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3 years ago

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A horizontal force is used to pull a 5.0 kilogram cart at a constant speed of 5.0 m/s across the floor. The force of friction be

tween the cart and the floor is 10.0 newtons. What is the magnitude of the horizontal force along the handle of the cart?
5 N
10 N
15 N
20 N

1 answer:

AlekseyPX3 years ago

6 0

Answer:10N

Explanation:

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A 645-turn coil with a 20.250 m 2 area is spun in Earth’s 5.00×10 −5 T magnetic field, producing a 1.25-V maximum emf. A

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Answer:

$\omega = 1.914\ rad/s$

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

$\varepsilon = NAB\omega$

$\omega= \dfrac{\varepsilon}{NAB}$

$\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}$

$\omega = 1.914\ rad/s$

Angular velocity of the coil = $\omega = 1.914\ rad/s$

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4 years ago

A hair dryer uses 1100 W of power

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Answer:10

Explanation:acellus

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3 years ago

Read 2 more answers

What is the electric force acting between two charges of -0.0050 C and

love history [14]

The electric force is $-3.6\cdot 10^8 N$ (attractive)

Explanation:

The magnitude of the electric force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have the following:

$q_1 = -0.0050 C$ (charge 1)

$q_2 = +0.0050 C$ (charge 2)

r = 0.025 m (distance)

Substituting, we find the electric force between the two charges:

$F=(9\cdot 10^9) \frac{(-0.0050)(0.0050)}{(0.025)^2}=-3.6\cdot 10^8 N$

And the negative sign means the force is attractive, since the two charges have opposite sign.

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3 years ago

A horizontal row on the periodic table is usually referred to as

liq [111]

Each horizontal row is called a period.

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4 years ago

A uniform meter stick is pivoted at the 50.00 cm mark on the meter stick. A 400.0 gram object is hung at the 20.0 cm mark on the

babunello [35]

Answer:C

Explanation:

Given

mass $m_1=400\ gm$ is at $x=20\ cm$ mark

mass $m_2=320\ gm$ is at $x=75\ cm$ mark

Scale is Pivoted at $x=50\ cm mark$

For scale to be in equilibrium net torque must be equal to zero

Taking ACW as positive thus

$T_{net}=0.4\times g\times (0.5-0.2)-0.32\times g\times(0.75-0.50)$

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Therefore a net torque of 0.04 g is required in CW sense which a mass $400\ gm$ can provide at a distance of $x_o$ from pivot

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$x_o=0.1\ m$

therefore in meter stick it is at a distance of $x=60\ cm$

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3 years ago