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3 years ago

When current increases and resistance remains constant what happens to voltage?

1 answer:

4 0

According to Ohm's law, V=IR (voltage equals current times resistance). So if thevoltage increases, then the current increases provided that the resistance remains constant

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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

poizon [28]

Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$

=> $b = 0.08 - 0.05 9$

=> $b = 0.021 \ m$

Generally at equilibrium position the net force acting on the spring is

$k * b - mg = 0$

=> $k * 0.021 - 2 * 9.8 = 0$

=> $k = 933 \ N/m$

$A = 0.30 * \sqrt{\frac{2}{933} }$

=> $A = 0.014 \ m$

8 0

3 years ago

An airplane has a starting velocity of 300m/s. It then accelerates at a rate of 45m/s2 for a time of 10s. What is it's final vel

Olenka [21]

A = (v - u) / t

a = acceleration
v = final velocity
u = initial velocity
t = time

45 = (v - 300) / 10

45 × 10 = v - 300

450 + 300 = v

v = 750 m/s

Hope this helps!

P.S. Let me know if you need an explanation

8 0

2 years ago

Please help! you are amazing! brainliest too!

german

I think it’s B hope it helps:)

6 0

2 years ago

Read 2 more answers

The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulu

Gemiola [76]

Answer:

$\displaystyle w=3.478\ rad/sec$

$M=0.0182\ J$

$v=0.398\ m/s$

Explanation:

Simple Pendulum

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

$\displaystyle w=\sqrt{\frac{g}{L}}$

We have the length of the pendulum is L=0.81 meters, then we have

$\displaystyle w=\sqrt{\frac{9.8}{0.81}}$

$\displaystyle w=3.478\ rad/sec$

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

$U=mgh$

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

$H+Y=L$

And

$H=L\ cos\alpha$

Solving for Y

$Y=L(1-cos\alpha )$

$Since\ \alpha=8.1^o, L=0.81\ m$

$Y=0.0081\ m$

The potential energy is

$U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)$

$U=0.0182\ J$

The mechanical energy is, then

$M=K+U=0+U=U$

$M=0.0182\ J$

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

$\displaystyle K=\frac{mv^2}{2}$