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3 years ago

What is the equivalent resistance of a circuit that contains two 50.0 32

Physics

2 answers:

Alex Ar [27]3 years ago

6 0

82ohms

Explanation:

The equivalent resistance in the circuit is 82ohms

Given parameters:

R1 = 50ohms

R2 = 32ohms

Unknown:

Equivalent resistance = ?

Solution:

A resistor is an body in circuit that opposes the flow of electric current.

Resistors are usually connected in circuit and in series arrangement.

When resistors are connected in series, they have the same current passing through them.

Equivalent resistance is the sum of each of the connected resistors

Equivalent resistance = R1 + R2 = 50 + 32 = 82ohms

learn more:

Circuits brainly.com/question/2364338

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oksian1 [2.3K]3 years ago

6 0

Answer:

100.0

Explanation:

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Physics double pivot question

andriy [413]

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

(a) Calculate the reaction force at the wall.

Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

Rw = (10 kg) (9.8 m/s²) / (2√3)

Rw = 28 N

(b) State the friction at the foot of the ladder.

Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

Take the sum of the forces in the y direction.

∑F = ma

Rf − mg = 0

Rf = mg

Rf = 98 N

3 0

3 years ago

rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon

Gekata [30.6K]

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m$

The distance Rickey slides across the ground before touching the base is 15.825 m

4 0

3 years ago

2. 3000. kg car is moving across level ground at 5.0 m/s when it begins an

gizmo_the_mogwai [7]

Answer:

yes

Explanation: Work is done when there is movement. Therefore it was work was being done.

7 0

2 years ago

Anyone knows this please help me

olasank [31]

Answer:

TRUE

Explanation:

3 0

3 years ago

Read 2 more answers

If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c

Lemur [1.5K]

Answer:

$\mu_s \geq 0.27$

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

$f_s=\frac{mv^{2}}{R}$

But we know that:

$f_s\leq \mu_s N$

And the normal force is given by the sum of the forces in the vertical direction:

$N-mg=0 \implies N=mg$

Finally, we have:

$f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27$

So, the minimum value for the coefficient of friction is 0.27.

4 0

3 years ago