A 55kg cart is pushed by a force of 225 n. what is the carts acceleration?

Planting along the natural slip of the land to reduce soil erosion is called ____

kirill [66]

Why the answer to your question is Contour farming I hope this helps add me as a friend and I can help more

6 0

3 years ago

Which property of light remains unchanged when it enters a different medium?

Amiraneli [1.4K]

The FREQUENCY of light remains unchanged once it leaves the source.

3 0

3 years ago

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Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.

babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

For constructive interference

$d\sin \theta =m\lambda$

For Destructive interference

$d\sin \theta =(m+\frac{1}{2})\lambda$

where $\lambda =wavelength$

$d=slit\ width$

m=order of maxima and minima

for second order maxima i.e. $m=2$

For smallest separation taking $\lambda =400 nm, \theta =90^{\circ}$

$d\sin 90=2\times 400\times 10^{-9}$

$d=0.8\times 10^{-6}$

$d=0.8\mu m$

6 0

3 years ago

Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to

zzz [600]

Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5 = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

4 0

4 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

7 0

4 years ago