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Alenkasestr [34]

3 years ago

8

Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 52.6 N to the north. Jenny’s foot exerts

a force of 112.8 N to the east. a) What is the magnitude of the resultant force on the ball? Answer in units of N.

2 answers:

Harman [31]3 years ago

8 0

Answer:

124N

Explanation:

Karolina [17]3 years ago

6 0

Using the addition of forces using right angled triangles. The resultant force sqaured. = 112.8 sqaured + 52.6 squared. So resultant force sqaured is 15490.6. So the resultant force is the sqaure root of this which is 124N to 3 significant figures

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A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

6 0

3 years ago

A 10 kg box is 1.3 m above the ground. How much potential energy does it have? (g on Earth of 9.8 m/s?

Volgvan

Potential energy = mgh
Potential energy = 10 x 9.8 x 1.3
Potential energy = 127.4 J

8 0

3 years ago

Are worm holes a real phenomenon?

Katena32 [7]

It depends, because worm holes are theoretical construed of space and time. It hasn’t been proven to exist but mathematically it hasn’t been spotted but we also haven’t been very far in our universe.

So to cut the story short, it is not a proven phenomenon only theoretical.

6 0

3 years ago

Read 2 more answers

A right-circular cylindrical tank of height 8 ft and radius 4 ft is laying horizontally and is full of fuel weighing 52 lb/ft3

tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

= $832\pi \int_{0}^{8} (21 - y)dy$

= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

7 0

3 years ago

1) An object travels 15 m in 3 s. What is its' speed?

Sav [38]

Answer:

5m/s

Explanation:

Distance/Time= Speed

D/T=S

15/3=S

5=S

8 0

3 years ago