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kogti [31]
2 years ago
3

Suppose you defined ""positive"" motion as motion westward. What would be the persons total velocity?

Physics
1 answer:
Finger [1]2 years ago
4 0

Answer: I think that you are asking about an interactive reader which talks about an exercise which asked for the person's velocity who were on the bus.

The answer is 14 m/s

Explanation: When you have to do an exercise about velocity, you can do this exercise knowing the velocity from the bus in relative the street and you ALWAYS have to define your positive or negative motion about the direction in which the person were walking to know if you have to add or subtract

In this case, the bus were travelling east at 15 m/s and the person walked toward the back of the bus at 1 m/s.

It doesn't matter if the positive is the west or the east. While you only know the positive and knowing that the person was walking toward back the bus you can deduce this:

The person's total velocity is (+15 m/s) - (+1 m/s) = +14 m/s

You might be interested in
What is Otter's average velocity over his entire trip when it takes him 2 minutes to walk 100 meters north and another 1 minute
Leya [2.2K]

Answer:

0.50m/s

Explanation:

Average velocity is the change in displacement of a body with respect to time.

Velocity = ∆S/∆t

∆S = 100m - 70m

∆S = 30m

∆t = 2min - 1 min

∆t = 1min = 60secs

Substitute the given parameters into the formula for velocity

Velocity = 30m/60s

Velocity = 1/2 m/s

Average Velocity = 0.5m/s

7 0
2 years ago
Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.
Stels [109]
The hoop is attached.

Consider that the friction force is given by:
F = μ·N
   = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ

Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
   = </span>(1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
   = tan⁻¹(2·0.9)
   = 61°

Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>

5 0
2 years ago
If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?  (
valkas [14]
The gravitational force between two masses m₁ and m₂ is
F=G \frac{m_{1} m_{2}}{d^{2}}
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d =  distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
     = 0.0222
d = 0.1492 m = 149.2 mm

Answer: 149.2 mm
8 0
2 years ago
The movie "The Gods Must Be Crazy" begins with a pilot dropping a bottle out of an airplane. A surprised native below, who think
Sergio039 [100]

Answer:

⇔⇔⇔↑∑∑∩∅¬⊕║⊇↔∴∉∵

Explanation:

8 0
2 years ago
At room temperature, a typical person loses energy to the surroundings at the rate of 62 W. If this energy loss has to be made u
Alex_Xolod [135]

To solve this problem it is necessary to use the given proportions of power and energy, as well as the energy conversion factor in Jules to Calories.

The power is defined as the amount of energy lost per second and whose unit is Watt. Therefore the energy loss rate given in seconds was

P = \frac{E}{t} \rightarrow E= Energy, t = time

P = 62W = 62 \frac{J}{s}

The rate of energy loss per day would then be,

P = 62\frac{J}{s} (\frac{86400s}{1day})

P = 5356800 \frac{J}{day}

That is to say that Energy in Jules per lost day is 5356800J

By definition we know that 1KCal = 4.184*10^{6}J

In this way the energy in Cal is,

E = 5356800J \frac{1KCal}{4.184*10^{6}J}

E = 1279.694 KCal

The number of kilocalories (food calories) must be 1279.694 KCal

4 0
2 years ago
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