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3 years ago

12

A 10-kg wagon is pulled by a force of 70 Newtons while friction provides 20 Newtons of force. How much acceleration will the wag

on have?

1 answer:

kkurt [141]3 years ago

5 0

Answer:

5m/s²

Explanation:

Given parameters:

Mass of wagon = 10kg

Force of pull = 70N

Frictional force = 20N

Unknown:

Acceleration of the wagon = ?

Solution:

Frictional force is a force that opposes motion.

The net force is given as:

Net force = mass x acceleration

Force of pull - Frictional force = mass x acceleration

Insert the parameters and solve;

70 - 20 = 10 x acceleration

50 = 10 x acceleration

Acceleration = 5m/s²

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The wavelength of the light is 0.63 micrometers. How much of this length stays in 1 centimeter

bazaltina [42]

11,066,669.
hope it help.

7 0

3 years ago

Machmer Hall is 400 m North and 180 m West of Witless.

yan [13]

Answer:

The distance from Witless to Machmer is 438.63 m.

Explanation:

Given that,

Machmer Hall is 400 m North and 180 m West of Witless.

We need to calculate the distance

Using Pythagorean theorem

$D = \sqrt{(d_{m})^2+(d_{w})^2}$

Where, $d_{m}$ =distance of Machmer Hall

$d_{w}$ =distance of Witless

Put the value into the formula

$D = \sqrt{(400)^2+(180)^2}$

$D=438.63\ m$

Hence, The distance from Witless to Machmer is 438.63 m.

5 0

3 years ago

How much pressure is applied to the ground

statuscvo [17]

Answer:

Pressure applied by the man= 285103.125 $Pa$ or 41.35 $lb/in^{2}$

Explanation:

Pressure is defined as the perpendicular force applied per unit area.

i.e. $Pressure=\frac{Force}{Area}$

Now, $Force= mg$

where, $m$ = mass of the body(man) = 93 kg

$g$ = acceleration due to gravity of Earth = 9.81 $m/{s^{2}}$

$Area$ covered is equal to the area of both stilts(a man generally stands on two feet)

therefore $Area=2(0.04)^{2}$ $m^{2}$

and putting in the values, we get,

$Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}$

Now we need to convert to our required units:

$1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}$

(We can get the above result by individually converting kg to lb and meters to inches respectively)

Using the above relations we get,

$Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}$

7 0

3 years ago

Add the vectors:

Anettt [7]

Vector 1 has components

$x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m$

$y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m$

and vector 2 has

$x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m$

$y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m$

Add these vectors to get the resultant, which has components

$x_{\rm total}\approx11.133\,\mathrm m$

$y_{\rm total}\approx13.268\,\mathrm m$

The magnitude of the resultant is

$\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m$

with direction $\theta$ such that

$\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ$

or about 50º N of E.

8 0

3 years ago

Is velocity related to momentum

enot [183]

Yes
momentum=mass x velocity
or you could rearrange the formula to find velocity = momentum/mass (you get this by dividing both sides in the original equation by mass)
you could also get that mass= momentum/velocity (you get this by rearranging both sides of the original equation by velocity)

6 0

3 years ago

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