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VladimirAG [237]
2 years ago
15

(i) 10 m (ii) 20 m (iii) 40 m (iv) 80 m

Physics
1 answer:
IRINA_888 [86]2 years ago
6 0

Answer:

20m

420=80m

100

increases

increases then decreases

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When ultra violets lights shine on glass what does it do to electrons in the glass structure?
andreev551 [17]

Answer:

No

Explanation:

7 0
2 years ago
determine the loudness (in decibels) of the sound at a rock concert if the intensity of the sound is 1 x 10–1 w/m2. remember, th
EleoNora [17]

The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is  110 dB.

Here we are dealing with loudness which is the perception of the Intensity of the sound.

The formula  to refer to in order  to  find the value of the loudness of a sound is ,

  db= 10log(I/I₀)

As we are provided with the current intensity which is  1 x 10⁻¹ Wm⁻². and the initial intensity which is  1 x 10⁻¹² Wm⁻².

So, by substituting the required values in the formula we get

db= 10 * log( 1 x 10⁻¹ /1 x 10⁻¹²)

 = 10 * 11 log(10)

 = 110

So, the result is 110 dB.

To know more about the intensity of sound refer to the link brainly.com/question/9323731?referrer=searchResults.

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4 0
1 year ago
An object glides on a horizontal tabletop with a coefficient of kinetic friction of 0.5. If its initial velocity is 4.3 m/s, how
Shkiper50 [21]

Answer:

Time, t = 0.87 seconds

Explanation:

Given that,

Initial velocity of the object, u = 4.3 m/s

The coefficient of kinetic friction between horizontal tabletop and the object is 0.5

We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.

Using first equation of motion to find it as :

v=u+at

a is the acceleration, here, a=\mu g

0=u+\mu gt

t=\dfrac{u}{\mu g}\\\\t=\dfrac{4.3}{0.5\times 9.8}\\\\t=0.87\ s

So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.

8 0
3 years ago
Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso
TEA [102]

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

3 0
3 years ago
A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi
Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block m = 6.60 kg

Initial speed of block v _{i} = 1.56 \frac{m}{s}

Final speed of block v_{f} = 0 \frac{m}{s}

Coefficient of kinetic friction \mu _{k} = 0.62

Ramp inclined at angle \theta = 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

  \mu _{k} mgd \cos 28.4 =  \Delta K - \Delta U

Where \Delta U = mg d\sin 28.4

\mu _{k} mgd \cos 28.4 =  \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4

\mu _{k} mgd \cos 28.4 +mgd\sin 28.4  =  \frac{1}{2}mv_{i} ^{2}

d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}

 d = 0.122 m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0
3 years ago
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