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3 years ago

14

A wagon is pulled at a speed of 0.40metets/seconds by a horse exerting an 1,800 Newton's horizontal force. what is the power of

this horse?

1 answer:

Tatiana [17]3 years ago

6 0

Given:

speed of 0.40meters/seconds

1,800 Newton's horizontal force

Required:

Power of the horse

Solution:

P = F(D/T) where P is power in watts, F is the force, D is the distance and T is time

P = (1,800N) (0.40 meters/seconds)

P = 720 Watts

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A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency

Mice21 [21]

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

$\mu = \frac{m}{L}$

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

$\mu = \frac{0.09\ kg}{1\ m}$

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

<u>T = 712.9 N</u>

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3 years ago

At what point will the electric field of a charged object be strongest?

puteri [66]

Answer:

The answer is C because they have to be close to be able to interact

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3 years ago

3.) An engineer is designing the runway for an airport. Of the planes that will use the airport,

scoray [572]

Answer: 704

Explanation:Vi = 0 m/s

vf = 65 m/s

a = 3 m/s2

d = ??

vf2 = vi2 + 2*a*d

(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d

(4225 m 2/m2)/(6 m/s2) = d

d = 704 m

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3 years ago

Read 2 more answers

A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en

Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be $a = 0.003* 6 =0.018\ m/s^2$

Explanation:

The objective of this solution is to obtain the acceleration of the particle

Now looking at Newton law which is mathematically represented as

$F = ma$

Where F is the force experience by a particle

m is the mass of the particle

a is the acceleration of the particle

And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

$F = qvB$

Where q is the charge of the particle

v is the velocity of the charge

B is the magnetic field the charge is under it influence

Now equating this two formulas

$ma = qvB$

Making a the subject we have

$a = \frac{qvB}{m}$

In the question the direction of the is in the positive x-axis which is $i$ hence the direction would be in the $i$ direction

So substituting $(2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T$ for B

$a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}$

Substituting $3.0 Km /s = 3.0*10^{3}\ m/s$ for v and $-6.0 \muC = -6.0*10^{-6} C$ for q

$a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}$

$a = 0.003 * 3i(2i+3j+4k)$

$a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j \ + (3*4)i \ \cdot \ k)$

According to vector multiplication

$i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k = 0$

So

$a = 0.003* 6 =0.018\ m/s^2$

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3 years ago

An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following stat

Ivahew [28]

Answer:

The electric field is directed toward the electron and has a magnitude of $E=\frac{ke}{r^{2} }$ .

Explanation:

An electric field is define as the surrounding of charges which exert a force on each other and this force can be attractive or repulsive depends on the charge.

In the given case electron is given and the magnitude of charge on electron is $e=1.6\times 10^{-19}C$

Electric field can be represented as,

$E=\frac{kQ}{r^{2} }$

Here, r is the distance between the point ande charge, k is the electric field constant and Q is the charge.

In the given question an electron is given so electric field will be,

$E=\frac{ke}{r^{2} }$

As we know that electric field start from the positive charge and vanish in the negative charge.

So, here the electric field will be $E=\frac{ke}{r^{2} }$ and it is directed toward the electron because of negative charge on the electron.

6 0

3 years ago