The burning of wood in a campfire is which type of reaction?

A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni

dsp73

Answer:

Explanation:

Theorem of Binomial Distribution will apply here.

n = 29 , p = .67 , q = 0.33

mean = np = 29 x .67 = 19.43

Standard Deviation = √npq

= √29 x .67 x .33

= √6.4

= 2.53

=

4 0

3 years ago

The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag

Deffense [45]

The answer is D using the work formula
W= F•d but if it was against gravity, it would be 0 if gravity is exerting the same amount, I would pick D using the formula, but I'm not so sure sorry

7 0

3 years ago

A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult

Dvinal [7]

Answer:

Explanation:

Given

time taken $t=2\ s$

Speed acquired in 2 sec $v=42\ m/s$

Here initial velocity is zero $u=0$

acceleration is the rate of change of velocity in a given time

$a=\frac{v-u}{t}$

$a=\frac{42-0}{2}=21\ m/s^2$

Distance travel in this time

$s=ut+0.5at^2$

where

s=displacement

u=initial velocity

a=acceleration

t=time

$s=0+\0.5\times 21\times (2)^2$

$s=42\ m$

so Jet Plane travels a distance of 42 m in 2 s

5 0

3 years ago

An ac generator with Em = 223 V and operating at 399 Hz causes oscillations in a series RLC circuit having R = 222 Ω, L = 147 mH

Doss [256]

Answer:

Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A

Explanation:

Em = 223 V

f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF

a)

Capacitive reactance = Xc=?

Xc= $\frac{1}{2\pi fC}$

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$

Xl= 2π * f * L

Xl= 2π * 399 * 147 * $10^{-3}$

Xl= 368.5 Ω

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$ = $\sqrt{222^2 + (368.5 - 17.267)^2}$

Z= 415.5 Ω

c).

Current:

I= V / Z= Em / Z

I= 223/415.5

I= 0.537 A

3 0

3 years ago

An asteroid with a mass of 5.0 x 105 kg collides with the Earth and slides horizontally along the ground without bouncing (not a

V125BC [204]

The strength of the friction doesn't matter. Neither does the distance or the time the asteroid takes to stop. All that matters is that the asteroid has

1/2 (mass) (speed squared)

of kinetic energy when it lands, and zero when it stops.
So

1/2 (mass) (original speed squared)

is the energy it loses to friction in order to come to rest.

8 0

3 years ago