What is the medium for this wave?

A boy is standing in the middle of a field when his friend sprints past him with a speed of 4 m/s. At the moment his friend pass

algol13

Answer:

Explanation:

1) ½(0.8)t² = 4t

0.4t² = 4t

0.4t = 4

t = 10 s

2) d = vt = 4(10) = 40 m

d = ½atⁿ = ½(0.8)10² = 40 m

3) v = at = 0.8(10) = 8 m/s

5 0

3 years ago

James went to the doctor for a check-up and found his weight to be 1,000 N. How much mass does James have?

Dmitry_Shevchenko [17]

Answer:

102.04kg

Explanation:

(1000kg*m/s^2)/(9.8 m/s^2)

5 0

3 years ago

The magnetic field on the Sun is created by _____.

ExtremeBDS [4]

It would be the flowing charged plasma blowing off the sun's surface as solar wind that creates the magnetic field.

7 0

3 years ago

Read 2 more answers

A 0.0780 kg lemming runs off a

attashe74 [19]

Answer:

Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J

Explanation:

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its potential energy (PE) when it lands

The potential energy PE, relative to the ground, will be zero, because the lemming is at the ground level.

HOWEVER, a much better question would be:

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its kinetic energy (KE) when it lands?

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, u = 0, a = 9.81m/s^2, s = 5.36m

So we find v using equation (2)

v^2 = u^2 + 2as

v^2 = 0 + 2(9.81)(5.36) = 105.1632

So the kinetic energy resulting from the vertical drop is Ekv = ½mv^2

Ekv = ½(0.078)(105.16) = 4.101J

BUT we need to add in the kinetic energy resulting from the horizontal velocity, which did not change during the vertical drop.

Ekh = ½(0.078)(4.84^2) = 0.914J

So the total kinetic energy is Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J

5 0

3 years ago

cylindrical water tank is 20.0 m tall, open to the atmosphere at the top, and is filled to the top. It is noticed that a small h

Naya [18.7K]

Answer:

a

The velocity is $v =17.98 \ m/s$

b

The diameter is $d = 0.00184m$

Explanation:

The diagram of the set up is shown on the first uploaded image

From the question we are told that

The height of the water tank is $h = 20.0 \ m$

The position of the hole $p_h = 16.5m$ below water level

The rate of water flow $\r V = 2.90 *10^{-3} m^3 /min = \frac{2.90 *10^{-3}}{60} = 0.048*10^{-3} m^3/s$

According to Bernoulli's theorem position of the hole

$\frac{P_o + h \rho g}{\rho} + \frac{1}{2} u^2 = \frac{P_o}{\rho } + \frac{1}{2} v^2$

Where u is the initial speed the water through the hole = 0 m/s

$P_o$ is the atmospheric pressure

$\frac{P_o }{\rho} + \frac{ h \rho g}{\rho} + 0 = \frac{P_o}{\rho } + \frac{1}{2} v^2$

$v = \sqrt{2gh}$

Substituting value

$v = \sqrt{2 * 9.8 * 16.5 }$

$v =17.98 \ m/s$

The Volumetric flow rate is mathematically represented as

$\r V = A * v$

Making A the subject

$A = \frac{\r V}{v}$

substituting value

$A = \frac{0.048 *10^{-3}}{17.98}$

$= 2.66*10^{-6}m^2$

Area is mathematically represented as

$A = \frac{\pi d^2}{4}$

making d the subject

$d = \sqrt{\frac{4*A}{\pi} }$

Substituting values

$d = \sqrt{\frac{4 * 2.67 *10^{-6}}{3.142} }$

$d = 0.00184m$

6 0

3 years ago