All categories

3 years ago

How much force is required to move a sled 5 meters if a person uses 60 J of work?

1 answer:

6 0

W=F*S

W - Work
F - Force
S - Distance (from latin word 'spatium)

so...
S= 5 (m)
W=60 (J)

W=F*S
F=W/S
F=60/5=12 J/m = 12 N (Newtons)

You might be interested in

A lens of focal length 10.0 cm

fgiga [73]

Answer:

30cm

Explanation:

7 0

3 years ago

A technique in which the muscles are stretched by an outside force is called _____.

geniusboy [140]

A technique in which the muscles are stretched by an outside force is called Passive Stretching

6 0

3 years ago

find a magnitude of the force such that if the act at right angle there resultant is √10N but if the act of 50° the resultant is

Readme [11.4K]

Explanation:

Let magnitude of the two forces be x and y.

Resultant at right angle R1= √15N) and at

60 degrees be R2= √18N.

Now, R1 = √(x² + y²) = √15,

R2= √(x² + y² +2xycos50) = √18.

So x² + y² = 15,

and x² + y² + 1.29xy = 18,

therefore 1.29xy = 3,

y = 3/1.29x.

y = 2.33/x

Now, x2 + (2.33/x)2 = 15,

x² + 5.45/x² = 15

multiply through by x²

x⁴ + 5.45 = 15x²

x⁴ - 15x2 + 5.45 = 0

Now find the roots of the equation, and later y. The two values of x will correspond to the

magnitudes of the two vectors.

Good luck

7 0

3 years ago

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT.

Tems11 [23]

Ultraviolet light with a wavelength shorter than visible light and a higher radiant energy than visible light.

The shorter the wavelength, the higher the frequency and energy.

6 0

3 years ago

Read 2 more answers

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$