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3 years ago

A Plane crashes into the side of a mountain. If the crash took place over 9

Physics

1 answer:

Julli [10]3 years ago

6 0

Given parameters:

Duration of crash = 9min

Velocity before crash = 2000 yards per second

Unknown:

Acceleration of the plane = ?

Solution:

Acceleration is the change of velocity with time.

Acceleration = $\frac{V - U}{t}$

where v is the final velocity

u is the initial velocity

t is the time taken

We need to convert the time into seconds from min;

60s = 1 min

So, 9min = 60 x 9 = 540s

So input the parameters and solve;

Acceleration = $\frac{2000 - 0 }{540}$ = 3.7yards/s²

The acceleration of the plane is 3.7yards/s²

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If it takes 100 N to move a box 5 meters, what is the work done on the box?

jeyben [28]

Answer: D. 500 J

=========================================================

Explanation:

To find the amount of work done, we multiply the force by displacement

work = force*displacement

work = (100 N)*(5 m)

work = (100*5) Nm

work = 500 J

In this case, "Nm" refers to "Newton meters" and not "nanometers"

1 newton meter is equal to 1 joule

7 0

3 years ago

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH = 1.06

Slav-nsk [51]

Answer:

$1.06085\times 10^{-10}\ m$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

e = Charge of electron = $1.6\times 10^{-19}\ C$

n = 1 (ground state)

Angular momentum is given by

$L=mvr$

From Bohr's atomic model we have

$L=\dfrac{nh}{2\pi}$

$mvr=\dfrac{nh}{2\pi}\\\Rightarrow v=\dfrac{nh}{2\pi mr}$

The centripetal force will balance the electrostatic force

$\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow \dfrac{ke^2}{r}=mv^2\\\Rightarrow \dfrac{ke^2}{r}=m(\dfrac{nh}{2\pi mr})^2\\\Rightarrow r=\dfrac{n^2h^2}{4\pi^2mke^2}\\\Rightarrow r=\dfrac{1^2\times (6.626\times 10^{-34})^2}{4\pi^2 \times 9.11\times 10^{-31}\times 8.99\times 10^{9}\times (1.6\times 10^{-19})^2}\\\Rightarrow r=5.30426\times 10^{-11}\ m$

The diameter is $2\times 5.30426\times 10^{-11}=1.06085\times 10^{-10}\ m$

7 0

3 years ago

Velocity is the

ale4655 [162]

Okay, hun. Velocity is a vector quantity that measures displacement over a period of time. Velocity = Speed/Time (v=s/t). Hope this helped you. I took physics over 4 years ago. I'm more of a biology/chemistry person. (I major in those)

4 0

3 years ago

Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall,

Genrish500 [490]

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

6 0

3 years ago

Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan

Drupady [299]

Answer:

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

Explanation:

We are given that

Gravitational force= $F_g=\frac{Gm_Emr}{R^3_E}$

r=0,U(0)=0

We know that

Gravitational potential energy= $-\int F_gdr$

$U(r)=-\int\frac{Gm_Emr}{R^3_E}dr$

$U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C$

Substitute r=0 ,U(0)=0

$0=0+C$

$C=0$

Substitute the value

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

4 0

3 years ago