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maks197457 [2]
3 years ago
8

1 . A farmer moves along the boundary of his rectangular field of side 10m x 20m . farmer covers 1m in 1sec . what will be the m

agnitude of displacement at the total distance travelled.
2 . Rahul pedals his cycle from rest. Pedals his cycle to get a
velocity of 10 ms-1  in 30 seconds.
then he applies brakes in such that the velocity of the cycle comes does to 6
ms-1  in next 5 seconds . calculate
the acceleration and deceleration .
Physics
1 answer:
SSSSS [86.1K]3 years ago
8 0
1). He's walking the 'perimeter' of his field, meaning 'the distance all the way around'.
If he stays right along the fence all the way, then the total distance he walks is
(10 + 20 + 10 + 20) = 60 meters. (It's a small field ... about 5% of an acre.)
The magnitude of displacement is the distance between the start point and end point,
regardless of the route.  After he finishes one whole trip around, his displacement is
zero, because he ends up at the same place where he started.
Notice the big fat red herring in the question: None of this depends on the speed
at which he walks.

2).  Acceleration = (change in speed) / (time for the change)

While speeding up, acceleration = (10 - 0) / 30 = 1/3 meter per second².

While slowing down, deceleration = (6 - 10) / 5 = -4/5 = -0.8 meter per second²

You might be interested in
The instantaneous velocity of an object is the blank of the object with a blank
miss Akunina [59]
Instantaneous velocity is the velocity at a specific instant in time. I bet you are taking Honors Physics. 
3 0
3 years ago
A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear
Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

f_o = f_s\frac {(v+vs)}{v} and f_o = f_s\frac {(v-vs)}{v} when moving towards observer

With f_o of 76 then taking speed in air as 343 m/s we have

76 = f_s\times\frac {(343-vs)}{343}

f_s=\frac {343\times 76}{343-v_s}

Similarly, with f_o of 65 we have

65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}

Now

f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}

v_s=27.76 m/s

Substituting the above into  any of the first two equations then we obtain

f_s=70.07 Hz

4 0
3 years ago
A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
3 years ago
What is the relationship between Newton's first law of motion and inertia?​
adelina 88 [10]

Explanation: Newton's first law of motion states that a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force. This is also known as the law of inertia. Inertia is the tendency of an object to remain at rest or remain in motion.

4 0
3 years ago
You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The
LuckyWell [14K]

Answer: 4.65\ m/s

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

v^2-u^2=2as

\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s

So, the velocity of putty just before hitting is 4.65\ m/s

5 0
3 years ago
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