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Pie
2 years ago
11

The lawn outside your neighbor's house has an approximate area of 175 m2One night it snows so that the snow on the lawn has a un

iform depth of 25.5 cm. What volume of snow is on the lawn, in cubic m
Physics
1 answer:
allsm [11]2 years ago
4 0
First we must convert 25.5 cm to meters by moving the decimal two places to the left

25.5 cm —> .255 m

volume = area x depth

volume = 175 x .255

volume = 44.63 m^3
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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
2 years ago
Use the diagram below to answer the following question:
d1i1m1o1n [39]

Answer:

3.0 cm

Explanation:

We can solve this problem by using the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length of the mirror

p is the distance of the object from the mirror

q is the distance of the image from the mirror

In this problem we have:

f = 1.5 cm is the focal length of the mirror (positive for a concave mirror)

p = 3.0 cm is the distance of the object from the mirror

Therefore, the distance of the image is:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.5}-\frac{1}{3.0}=\frac{1}{3.0}\\\rightarrow q=3.0 cm

And the positive sign means that the image is real.

(The second part of the exercise is just the description of the image of the first exercise).

5 0
2 years ago
Is a single property such as density sufficient to always correctly identify a substance?
bazaltina [42]
Not at all. Density on its own is not sufficient
7 0
2 years ago
you decide to work part time at a local supermarket. The job pays eight dollars and 60 per hour and you work 20 hours per week.
aleksley [76]

Answer:

I guess that we want to find how much money you get each week.

We know that the job pays $8.60 per hour.

We know that you work 20 hours per week.

Then the gross pay (the total money that you earn) in a week is 20 times $8.60, or:

20*$8.60 = $172.

Now we know that your employer witholds:

10% + 7.65% + 5% = 22.65%

Then your employer withholds 22.65% of your gross pay.

if the 100% of your gross pay is $172

Then the 22.65% will be:

(22.65%/100%)*$172 = 0.2265*$172 = $38.96

This means that your employer withholds $38.96 of your weekly gross pay.

Then each week you get:

$172 - $38.96 = $133.04

4 0
2 years ago
A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B
Novay_Z [31]

Answer:

Wavelength, \lambda=1.28\times 10^{-14}\ m

Explanation:

Given that,

Mass of the particle, m=4.3\times 10^{-28}\ kg

Acceleration of the particle, a=2.4\times 10^7\ m/s^2

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

\lambda=\dfrac{h}{mv}

v = a × t

\lambda=\dfrac{h}{mat}

\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}

\lambda=1.28\times 10^{-14}\ m

Hence, this is the required solution.

4 0
3 years ago
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