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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.
You might be interested in
Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?
irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

1)

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

M=Fd

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

M_c = W_2 d_2

where W_2 = 500 N is the weight of the person and d_2 = 2 m is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

M_a = W_1 d_1

where W_1 is his weight and d_1 = 4 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the person on the left:

W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N

2)

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

M_c = W_2 d_2

where W_2 = 400 N is the weight of the person and d_2 is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 300 N is his weight and d_1 = 4 m is the distance from the fulcrum.

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the distance of the person on the right:

W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m

3)

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

M_c = W_2 d_2

where W_2 is the weight of the seesaw and d_2 = 3 m is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 600 N is his weight and d_1 = 1 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the seesaw:

W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N

#LearnwithBrainly

8 0
3 years ago
Hello Peeps can y'all PLEASE HELP ME? I need to graduate!
Ainat [17]
0A: accelerating
AB: constant
BC: decelerating
CD:at rest
DE:accelerating
EF: constant


hope this helps
3 0
3 years ago
What is the number N0 of 99mTc atoms that must be present to have an activity of 15mCi?
svetoff [14.1K]
Base on my research, within 2 hours you have a number of atoms which remain. 
N= N0*2^(-t/6.020 = N= N0*2^-0.33223= 07943 N0

So, the number of atoms that are being disintegrated is N0-N=N0*(1-0.79430)=0.2057 N0

It must be equal to 15 mCi = 15*3.7*10^7= 5.55*10^8 atoms 

N0= 5.55*10*8/0.2057  = 2.698*10^9 atoms

Therefore, 2.698*10^9 atoms is the number of N0
4 0
3 years ago
If the net force acting on an object is zero, its inertia is also zero.<br> True or false
S_A_V [24]

Answer:

its false

Explanation:

5 0
3 years ago
How to calculate kinetic energy given mass and velocity
sashaice [31]

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

3 0
2 years ago
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