How much work is it to lift a 20. kg sack of potatoes vertically 6.5 m?

In general, fitness evaluations are meant to help you measure your physical fitness against those of similar age and gender

kirill [66]

Answer:

true

Explanation:

edge2020

8 0

3 years ago

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Jared has a nut sack that weighs 14 ounces when weighed. If Becky puts an amount of force on them, x , what is the equation that

topjm [15]

X+14 bc that’s what it will equal.

6 0

3 years ago

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle

expeople1 [14]

Explanation:

Given that,

Wavelength of the light, $\lambda=691\ nm=691\times 10^{-9}\ m$

(a) Slit width, $a=3.8\times 10^{-4}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}$

$\theta=0.104^{\circ}$

(b) Slit width, $a=3.8\times 10^{-6}\ m$

The angle that locates the first dark fringe is given by :

$sin\theta=\dfrac{\lambda}{a}$

$sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}$

$\theta=10.47^{\circ}$

Hence, this is the required solution.

7 0

3 years ago

How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do

sleet_krkn [62]

Answer:

h~=371.26m

Explanation:

when an object falls we use the equations of accelerated motion. There is only one that gives distance.

$x = ut + \frac{1}{2} a {t}^{2}$

Since we have no initial velocity (started from rest) we can get rid of the (ut) term

where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).

$h = \frac{1}{2} g {t}^{2} = \frac{1}{2} \times 9.81 \times {8.7}^{2} = 371.26m$

4 0

3 years ago

A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(

Valentin [98]

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

$\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}$

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

7 0

3 years ago