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spayn
1 month ago
12

Lucy took 3 hours to cover 2/3 of a journey. She covered the remaining 60 miles in 2 hours. What was the average speed for the w

hole journey?
Physics
1 answer:
serg [3.5K]1 month ago
5 0

The accurate answer is 36 miles per hour

Explanation:

To calculate the average speed, start by determining the total distance and total time Lucy traveled, as average speed is the total distance divided by the total time.

Total Distance

Lucy is known to have covered 2/3 of the total distance initially, later finishing the journey with 60 miles. Therefore, the last segment of 60 miles corresponds to 1/3 of the total distance.

\frac{2}{3} + \frac{1}{3} = \frac{3}{3} (Total distance)

Consequently, since 60 miles equals one-third, multiplying this by 3 gives the total distance: 180 miles.

This indicates the total distance traversed is 180 miles

Total Time

Lucy took 3 hours for the first part of her journey and 2 hours for the second, leading to a total time of 5 hours (3 + 2 = 5)

Determining Average Speed

Average speed =\frac{Distance}{time}

Average speed = \frac{180 m}{5 h}

Average speed = 36m/h

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At one instant of time a rocket is traveling in outer space at 2500m/s and is exhausting fuel at a rate of 100 kg/s. If the spee
Keith_Richards [3271]

Thrust is quantified as a reaction force, in accordance with Newton's third law. When a system accelerates or expels mass in one direction, this resulting mass generates a force of equal strength but in the opposite direction on that system. This relationship can be expressed mathematically as:

T = v\frac{dm}{dt}

Where:

v = velocity of the exhaust gases as perceived from the rocket.

\frac{dm}{dt}= Change in mass over time

The provided data is as follows:

v = 1500m/s

\frac{dm}{dt} = 100kg/s

After substitution, we obtain:

T = 1500*100

\therefore T = 1.5*10^5N

7 0
1 month ago
A BMX bicycle rider takes off from a ramp at a point 2.4 m above the ground. The ramp is angled at 40 degrees from the horizonta
Ostrovityanka [3204]

Answer:

The BMX rider lands 5.4 meters horizontally away from the ramp's end.

Explanation:

The BMX position vector is represented as:

r = (x0 + v0 × t × cos α, y0 + v0 × t × sin α + ½ × g × t²)

Where:

r = position at time t

x0 = initial horizontal position

v0 = initial speed

α = angle of jump

y0 = initial vertical height

g = gravitational acceleration (-9.8 m/s², upward positive)

Refer to the diagram for clarity. At the landing time, the vertical coordinate of the position vector is -2.4 m, measured from the ramp's edge as the origin. Using the vertical component equation for y, one can solve for t, then substitute t to find the horizontal distance.

The vertical position equation:

-2.4 m = 0 + 5.9 m/s × t × sin 40° - ½ × 9.8 m/s² × t²

Rearranged:

0 = -4.9 t² + 5.9 t × sin 40° + 2.4

Solving this quadratic yields:

t = 1.2 seconds

Then, calculate horizontal distance:

x = 0 + 5.9 m/s × 1.2 s × cos 40° = 5.4 m

This means the BMX lands 5.4 meters from the ramp's edge.

Have a great day!

8 0
2 months ago
A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts
inna [3103]
Refer to the diagram shown below.

m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.

Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.

Answer: The total resistive force amounted to 910 N.

Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.

Answer: The tension in the hitch is 1113 N.

3 0
1 month ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ \frac{du}{dy}

therefore

= µ \frac{v}{h}............1

substituting the values

= 1.75 ×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

the area between the air and puck is given by

Area = \frac{\pi }{4} d^{2}

area = \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v

now applying Newton's second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

solving for t = \frac{0.1 v * 0.03}{1.37*10^{-3} v}

the time needed after impact for the puck is 2.18 seconds

3 0
1 month ago
1. A 930-kg car traveling 56 km/h comes to a complete stop in 2.0 s. What is the
kicyunya [3294]
The force exerted on the car during the stop measures 6975 N. Explanation: Given that the mass (m) is 930 kg, speed (s) at 56 km/h converts to 15 m/s, and the stopping time (t) is 2 s, we compute the force using F = m * a. Here, acceleration (a) can be obtained through a = s/t. The total force calculation confirms that F = 930 kg * (15 m/s) / 2 s results in 6975 N.
6 0
20 days ago
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