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2 years ago

A wave has a wavelength of 3.3 m and a speed of 5.6 m/s. What is the frequency of this wave

Physics

1 answer:

nikitadnepr [17]2 years ago

5 0

Answer: λ (wavelength) = 3.3 m → ΔS = 3.3 m

v (speed) = 5.6 m/s → ΔV = 5.6 m/s

T (period) → ΔT = ?

f (frequency) = ?

If:

Now: (if, ΔT = T), Using the formula of the period of a wave, find the frequency:

Explanation: Your Welcome u.u

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A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is

Sindrei [870]

Answer:

d = 0.05 [m] = 50 [mm]

Explanation:

We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.

$E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm]$

5 0

2 years ago

A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.

vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

F = B i L

Rod is not necessarily vertical

$F_x =i L B_d$

$F_y= i L B_w$

the normal reaction N = mg-F y

static friction f = μ_s (mg-F y )

horizontal acceleration is zero

$F_x-f = 0$

$iLBd = \mu_s(mg-F_y )$

B_w = B sinθ

B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

$B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}$

$\theta =tan{-1}{\mu_s}$

$\theta =tan{-1}{0.6}$

$\theta = 31^0$

$B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}$

B = 0.1 T

4 0

3 years ago

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

3 years ago

A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electr

Katen [24]

Answer:

ФE = 9.403W

Explanation:

In order to calculate the magnitude of the electric flux trough the sheet, you use the following formula:

$\Phi_E=\vec{A}\cdot \vec{E}=AEcos\theta$ (1)

A: area of the rectangular sheet = (0.400m)(0.600m) = 0.24m^2

E: magnitude of the electric field = 95.0N/C

θ: angle between the direction of the electric field and the normal to the surface of the sheet

You replace the values of the parameters in the equation (1):

$\Phi_E=(0.24m^2)(95.0N/m)cos(20\°)=9.304W$

The magnitude of the electric flux is trough the sheet is 9.403W

5 0

2 years ago

The atomic number of magnesium is 12. This means that its nucleus must contain

nasty-shy [4]

12 protons in the nucleus

5 0

2 years ago