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3 years ago

How much current will a 500 W vacuum cleaner draw if it has a resistance of 30Ω?

Physics

1 answer:

Strike441 [17]3 years ago

3 0

Answer:

Electric current, I = 4.08 A

Explanation:

We have,

Power of vacuum cleaner is 500 W

Resistance of vacuum cleaner is 30 ohms

It is required to find the current flowing through the vacuum cleaner. It can be calculated using the definition of power delivered. Its formula is given by :

$P=I^2R$

I is electric current

$I=\sqrt{\dfrac{P}{R}} \\\\I=\sqrt{\dfrac{500}{30}} \\\\I=4.08\ A$

So, the electric current flowing through the vacuum cleaner is 4.08 A.

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A vector has a component of 10 m in the x-direction, a component of 10 m in the y-direction, and a component of 5 m in the z dir

Kobotan [32]

The magnitude of this vector is 15

A vector is a quantity or phenomenon that has two independent properties: magnitude and direction. The term also denotes the mathematical or geometrical representation of such a quantity. Examples of vectors in nature are velocity, momentum, force, electromagnetic fields, and weight.

The magnitude of a vector formula is used to calculate the length for a given vector (say v) and is denoted as |v|. So basically, this quantity is the length between the initial point and endpoint of the vector.

Let vector be = a

component of vector in x direction = 10 i

component of vector in y direction = 10 j

component of vector in z direction = 5 z

vector a = 10 i + 10 j + 5 z

magnitude of vector a = |a| = $\sqrt{10^{2} +10^{2} + 5^{2} }$

= 15

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2 years ago

Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.

balu736 [363]

Answer:

1. False

2. True

3. False

4. True

Explanation:

Conservation of Momentum

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

$p_a=m_a.v_a$

The total momentum of both pucks at the initial condition is

$p_1=m_a.v_a+m_b.v_b$

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

$m_a=m$

$m_b=2m$

We are given

$v_a=6\ m/s\\v_b=2\ m/s$

The total initial momentum is

$p_1=6m+2(2m)=10m$

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

$p_2=3m.v'$

Equating the initial and final momentum

$10m=3m.v'$

Solving for v'

$v'=10/3\ m/s=3.33\ m/s$

1. Compute the initial kinetic energy:

$\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2$

$\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2$

$K_1=18m+4m=22m$

The final kinetic energy is

$\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2$

$\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2$

$K_2=16.63m$

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

$p_1=10m$

$p_2=3m.v'=3m(10/3)=10m$

Proven

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Earth is like a bar magnet. What does this mean about its magnetic poles?

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B a bar magnet has a north and a south pole.

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Here's an interesting challenge you can give to a friend. Hold a $1 bill by the upper corner. Have a friend prepare to pinch the

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Mass m moves to the right with speed =v along a frictionless horizontal surface and crashes into an equal mass m initially at re

Amiraneli [1.4K]

After the collision the magnitude of the momentum of the system is Mv

Given:

mass of 1st object = M

speed of 1st object = v

mass of 2nd object = M

speed of 2nd object = 0

To Find:

magnitude of the momentum after collision

Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.

Applying conservation of linear momentum

Mv + M(0) = 2MV

Mv = 2MV

V = v/2

So, after collision momentum is

p = 2MV = 2xMxv/2 = Mv

So, after collision momentum is Mv

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