A driver drives for 30.0 minutes at 80.0 km/h, then 45.0 minutes at 100 km/h. She then stops 30 minutes for lunch. She then travels for 30 minutes at 80 km/h. (a) Sketch a plot of her displacement versus time and speed versus time. (b) Calculate her average speed.

Answers

Answer 1

Answer:

Answer:

b) 68,9 km/h a) picture

Explanation:

In this problem, since velocity is expressed in km/h and time in minutes, we have to convert either time to hours or velocity to km/min. It is easier to use hours.

Using this formula we pass time to hours:

$t_(hours)=t_(min)*(1 h)/(60 min)\n30min*(1 h)/(60 min)=0,5h\n45min*(1 h)/(60 min)=0,75h$

Now we can plot speed vs time (image 1). The problem says that the driver uses constant speed, so all lines have to be horizontal.

Using the values of the speed we calculate the distance in each interval

$d=v*t\n80km/h*0.5h=40km\n100km/h*0.75h=75km$

Using these values and the fact that she was having lunch in the third one (therefore stayed in the same position), we plot position vs time, using initial position zero (image 2, distance is in km, not meters).

Finally, we compute the average speed with the distance over time:

$v_(average)=(155km)/(2.25h)=68.9km/h$

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An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?

Answers

I assume the graph is looking like in the picture bellow.

North component:
cos(41.5) * 835 = 625.37 km/h

West component of speed:
sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:
2.2*625.37 = 1375.81 km north
2.2*553.29 = 1217.23 km west

Final answer:

To find the components of the velocity vector, you can use trigonometry. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.

Explanation:

To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: North component = velocity * sin(angle). To find the west component, we can use the cosine function: West component = velocity * cos(angle).

After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: Distance north = North component * time and Distance west = West component * time.

Let's calculate the values:

North component = 835 km/h * sin(41.5°)
West component = 835 km/h * cos(41.5°)
Distance north = North component * 2.20 h
Distance west = West component * 2.20 h

Learn more about Velocity components here:

brainly.com/question/14478315

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John and Suzie are trying to improve the speed of their race car by adjusting the angle of the rear spoiler. Which choice is the controlled variable(s)?the maximum speed and the angle of the rear spoiler

the type of tire and the type of fuel

only the maximum speed

only the angle of the rear spoiler

Answers

Answer:

the type of tire and the type of fuel

Explanation:

If Jim could drive a Jetson's flying car at a constant speed of 490 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 4.5 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days). unanswered

Answers

Answer:

109.5 million years

Explanation:

The question asked us to find the time.

Remember that

Rate of velocity = distance / time, and this,

time taken = distance/rate

Due to the confusing nature of the units, we would have to be converting them to a more uniform one.

1 km is equal to 9.461*10^12 km/light-year, that's if we try to convert km to light year.

Since the speed is in km, the distance has to be in km also, and therefore, we convert ly to km:

4.5 light-years = 9.461*10^12 km/light-year) = 42.57*10^13 km

We that this value as our distance, in km.

Also,

Time = distance/speed

Time = 45.57*10^13 km / 490 km/hr = 9.3*10^11 hr

Now the next step is to convert hours to years, using the conversion factor 8766 hr/yr.

time (in years) = 9.6*10^11 hr / 8766 hr/yr) = 10.95*10^7 years

the final step is to divide the time in years by 10^6 years/million years, which gives the final answer as the trip takes 109.5 million years.

A mysterious object with a surface area of 0.015 m2, volume of 0.000125 m3, density of 100 kg/m3, specific heat of 100 J/(kgK), thermal conductivity of 2 W/(mK), with an unknown initial temperature was placed in a fluid with a density of 50 kg/m3, specific heat of 70 J/(kgK), thermal conductivity of 0.1 W/(mK), at a temperature of 400K. The heat transfer coefficient is given to be 10 W/(m2K). After 10 seconds, the temperature of the object is measured to be 380K. Determine the object's initial temperature.

Answers

Answer:

The object's initial temperature is 333.6 K

Explanation:

We first assume that the liquid can only transfer heat to the object through convective heat transfer method.

Let T₀ = the initial temperature of the object

T = temperature of the object at anytime.

The rate of heat transfer from the liquid to the object is given as

Q = -hA (T∞ - T)

T∞ = temperature of the fluid = 400 K

A = Surface area of the object in contact with the liquid = 0.015 m²

h = Convective heat transfer coefficient is given to be = 10 W/(m²K)

The rate of heat gained by the object is given by

mC (d/dt)(T∞ - T)

m = mass of the object = ρV

ρ = density of the object = 100 kg/m³

V = volume of the object = 0.000125 m³

m = ρV = 100 × 0.000125 = 0.0125 kg

C = specific heat capacity of the object = 100 J/(kgK)

The rate of heat loss by the liquid = rate of heat gain by the object

-hA (T∞ - T) = mC (d/dt)(T∞ - T)

(d/dt)(T∞ - T) = - (dT/dt) ( Since T∞ is a constant)

- mC (dT/dt) = -hA (T∞ - T)

(dT/dt) = (hA/mC) (T∞ - T)

Let s = (hA/mC)

(dT/dt) = -s (T - T∞)

dT/(T - T∞) = -sdt

Integrating the left hand side from T₀ (the initial temperature of the object) to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -st

(T - T∞)/(T₀ - T∞) = e⁻ˢᵗ

(T - T∞) = (T₀ - T∞)e⁻ˢᵗ

s = (hA/mC) = (10 × 0.015)/(0.0125×100) = 0.12

T = 380 K at t = 10 s

T₀ = ?

T∞ = 400 K

st = 0.12 × 10 = 1.2

(380 - 400) = (T₀ - 400) e⁻¹•²

(-20/0.3012) = (T₀ - 400)

(T₀ - 400) = - 66.4

T₀ = 400 - 66.4 = 333.6 K

Hope this Helps!!!

A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Answers

The man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip.

A - point at the top of the ladder

B - point at the bottom of the ladder

C - point where the man is positioned in the ladder

L- the length of the ladder

α - angle between ladder and ground

x - distance between C and B

The forces act on the ladder,

Horizontal reaction force (T) of the wall against the ladder

Vertical (upward) reaction force (R) of ground against the ladder.

Frictionalhorizontal ( to the left ) force (F)

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N

in static conditions,

∑Fx = T - F = 0 Since, T = F

∑Fy = mg - R = 0 Since, 735 - R = 0, R = 735

∑ Torques(b) = 0

In point B the torque produced by forces R and F is Zero

Then:

∑Torques(b) = 0

And the arm lever for each force,

mg = 735

Since, ∑Torques(b) = 0

$\bold {735* x* cos\alpha = F* L* sin\alpha }$ Since,T = F

$\bold {F = \frac {735* x* cos\apha }{L* sin\alpha }}$ $\bold { \frac {cos \alpha } { sin\alpha }= cot\alpha =\frac 1{tan\alpha}}$

$\bold {F = \frac {735* x* cos\apha }{L }}$

$\bold {F = 735* x* tan\alpha }}$

F < μR the ladder will starts slipping over the ground

μ(s) = 0.25

$\bold { X (max) = 0.25* L* tan \alpha }$

Therefore, the man can climb $\bold { X (max) = 0.25* L* tan \alpha }$ , before the ladders starts to slip. \

To know more about Torque,

brainly.com/question/6855614

Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right) reaction (T) of the wall against the ladder

Point B : Vertical (upwards) reaction (R) of ground against the ladder

frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s² mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0 ⇒ T = F

∑Fy = mg - R = 0 ⇒ 735 - R = 0 ⇒ R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0

And the arm lever for each force is:

mg = 735

d₁ for mg and d₂ for T

cos α = d₁/x then d₁ = x*cosα

sin α = d₂ / L then d₂ = L*sinα

Then:

∑Torques(b) = 0 ⇒ 735*x*cosα - T*L*sinα = 0

735*x*cosα = T*L*sinα

T = F then 735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L or F = (735)*x/L*tanα

When F < μ* R the ladder will stars slippering over the ground

μ(s) = 0,25 0,25*R = 735*x/L*tanα

x = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then x (max) = 0,25*L*tanα

What two statements are true about energy transformations

Answers

Answer:

First statement:

Energy can neither be created nor destroyed.

Second statement:

Energy can be converted from one form to another.

Explanation:

According to the law of conservation of energy:energy can neither be created nor destroyed but can be converted from one form to another

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