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egoroff_w [7]
2 years ago
15

Please help me with this i need your help

Physics
1 answer:
krek1111 [17]2 years ago
3 0
Answer:
Speed of the wave is 7.87 m/s.
Explanation:
It is given that, tapping the surface of a pan
of water generates 17.5 waves per second
We know that the number of waves per
second is called the frequency of a wave.
So, f= 17.5 HZ
Wavelength of each wave,
A = 45 cm = 0.45 m
Speed of the wave is given by:
175 × 0.45
V= 7.87 m/s
So, the speed of the wave is 7.87 m/s
Hence, this is the required solution.
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Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this,
insens350 [35]

Answer:

The distance travel before stopping is 1.84 m

Explanation:

Given :

coefficient of kinetic friction \mu_{k} = 0.250

Zak's speed v = 3 \frac{m}{s}

Gravitational acceleration g = 9.8 \frac{m}{s^{2} }

Work done by frictional force is given by,

  W = \Delta K

 \mu _{k} mg d = \frac{1}{2} m v^{2}

  d = \frac{v^{2}  }{2 g \mu _{k} }

  d = \frac{9}{2 \times 9.8 \times 0.250}

  d = 1.84 m

Therefore, the distance travel before stopping is 1.84 m

3 0
3 years ago
Can someone help me a bit on this? Will mark brainliest. ( no physical science option soooo)
Advocard [28]

Answer:

When a light wave goes through a slit, it is diffracted, which means the slit opening acts as a new source of waves. How much a light wave diffracts<em> (how much it fans out)</em> depends on the wavelength of the incident light. The wavelength must be larger than the width of the slit for the maximum diffraction. Thus, for a given slit, red light, because it has a longer wavelength, diffracts more than the blue light.

The corresponding relation for diffraction is

d sin(\theta) \approx \lambda,

where \lambda is the wavelength of light, d is the slit width, and \theta is the diffraction angle.

From this relation we clearly see that the diffraction angle \theta is directly proportional to the wavelength  \lambda of light—longer the wavelength larger the diffraction angle.

7 0
3 years ago
If a certain shade of blue has a frequency of 7.33 x 10^14Hz what is the energy of exactly one photon of this light. Is this as
geniusboy [140]
In calculating the energy of a photon of light, we need the relationship for energy and the frequency which is expressed as:
 
E=hv

where h is the Planck's constant (6.626 x 10-34 J s)and v is the frequency.
E = 6.626 x 10-34 J s (<span>7.33 x 10^14 /s) = 4.857 x 10^-19 J</span>
6 0
3 years ago
Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu
lora16 [44]

The initial force between the two charges is given by:

F=k \frac{q_1 q_2}{d^2}

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

q_1' = q_1\\q_2' = q_2\\d' = 2d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d

So, the new force is:

F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F

So the force has increased by a factor 6.

8 0
3 years ago
Read 2 more answers
is set into oscillatory motion with an amplitude of 23.195 cm on a spring with a spring constant of 15.2676 N/m. The mass of the
fredd [130]

Answer:

maximum speed of the bananas is 18.8183 m/s

Explanation:

Given data

amplitude A =  23.195 cm

spring constant K = 15.2676 N/m

mass of the bananas m = 56.9816 kg

to find out

maximum speed of the bananas

solution

we know that radial oscillation frequency formula that is = √(K/A)

radial oscillation frequency = √(15.2676/23.195)

radial oscillation frequency is 0.8113125 rad/s

so maximum speed of the bananas = radial oscillation frequency × amplitude

maximum speed of the bananas = 0.8113125 × 23.195

maximum speed of the bananas is 18.8183 m/s

8 0
3 years ago
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