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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

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A block pushed along the floor with velocity V0 slides a distance d after the pushing force is removed. a) if the mass of the bl
KengaRu [80]

Answer:

a) \ d_2=d_1\\b) \ d_2=4d_1

Explanation:

Assume that the distance travelled initially is d.

In order to stop the block you need some external force which is friction.

If we use the law of energy conservation:

E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}

a)

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.

6 0
3 years ago
How do scientist study the earth through plate tectonic plates and boundaries/
Serggg [28]

Answer:

By plotting the locations of earthquakes

Explanation:

When plotting the locations of earthquakes, scientists have been able to locate plate boundaries and also be able to determine plate characteristics and predict the movement of plates

5 0
3 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
3 years ago
A force has both______ and _______. Question 1 options: Speed and impact Magnitude and direction Motion and speed Direction and
Travka [436]

Answer:

magnitude

and direction

Explanation:

5 0
2 years ago
A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.​
disa [49]

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

6 0
2 years ago
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