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sattari [20]
3 years ago
6

Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

ast for 79 km. What is her total displacement from the office? Answer in units of km. \ At what angle is her displacement? (C
Physics
1 answer:
Natasha2012 [34]3 years ago
4 0

Answer:

The total displacement is 102 km 51^o north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

d_n=34km+79km*sin(35^o)=79km

and

d_e=79*cos(35^o)=65km

The total displacement is given by:

D=\sqrt{(79km)^2+(65km)^2}=102km

with an agle of:

\alpha =arctg(\frac{79km}{65km})=51^o

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Two capacitors have the same size of plates and the same distance 4 mm between the plates The potentials of the two plates in ca
OverLord2011 [107]

Answer:

    E = -4000 N / C

Explanation:

The potential and electric field are related

         V = - E s

          E = - V / s

we reduce the magnitudes to the SI system

          s = 4 mm (1 m / 1000 mm) = 0.004 m

we calculate

          E = - 16 /0.004

          E = -4000 N / C

8 0
3 years ago
Free Fall: A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses
Slav-nsk [51]

Answer:

v₀₁= 5.525 m / s

Explanation

Freefall Formulas :

The sign of acceleration due to gravity  (g) is positive if the object is going down and negative if the object is going up.

vf= v₀+gt  

vf²=v₀²+2*g*h

h= v₀t+ (1/2)*g*t²

Where:  

h: hight in meters (m)    

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Kinematics of the rock from the starting point with vo until it reaches its maximum height:

vf₁= v₀₁-gt₁  :vf₁ =0 to maximum height

0= v₀₁-gt₁

v₀₁ = g*t₁

t₁ =v₀₁ / g      Equation (1)

vf₁²= v₀₁²-2*g*h₁   : vf₁ =0 to maximum height

0 = v₀₁²-2*g*h₁

2*g*h₁ = v₀₁²

h₁ = (v₀₁²)/(2g)   Equation (2)

Kinematics of the rock when it falls from the maximum height until it touches the floor

h₂= v₀₂t+ (1/2)*g*t₂²  v₀₂=vf₁ =0

h₂= 0+ (1/2)*g*t₂²

h₂= (1/2)*g*t₂²   Equation (3)

Equation that relates h₁ to h₂

h₂=  h₁ + 56.3  ,  h₁ = (v₀₁²)/(2g)

h₂= (v₀₁²)/(2g) + 56.3  Equation (4)

Equation that relates t₁ to t₂

t₁ + t₂ =4 s

t₂ =4 -t₁

t₂ =4 -(v₀₁/g )

Calculation of v₀₁

We replace equation 4 and equation 5 in equation 3

(v₀₁²)/(2g) + 56.3 = (1/2)*g*(4 -(v₀₁/g ) )²

(v₀₁²)/(2g) + 56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g )+((v₀₁/g )²)

we eliminate (v₀₁²)/(2g) on both sides of the equation

56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g ))

56.3 = 78.4 - 4*v₀₁

4*v₀₁ =78.4-56.3

v₀₁= (78.4-56.3) / ( 4)

v₀₁= 5.525 m / s

7 0
3 years ago
A fairground ride consists of a large vertical drum that spins so
expeople1 [14]

Answer:

v = 3.84 m/s

Explanation:

In order for the riders to stay pinned against the inside of the drum the frictional force on them must be equal to the centripetal force:

Centripetal\ Force = Frictional\ Force\\\\\frac{mv^2}{r} = \mu R = \mu W\\\\\frac{mv^2}{r} = \mu mg\\\\\frac{v^2}{r} = \mu g\\\\v = \sqrt{\mu gr}

where,

v = minimum speed = ?

g = acceleration due to gravity = 9.81 m/s²

r = radius = 10 m

μ = coefficient of friction = 0.15

Therefore,

v=\sqrt{(0.15)(9.81\ m/s^2)(10\ m)}

<u>v = 3.84 m/s</u>

6 0
3 years ago
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