Which field in an 802.11a plcp frame are used to initialize part of the transmitter and receiver circuits?

Which of these habitats has a highly salinity

Llana [10]

The answer for APEX Learning is Ocean

7 0

3 years ago

A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact

Leno4ka [110]

Answer:

471392.4 N

Explanation:

From the question,

Just before contact with the beam,

mgh = Fd.................... Equation 1

Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.

make f the subject of the equation

F = mgh/d................ Equation 2

Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m

Constant: g = 9.8 m/s²

Substitute into equation 2

F = 1900(4)(9.8)/0.158

F = 471392.4 N

6 0

3 years ago

the dimensions of a cuboidal block are 2.5 m * 2m * 1.2m its weight 900 n what is the minimum presseure exerted by the block on

aleksandr82 [10.1K]

Answer:

your answer is here 180

Explanation:

900/2.5 * 2 = 180

6 0

2 years ago

Read 2 more answers

What must be the units for the gravitational constant G in order for gravitational force to have units of newtons?

babunello [35]

Answer:

m³/(kg⋅s²)

Explanation:

Hello.

In this case, since the involved formula is:

$F=G*\frac{m_1m_2}{r^2}$

By writing a dimensional analysis with the proper algebra handling, we obtain:

$N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}$

Thus, answer is:

m³/(kg⋅s²)

Note that the [=] is used to indicate the units of G.

Best regards

4 0

2 years ago

Multiple-Concept Example 6 reviews the concepts that play a role in this problem. A diver springs upward with an initial speed o

adell [148]

Answer:

Part a)

$v_f = 7.99 m/s$

Part b)

$y = 3.25 m$

Explanation:

Part a)

Since the diver is moving under gravity

so here its acceleration due to gravity will be uniform throughout the motion

so here we will have

$v_f^2 - v_i^2 = 2 a y$

here we have

$v_i = 2.22 m/s$

$y = -3 m$

$v_f^2 - (2.22)^2 = 2(-9.81)(-3)$

$v_f = 7.99 m/s$

Part b)

at highest point of his motion the final speed will be zero

so we will have

$v_f^2 - v_i^2 = 2 a (\Delta y)$

$0 - 2.22^2 = 2(-9.81)(y - 3)$

$y = 3.25 m$

7 0

3 years ago