set<br> 2.) 6.4 x 109 nm to cm

anyanavicka [17]

Answer:

$6.4\cdot 10^2 cm$

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

$1 nm = 10^{-9} m$

So we have:

$6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m$

Now we can convert from metres to centimetres, keeping in mind that

$1 m = 10^2 cm$

So, we find:

$6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm$

8 0

2 years ago

How is work calculated?

BartSMP [9]

Work is force times distance. If there's no distance, there's no work being done.

6 0

3 years ago

In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a

ohaa [14]

Answer: 10.58 C has flowed during the lightning bolt

Explanation:

Given that;

Time of flow t = 1.2 × 10⁻³

perpendicular distance r = 21 m

Magnetic field B = 8.4 x 10⁻⁵ T

Now lets consider the expression for magnetic field;

B = u₀I / 2πr

the current flow is;

I = ( B × 2πr ) / u₀

so we substitute

I = ( (8.4 x 10⁻⁵) × 2 × 3.14 × 21 ) / 4π ×10⁻⁷

= 0.01107792 / 0.000001256

= 8820 A

Hence the charge flows during lightning bolt will be;

q = It

so we substitute

q = 8820 × 1.2 × 10⁻³

q = 10.58 C

therefore 10.58 C has flowed during the lightning bolt

6 0

3 years ago