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2 years ago

A 0.2-kg ball moving at 2.0 m/s perpendicular to a wall rebounds from the wall at 1.5 m/s. The impulse given to the ball is:

Physics

1 answer:

kiruha [24]2 years ago

8 0

Answer:

J = 0.7 N-s away from wall

Explanation:

Given that,

Mass of the ball, m = 0.2 kg

Initial speed of the ball, u = 2 m/s

Final speed of the ball, v = -1.5 m/s (as it rebounds)

Impulse is equal to the change in momentum. So,

$J=m(v-u)\\\\J=0.2\times (-1.5-2)\\\\J=-0.7\ N-m$

So, the impulse given to the ball is 0.7 N-m and it is away from the wall. Hence, the correct option is (c).

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A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

2 years ago

What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and

Andrei [34K]

Answer:

$A. 8.29\times 10^{-18}\ J$

Explanation:

Given that:

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

k = Boltzmann constant = $9\times 10^{9}\ Nm^2/C^2$

r = distance between the two carbon nuclei = 1.00 nm = $1.00\times 10^{-9}\ m$

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is $q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C$

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

$U = \dfrac{kQq}{r}$

So, the potential energy between the two nuclei of carbon is as below:

$U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J$

Hence, the energy stored between two nuclei of carbon is $8.29\times 10^{-18}\ J$ .

8 0

2 years ago

A piano emits frequencies the range from a low of about 28 Hz to a high of about 4200 Hz. Find the range of wavelengths in air a

Step2247 [10]

V=wave velocity , f= frequency, λ=wavelength 
Use it to find corresponding wavelengths for f=28 Hz 
λ= v/f= 337/28=12.036 m

for f=4200 Hz 
λ= v/f=337/4200= 0.08 m 
So max. wavelength is 12.036 m and 
Min Wavelength is 0.08 m 
So the range is between .08 m and 12.036 m
Hope this helps.

4 0

3 years ago

A box of paper is labeled 24lb paper. This means that 500 sheets (counted number) of paper size 17in x 22in weighs 24 pounds (Th

Tresset [83]

Answer:

The mass of a single paper is approximately 0.047 lb/paper which in SI Units is approximately 21.77 g/paper

Explanation:

The given information on the size and the weight of paper are;

The mass of a box of 500 sheets of paper = 24 lb

The number of sheets in the paper = 500 sheets

The dimensions of the paper = 17 in. × 22 in., which is equivalent to 43.18 cm × 55.88 cm

The mass of a single paper = The mass of the box of paper/(The number of sheets of paper present in the box)