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ch4aika [34]
2 years ago
14

A football field lies so one endzone is to the east and one is to the west. A running back gets the ball and starts running 20o

north of east. After running 10 meters, he sees open field and starts running straight east for 60 meters at which point a safety closes in on him. He cuts so he is running 35o east of south. After running another 20 meters he gets in the end zone. Ultimately we want to know what is the vector that points from where he started to where he scored.
Physics
1 answer:
Jobisdone [24]2 years ago
8 0

Answer:

( 80.87 i - 12.96 j ) m

Explanation:

Running back movements :

20⁰ north of east  for  10 meters,  

Runs straight east for 60 meters

runs 35⁰ east of south for 20 meters

A) show vector pointing from the starting point to the end ( where he scored )

The final vector displacement : ( 80.87 i - 12.96 j ) m

which is : 81.90 m, 9.10⁰ south of east

attached below is the required  diagram

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A box of mass 8 kg slides across a frictionless surface at an initial speed 1.5 m/s into a relaxed spring of spring constant 69
Sav [38]

Answer:

1.1 sec

Explanation:

m = mass of the box = 8 kg

k = spring constant of the spring = 69 N/m

v = initial speed of the box = 1.5 m/s

t = time period of oscillation of box in contact with the spring

Time period is given as

t = \pi \sqrt{\frac{m}{k}}

Inserting the values

t = (3.14) \sqrt{\frac{8}{69}}

t = 1.1 sec

5 0
2 years ago
Carts A and B are identical and are moving toward each other on a track. The speed of cart A is v, while the speed of cart B is
borishaifa [10]

Answer: k= \frac{5mv^{2} }{2}

Explanation:

Recall that the formula for kinetic energy is given below as

k = \frac{mv^{2} }{2}

where k=kinetic energy (joules), m= mass of object (kg), v= velocity of object m/s)

For cart A

m_{a} = mass of cart A

v_{a} = v = velocity of cart A

K.E_{a} = kinetic energy of cart A

hence, K.E_{a} = \frac{m_{a}v^{2}  }{2}

For cart B

m_{b} = mass of cart B

v_{b} = 2v = velocity of cart B

K.E_{b} = kinetic energy of cart B

hence, K.E_{b} = \frac{m_{b}(2v^{2}) }{2} = 2m_{b} v^{2}

from the question, both cart are identical which implies they have the same mass i.e m_{a} = m_{b} = m which implies that

K.E_{a}= \frac{mv^{2} }{2} and K.E_{b}  =2mv^{2}

The total kinetic energy K is the sum of cart A and cart B kinetic energy

K=K.E_{a} + K.E_{b}

K=\frac{mv^{2} }{2} + 2mv^{2}

hence

K=\frac{5mv^{2} }{2}

6 0
2 years ago
You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interfer
boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

5 0
2 years ago
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
kipiarov [429]

Answer:

I=68.31\times 10^{-6}\ A

Explanation:

Given that

J(r) = Br

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dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

Given that

B= 2.35 x 10⁵ A/m³

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r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

By putting the values

I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A

I=68.31\times 10^{-6}\ A

7 0
2 years ago
Read 2 more answers
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In-s [12.5K]

This question deals with the volume of different shapes.

a) volume of the sphere is "33.51 m³".

b) volume of the cylinder is "25.13 m³".

a)

The volume of a sphere is given by the following formula:

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<u>Volume = 33.51 m³</u>

<u />

b)

The volume of a cylinder is given by the following formula:

Volume = \pi r^2l\\\\Volume =\pi (2\ m)^2(2\ m)

<u>Volume = 25.13 m³</u>

<u />

Learn more about <em>volume </em>here:

brainly.com/question/16686115?referrer=searchResults

The attached picture shows the formulae of the <em>volume</em> of different shapes.

7 0
2 years ago
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