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suter [353]
3 years ago
13

Help me please to do those problems...

Physics
1 answer:
Stells [14]3 years ago
3 0

Keeping in mind that:

- Distance is the length of the total path covered by the person, regardless of the directions of the different parts of motion

- Displacement is just the distance in a straight line between the final point and the initial point

Let's apply these concepts to solve the different parts of the problem:

1. 120 m, 80 m west

The total distance is the sum of the length of the different paths:

distance = 100 + 20 = 120 m

To find the displacement, we need to find the distance between the starting point and the ending point. Assuming the starting point as

x = 0

Michelle moved 100 m westward  and 20 m eastward, so the ending point is at

ending point = 100 - 20 = 80 m (westward)

So, the displacement is

displacement = 80 - 0 = 80 m (west)

2. 6400 m, 0

The distance is equal to the length of the track multiplied by the number of laps, so:

distance = 1600 \cdot 4 = 6400 m

The track is circular, and the runner completes exactly 4 laps: it means that at the end of the motion, the runner is at her starting point. So ending point and starting point coincide, and so the displacement

displacement = ending point - starting point = 0

3. 40 m; 10 m forward

The total distance is just the sum of the lengths of the different parts of the motion, so:

distance = 20 + 5 + 10 + 5 = 40 m

The find the displacement, we need to assign signs to every direction:

Forward --> positive along the forward-backward direction

Backard --> negative along the forward-backward direction

Right --> positive direction along the right-left direction

Left --> negative direction along the right-left direction

Along the forward-backward direction, the displacement is:

20 m forward and 10 m backward, so 20 - 10 = 10 (forward)

Along the left right direction, the displacement is:

5 m right and 5 m left, so 5 - 5 = 0

So the net displacement is 10 m (forward)

4. 16 km; 4 km south

Again, t total distance is the sum of the lengths of the different parts of the motion, so:

distance = 2 + 4 + 6 + 4 = 16 km

To find the displacement, we assign signs to every direction:

North --> positive along the north-south direction

south --> negative along the north-south direction

east --> positive direction along the east-west direction

west --> negative direction along the east-west direction

Along the north-south direction, the displacement is:

2 km north and 6 km south, so 2 - 6 = -4 km (4 km south)

Along the east-west direction, the displacement is:

4 km east and 4 km west, so 4 - 4 = 0

So the net displacement is 4 km south

5. 37.7 m; 24 m

The distance covered by the skater is the length of half circumference, so given the radius

r = 12 m

The distance is

distance= \pi r = \pi \cdot (12)=37.7 m

The displacement is the distance in a straight line between the ending position and the starting position. Since the skater ends her motion halfway around the circle, the distance between the initial and final point is equal to the diameter of the circle (two times the radius). So,

displacement = 2r = 2 \cdot 12 = 24 m

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3 years ago
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You need to repair a gate on the farm. The gate weighs 100 kg and pivots as indicated. A small diagonal bar supports the gate an
tekilochka [14]

Answer:

The force is  F = 3920 \ N

Explanation:

The diagram for this question is shown on the first uploaded image

   From the question we are told that

          The weight of the gate is G  =  100\  kg

 

The vertical component of F is  F_y =  F\ sin  \theta

   From the diagram , taking moment about the pivot we have  

                W_g  * 2 - F_y  * 1 =  0

Where W_g  is the weight of the gate evaluated as

             W_g  =  m_g * g =  100 * 9.8 =  980 \ N

=>        980 * 2 - Fsin(30)  * 1 =  0

=>         F = \frac{1960}{sin(30)}

=>      F = 3920 \ N

7 0
3 years ago
Imagine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and ra
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Here if we assume that there is no air friction on both balls then we can say

F = mg

now the acceleration is given as

F = ma = mg

a = g

so here both the balls will have same acceleration irrespective of size and mass

so we can say that to find out the time of fall of ball we can use

y = \frac{1}{2}gt^2

t = \sqrt{\frac{2y}{g}}

now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size

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7 0
3 years ago
Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa
tester [92]

Answer:

Part a)

t = 2.83 s

Part b)

Ball thrown downwards =v_f = 23.8 m/s

Ball thrown upwards =v_f = 23.8 m/s

Part c)

d = 22.24 m

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

\Delta y = 0 = v_y t + \frac{1}{2}at^2

0 = 13.9 t - \frac{1}{2}(9.81) t^2

t = 2.83 s

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

v_f^2 - v_i^2 = 2 a \Delta y

v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)

v_f^2 = 567.9

v_f = 23.8 m/s

Part c)

Relative speed of two balls is given as

v_{12} = v_1 - v_2

v_{12} = (13.9) - (-13.9) = 27.8 m/s

now the distance between two balls in 0.8 s is given as

d = v_{12} t

d = 27.8 \times 0.8

d = 22.24 m

7 0
3 years ago
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