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2 years ago

A truck covered 2/7 of a journey at an average speed of 40

Physics

1 answer:

sertanlavr [38]2 years ago

6 0

Answer:

The amount of time for the whole journey is 8 hours.

Explanation:

A truck covered 2/7 of a journey at an average speed of 40 mph. Representing 1 the total of the trip traveled, then the rest of the distance traveled is calculated as: $1-\frac{2}{7} =\frac{5}{7}$

So if the truck covered the remaining 200 miles at $\frac{2}{7}$ , this means that $\frac{5}{7}$ of the trip represents the 200 miles. So, to calculate the total distance traveled by the truck, you apply the following rule of three: if $\frac{5}{7}$ of the route represents 200 miles, the integer 1 (which represents the total of the route), how many miles are they?

$miles=\frac{1*200miles}{\frac{5}{7} } =\frac{7}{5} *200 miles$

miles= 280

So the total distance traveled is 280 miles. Since speed is the relationship between the space traveled by an object and the time used for it ( $speed=\frac{distance}{time}$ ), then if the average of the entire trip was 35 mph and the distance traveled 280 miles, the time is calculated as:

$time=\frac{distance}{speed}=\frac{280 miles}{35 mph}$

time= 8 h

 The amount of time for the whole journey is 8 hours.

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The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is c

puteri [66]

Answer:

64.59kpa

Explanation:

See attachment

6 0

1 year ago

When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th

earnstyle [38]

Answer: The spring of the spring is 25 N/m.

Explanation:

Mass of the body = 25 g= 0.025 kg (1 kg = 1000 g)

Oscillation is 4 sec = 20

Oscillation in 1 sec = $\frac{20}{4}=5$

Frequency of the vibration of the spring = $5 s^{-1}=5 Hz$

Force constant can be calculated bu using the relation between the frequency and, mass and spring constant 'k'

$Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}$

$5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}$

$k=24.649 N/m\approx 25 N/m$

The spring of the spring is 25 N/m.

3 0

2 years ago

Read 2 more answers

Select the areas that would receive snowfall because of the lake effect.

Studentka2010 [4]

Answer:

- Grand Marais

- Two Harbors

- Duluth

Explanation:

The places that would would get snowfall because of the lake effect are Grand Marais, Two Harbors, and Duluth. The reason for this is that these three places are located right on the shores of the Lake Superior. This lake is one of the biggest lakes in the world. It has enormous amount of water in it, having big impact on the regional climate because of that. The water from this lake creates a lot of humidity in the air, and there's a lot of evaporation as well, both causing the formation of clouds, and when it is cold enough, instead of precipitation, the region gets large amounts of snowfall.

8 0

2 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

2 years ago

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

marissa [1.9K]

The second problem requires a figure to be answered. For the first problem
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s2
The reaction of the ramp is
F = 8 kg (5.63 m/s2)
F = 45 N
Differentiate the kinematic equation involving time to get the rate of increase of the velocity.

4 0

2 years ago