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3 years ago

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A loop circuit has a resistance of R1 and a current of 1.8 A. The current is reduced to 1.6 A when an additional 3.8 Ω resistor

is added in series with R1. What is the value of R1? Assume the internal resistance of the source of emf is zero. Answer in units of Ω.

2 answers:

Allisa [31]3 years ago

6 0

Answer:

Value of $R_1=30.4ohm$

Explanation:

We have given

In first case resistance is $R_1$ and current is 1.8 A

Let the potential difference is v

So $1.8=\frac{v}{R_1}$ ----eqn 1

In second case resistance is $R_1+3.8$ and current is 1.6 A and potential difference will be as it is a series connection

So $1.6=\frac{v}{R+3.8}$ ----eqn 2

From eqn 1 and eqn 2

$1.8R_1=1.6R_1+6.08$

$R_1=30.4ohm$

alisha [4.7K]3 years ago

6 0

Answer:

30.4 ohm

Explanation:

Let V be the potential difference.

When R1 is in the circuit.

V = 1.8 x R1 ... (1)

Now new resistance is R = R1 + 3.8

V = 1.6 x (R1 + 3.8) .... (2)

By comparing (1) and (2), we get

1.8 R1 = 1.6 (R1 + 3.8)

1.8 R1 - 1.6 R1 = 6.08

R1 = 30.4 ohm

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If a net horizontal force of 0.8 N is applied to a toy whose mass is 1.2 kg what acceleration is produced?

julia-pushkina [17]

Answer:

<h2>0.67 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

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From the question we have

$a = \frac{0.8}{1.2} \\ = 0.666666...$

We have the final answer as

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3 years ago

The 6 kg block is then released and accelerates to the right, toward the 4 kg block. The surface is rough and the coefficient of

KatRina [158]

Answer:

v =3.41 m/s

Explanation:

given,

mass of block 1 = 6 Kg

mass of another block 2 = 4 Kg

coefficient of friction = 0.3

Assuming 6 Kg block is attached to the spring of spring constant 350 N/m

and distance between the two block is equal to 0.5 m

using formula

$U = \dfrac{1}{2}kx^2$

$U = \dfrac{1}{2}\times 350 \times 0.5^2$

U = 43.75 J

using conservation of energy

KE = U - f.d

where f is the frictional force acting

$\dfrac{1}{2}mv^2 = 43.75- \mu m g d$

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v =3.41 m/s

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3 years ago

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. (a) Determine the time necessary for the

jeka94

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

$y = h + \frac{1}{2}at^2$

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

$0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s$

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

$v_x = 23.4 m/s$

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

$d=v_x t = (23.4 m/s)(3.35 s)=78.4 m$

(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

$v_x = 23.4 m/s$

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

$v_y = u_y +at$

where

$u_y = 0$ is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

$v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s$

and the negative sign means it points downward.

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3 years ago

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Jet001 [13]

Answer:

The fourth graph is the answer

Explanation:

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<em>Looking at other choices, we see that the first two graphs do not even represent the graphs of our inequalities, and the third graph does represent the inequalities but shades the wrong region. </em>

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3 years ago

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