A wave has a wavelength of 1.5 meters and period of 0.083s. What is the waves speed?

Answer:

I think its d

Explanation:

I'm not sure I'm sorry if I'm wrong

Answer:

475.7 m/s

Explanation:

Given,

Frequency ( f ) = 67 Hz

Wavelength ( λ ) = 7.1 m

To find : Speed ( v ) = ?

Formula : -

v = f λ

v

= 67 x 7.1

= 475.7 m/s

Therefore,

the speed of the wave is 475.7 m/s.

Answer:

true it all changes

Explanation:

________________________

This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.

The minimum horizontal velocity required is "2.6 m/s".

First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.

[tex]h=v_it+\frac{1}{2}gt^2[/tex]

where,

h = height = 3 m

vi = initial vertical speed = 0 m/s

t = time interval = ?

g = acceleration due to gravity = 9.81 m/s²

therefore,

[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]

t = 0.78 s

Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:

[tex]s = vt\\\\v = \frac{s}{t}[/tex]

where,

s = horizontal distance = 2 m

t =0.78 s

v = minimum horizontal velocity = ?

Therefore,

[tex]v = \frac{2\ m}{0.78\ s}[/tex]

v = 2.6 m/s

Learn more about equations of motion here:  

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

Answer:

The answer is "15.56 cm".

Explanation:

[tex]v= 2.5 \ cm\\\\f= 2.154 \ cm[/tex]

Calculating object of length is x so:

[tex]u= -x[/tex]

Using formula:

[tex]\to \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to \frac{1}{x}=\frac{1}{2.154}-\frac{1}{2.5}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to x= 15.56 \ cm[/tex]

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A  = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]

Therefore, the length of the wire is 134.62 m.

Answer:

Distance travel by go-cart = 500 meter

Explanation:

Given:

Speed of go cart = 25 m/s

Time travel = 20 seconds

Find:

Distance travel by go-cart

Computation:

Distance = Speed x time

Distance travel by go-cart = Speed of go cart x Time travel

Distance travel by go-cart = 25 x 20

Distance travel by go-cart = 500 meter

Answer:

If light enters any substance with a higher refractive index (such as from air into glass) it slows down. The light bends towards the normal line. If light travels enters into a substance with a lower refractive index (such as from water into air) it speeds up. The light bends away from the normal line.

Answer:

question #1 is A

Question #2 is C

Explanation:

Answer:

Change in KE = +1.96×10^4 J while the change in ME = 0 J

Answer:

2

Explanation:

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          [tex]\frac{1}{f_1} = \frac{1}{q}[/tex]

           q = f₁

2) for an object located at p = 25 cm

            [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]

        [tex]\frac{1}{f_2}[/tex] = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = [tex]\frac{1}{f}[/tex]

where the focal length is in meters

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

Answer:

e = 0.0898m

v = 2.07m/s

Explanation:

a) According to Hooke's law

F = ke

e is the extension

k is the spring constant

Since F = mg

mg = ke

e = mg/k

Substitute the given value

e = 1.1(9.8)/120

e = 10.78/120

e = 0.0898m

Hence it is stretched by 0.0898m from its unstrained length

2) Total Energy = PE+KE+Elastic potential

Total Energy = mgh +1/2mv²+1/2ke²

Substitute the given value

5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²

Solve for v

5.0 = 2.156+0.55v²+0.48338

5.0-2.156-0.48338= 0.55v²

2.36 =0.55v²

v² = 2.36/0.55

v² = 4.29

v ,= √4.29

v = 2.07m/s

Hence the required velocity is 9.28m/s

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]

Also, at any instant t

[tex]\Rightarrow l^2=x^2+y^2[/tex]

differentiate w.r.t.

[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]

Answer:

D:#2 African Plate

Explanation:

The African Plate is flanked by the Red sea rift and the minor African plate.

The Red sea rift is a small part of a greater line of rifts known as the Great African Rift Valley. The rift valley is making several small lakes all through Africa and it will eventually split up the African continent.

The Red sea lift is the divergent boundary between the African plate and the Arabian plate. It means that the two plates are moving apart or spreading apart.

Answer:

B = 1.37 mT

Explanation:

Given that,

The magnitude of the electric field, E = 480 N/C

The speed of the proton, [tex]v=3.50 \times 10^5\ m/s[/tex]

We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,

[tex]qE=qvB[/tex]

Where

B is the magnetic field

[tex]B=\dfrac{E}{v}\\\\B=\dfrac{480}{3.5\times 10^5}\\\\B=1.37\times 10^{-3}\ T\\\\or\\\\B =1.37\ mT[/tex]

So, the magnetic field is equal to 1.37 mT.

Answer:

0.1 kg/s.

Explanation:

The density of water, d = 1000 kg/m³

Volume, V = 2 L

Time, t = 20 s

We need to find the mass flow rate from the faucet. We know that the density of an object is given by :

[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]

So, the mass flow rate is equal to 0.1 kg/s.

Answer:

B

Explanation:

The Lorentz force is the sum of the electric force and magnetic force.

F = qE + qvB

qE represents electric force and qvB represents magnetic force

Answer:

Some plains form as ice and water erodes, or wears away, the dirt and rock on higher land. Water and ice carry the bits of dirt, rock, and other material, called sediment, down hillsides to be deposited elsewhere. As layer upon layer of this sediment is laid down, plains form. Volcanic activity can also form plains.

Answer:

28.7%

Explanation:

efficiency = work output /work input × 100

Answer:

U=30cm

Explanation:

All you have to do is to put

Mirror formula , 1/f=1/u + 1/v

You should be careful in sign convention .

Virtual image is negative

we take focal length of convex lens negative even if its not given and so on...

Answer:

[tex]4.25\ \text{m/s}[/tex]

[tex]3391.22\ \text{N}[/tex]

Explanation:

y = Height of compression = 0.38 m

m = Mass of basketball player = 101 kg

h = Height of center of gravity after jump = 0.92 m

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Energy balance of the system is given by

[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]

The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]

[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]

The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].

Answer:

[tex]0.886[/tex] N buoyant force is exerted by air

Explanation:

My weight is [tex]75[/tex] Kg

Weight = mass * gravity

As we know

Buoyant Force is equal to the product of density * acceleration due to gravity and volume of the body

Assuming weight density is a bit less than that of water or equal to water i.e [tex]997.77[/tex] kg/m3

Volume is equal to mass / density

[tex]= 75[/tex] Kg * g/[tex]997.777[/tex]

[tex]= 0.0751[/tex] * g

Buoyant Force

= Volume * g * density

[tex]= 0.0751 * 9.8 * 1.2041[/tex]kg/m3

[tex]= 0.886[/tex] N

Answer:

W = 1/2 K x^2

x^2 = 2 * W / K = 2 * 49 J / (N/m)  = .218 / m^2

x =  .467 m