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3 years ago

A golfer collected data on the distance a golf cart traveled in a straight line and plotted it on a graph

Physics

1 answer:

devlian [24]3 years ago

5 0

Answer:

The cart moved away from the starting point between 8 s and 10 s.

Explanation:

Given that :

Which of these does NOT describe the cart’s motion on this graph?

The cart was at rest between 5 s and 7 s : From the distance of golf cart Vs time graph ; the car was at rest between 5s and 7s as the graph was flat with that time interval, meaning there was no change in distance.

The cart moved toward the starting point between 7 s and 12 s. : The graph depicts a negative slope at this time interval as the distance from starting point fell from about 26 m to 10 m

The cart moved away from the starting point between 8 s and 10 s. : At this time interval, the cart moved towards the starting not and not away. This could be seen in the decrease in Distance from starting point between the tune interval.

The cart moved away from the starting point between 2 s and 5 s. - - > The cart moved away from the starting point, with the positive slope signifying an increase in distance.

Therefore, The cart moved away from the starting point between 8 s and 10 s does not describe the motion of the cart on the graph.

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Explain how energy is transferred in an impact situation such as a car crash.

DaniilM [7]

During a car crash, energy is transferred from the vehicle to whatever it hits, be it another vehicle or a stationary object. ... The object that was struck will either absorb the energy thrust upon it or possibly transfer that energy back to the vehicle that struck it.

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8 0

3 years ago

The pilot of an aircraft wishes to fly due west in a 33.9 km/h wind blowing toward the south. The speed of the aircraft in the a

diamong [38]

Answer: $\theta =9.96^{\circ}$ North of west

Explanation:

Given

Plane wishes to fly in west

but wind with speed 33.9 km/h towards south obstructing its path

so plane must fly at an angle of \theta w.r.t west such that it final velocity is towards west

Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

thus $195sin\theta =33.9$

$\theta =9.96^{\circ}$

so Plane should head towards $9.96^{\circ}$ North of west in order to fly in west.

So plane

actual velocity is

$v=-195cos9.96\hat{i}+195sin9.96\hat{j}$

5 0

3 years ago

A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance

seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

4 0

3 years ago

Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s

Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density = 0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

= -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level ) = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = $\sqrt{CRT}$ = $\sqrt{1.4*287*200.5 }$ = 283.8 m/s

hence it is a subsonic aircraft

4 0

3 years ago

A water bath in a physical chemistry lab is 1.55 m long, 0.710 m wide, and 0.570 m deep (high). If it is filled to within 3.55 i

Lesechka [4]

Answer:

528 liter.

Explanation:

Volume of the tank(cuboid) = l*b*h

But volume of the water = l*b*h

Where

l= length of the tank

b = width of the tank

h = the length from the bottom of the tank,

3.55 in to m,

0.09017m

Length of the water in the tank = 0.570 - 0.09017

= 0.47983 m.

Volume = 0.47983*0.710*1.55

= 0.528 m3.

1 m3 = 1000 liter.

0.528 m3 = 0.528*1000

= 528 liter

7 0

3 years ago