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2 years ago

If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?

Physics

1 answer:

vodomira [7]2 years ago

5 0

Answer:

Option A is the correct answer.

Explanation:

Let the east point towards positive X-axis and north point towards positive Y-axis.

First walking 1.2 km north, displacement = 1.2 j km

Secondly 1.6 km east, displacement = 1.6 i km

Total displacement = (1.6 i + 1.2 j) km

Magnitude = $\sqrt{1.2^2+1.6^2} = 2 km$

Angle of resultant with positive X - axis = $tan^{-1}(1.2/1.6)=36.87^0$ = 36.87⁰ east of north.

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A force vector F1 points due east and has a magnitude of 200N. A second force F2 is added to F1. The resultant of the two vector

PilotLPTM [1.2K]

Answer:

The second vector $\vec{F_2}$ points due West with a magnitude of 600N

Explanation:

The original vector $\vec{F_1}$ points with a magnitude of 200N due east, the Resultant vector $\vec{R}$ points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of 400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:

$\vec{F_1}+\vec{F_2}=\vec{R}\implies\vec{F_2}=\vec{R}-\vec{F_1}=-400N-200N=-600N$

With the negative sign signifying that the vector points west.

3 0

3 years ago

Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from

kodGreya [7K]

Answer:

$ax = 2.60m/s^{2}$ , $t = 26.92s$

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

$(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x$ (1)

Where $(vx)f^{2}$ is the final velocity, $(vx)i^{2}$ is the initial velocity, $ax$ is the acceleration and $\Lambda x$ is the distance traveled.

Equation (1) can be rewritten in terms of ax:

$(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x$

$2ax \Lambda x = (vx)f^{2} - (vx)i^{2}$

$ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}$ (2)

Since the plane starts from rest, its initial velocity will be zero ( $(vx) = 0$ ):

Replacing the values given in equation 2, it is gotten:

$ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}$

$ax = \frac{4900m/s}{2(940m)}$

$ax = \frac{4900m/s}{1880m}$

$ax = 2.60m/s^{2}$

So, The acceleration of the plane is $2.60m/s^{2}$

Now that the acceleration is known, the next equation can be used to find out the time:

$(vx)f = (vx)i + axt$ (3)

Rewritten equation (3) in terms of t:

$t = \frac{(vx)f - (vx)i}{ax}$

$t = \frac{70m/s - 0m/s}{2.60m/s^{2}}$

$t = 26.92s$

Hence, the plane takes 26.92 seconds to reach its take-off speed.

5 0

2 years ago

A box exerts 10,000 Pa of pressure on the ground. If the box weighs 1000 N, how much area is in contact with the ground?

lara [203]

Pressure = (total force) / (Area)

10,000 Pa = (1,000 N) / (Area)

Multiply each side by (Area) :

(10,000 Pa) x (Area) = 1,000 N

Divide each side by (10,000 Pa) :

Area = (1,000 N) / (10,000 Pa)

Area = 0.1 m²

4 0

2 years ago

Danny measures the temperature of a 1.0-kg sample of sand and a 1.0-kg sample of soil. He then leaves both samples in the Sun fo

slavikrds [6]

Answer:

The sand absorbed more heat than the soil.

Explanation:

your welcome

5 0

2 years ago

Two lifeguards pull on ropes attached to a raft. If they pull in the same direction, the raft experiences a net external force o

makkiz [27]

$F_1=207.5\ N$

$F_2=92.5\ N$

Explanation:

Let $F_1\ and\ F_2$ is the two forces acting on the raft. If they pull in the same direction, the net force is given by :

$F_1+F_2=300$ ...............(1)

If they pull in opposite direction, net force is given by :

$F_1-F_2=115$ ...............(2)

It is required to find the magnitude of both forces. On solving equation (1) and (2) we get :

$F_1+F_2=300$

$F_1-F_2=115$

$F_1=207.5\ N$

$F_2=92.5\ N$

So, the forces are 207.5 N and 92.5 N. Hence, this is the required solution.

6 0

3 years ago