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Natasha2012 [34]
3 years ago
12

You pull a wagon carrying your younger brother 1000 meters (1 km) to the store using 20 newtons of force the whole way. It takes

you 15 minutes to get there. How much power did you generate, in joules per minute? In watts?
Physics
1 answer:
Veseljchak [2.6K]3 years ago
8 0

Lifting Weights: On earth it takes about 10 Newton-meters (N-m) of energy to raise a 1 kilogram mass to a height of 1 meter. Since 1 N-m equals 1 Joule, that's 10 Joules. If it takes 1 second to lift the weight 1 meter, than you have converted 10 Joules of energy to potential energy in one second.


Hope this helps!

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A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of
lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0
3 years ago
Read 2 more answers
A car travels at a steady 40.0 m/s around a horizontal curve of radius 200 m. What is the minimumcoefficient of static friction
zhenek [66]

Answer:

c. 0.816

Explanation:

Let the mass of car be 'm' and coefficient of static friction be 'μ'.

Given:

Speed of the car (v) = 40.0 m/s

Radius of the curve (R) = 200 m

As the car is making a circular turn, the force acting on it is centripetal force which is given as:

Centripetal force is, F_c=\frac{mv^2}{R}

The frictional force is given as:

Friction = Normal force × Coefficient of static friction

f=\mu N

As there is no vertical motion, therefore, N=mg. So,

f=\mu mg

Now, the centripetal force is provided by the frictional force. Therefore,

Frictional force = Centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu=\frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu = \frac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}\\\\\mu=\frac{1600\ m^2/s^2}{1960\ m^2/s^2}\\\\\mu=0.816

Therefore, option (c) is correct.

7 0
3 years ago
Why does gravitational energy increase away from earth
aliina [53]
Because of the rule of mask.
4 0
3 years ago
Which statement best describes beaches? Beaches change over time as waves move further up the shore. Beaches are always changing
NemiM [27]

Answer:

I think the 1st statement is right.

Explanation:

Wind patterns doesn't stay the same.

Waves don't follow the same patterns.

Waves move further up the shore.

I didn't hear about "waves adding" before..so i guess 1st statement is right.

3 0
3 years ago
Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of
Radda [10]

According to the information provided to define an average density, it is necessary to use the concepts related to mass calculation based on gravitational constants and radius, as well as the calculation of the volume of a sphere.

By definition we know that the mass of a body in this case of the earth is given as a function of

M = \frac{gr^2}{G}

Where,

g= gravitational acceleration

G = Universal gravitational constant

r = radius (earth at this case)

All of this values we have,

g = 9.8m/s^2\\G  = 6.67*10^{-11} m^3/kg*s^2\\r = 6378*10^3 m

Replacing at this equation we have that

M = \frac{gr^2}{G} \\M = \frac{(9.8)(6378*10^3)^2}{6.67*10^{-11}} \\M = 5.972*10^{24}kg

The Volume of a Sphere is equal to

V = \frac{4}{3}\pi r^3\\V = \frac{4}{3} \pi (6378*10^3)^3\\V = 1.08*10^{21}m^3

Therefore using the relation between mass, volume and density we have that

\rho = \frac{m}{V}\\\rho = \frac{5.972*10^{24}}{1.08*10^{21}}\\\rho = 5.52*10^3kg/m^3

6 0
3 years ago
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