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Oksana_A [137]
2 years ago
12

A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much was done?

Physics
2 answers:
bezimeni [28]2 years ago
7 0

Answer:

45 J

Explanation:

The equation for work is:

Work=Force*Distance

We can substitute the given values into the equation:

Work=15N*3m\\Work=45Nm\\Work=45J

IRISSAK [1]2 years ago
6 0

Answer:

45 J

Explanation:

w =fd

w = 15(3)

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a 46 kilogram student climbs 11 meter up a rope at a constant speed if the student power output is 230 watts how long in seconds
Bezzdna [24]
230×46=10580÷11=961 second
3 0
3 years ago
Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu
Elenna [48]

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)²  = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424  x 10²⁵

= 6.4 x 10²³ kg .

4 0
2 years ago
A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st
Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r

where,

\tau= shear stress at a distance 'r' from the center

T = is the applied torque

I_{p} = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

\tau _{max}=\frac{4}{3}\times \frac{V}{A}

Applying values we get

\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa

3 0
3 years ago
A person is pulling a freight cart with a force of 58 pounds. how much work is done in moving the cart 70 feet if the cart's han
Kobotan [32]

<span>The person is dragging with a force of 58 lbs at an angle of 27 degrees relating to the ground. We want to use cosine function to look for the horizontal force component. And then we can compute for W = (Horizontal Force) x (Distance). We want the horizontal force component since that is the component that is parallel to the direction the cart is moving. </span><span>

(cos 27 degrees)(58 lbs) = 51.69 lbs (This is the horizontal force component.) 
W = (51.69 lbs) x (70 ft) = 3618.3 ft*lbs</span>

6 0
3 years ago
Correct me if im wrong
White raven [17]
Your answer is correct. No problem and Have a nice day
6 0
3 years ago
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