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Thepotemich [5.8K]

3 years ago

12

ANYONE help me with this

1 answer:

ycow [4]3 years ago

8 0

Displacement is simply the change in position, or the difference in the final and initial positions:

$\Delta d = d_f - d_i$

Then

(a) ∆<em>d</em> = 5 m - 0 m = 5 m

(b) ∆<em>d</em> = 1 m - (-2 m) = 1 m + 2 m = 3 m

(c) ∆<em>d</em> = 2 m - (-2 m) = 2 m + 2 m = 4 m

(d) ∆<em>d</em> = 6 m - 2 m = 4 m

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Look at your data. With three foods, Flock v

wolverine [178]

Answer:the first one was x

the second one is y

Explanation:

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3 years ago

A horse runs in a straight line, moving 38 m away in 9.0 s. It then turns back and runs halfway back in 1.8 s. Find its average

podryga [215]

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7 0

3 years ago

Can we call centre of gravity centre of weight?

algol [13]

Answer:

Centre of gravity is a theoretical point in the body where the total weight of the body is thought to be concentrated. In a uniform gravitational field, the centre of gravity is identical to the centre of mass. Yet, the two points do not always coincide.

3 0

3 years ago

A motorboat travels 92 km in 2 hours going upstream. It travels 132 km going downstream in the same amount of time. What is the

taurus [48]

Answer:

The speed on boat in still water is $56 \frac{km}{h}$ and the rate of the current is $10 \frac{km}{h}$

Explanation:

Since speed , $v= \frac{Distance\, traveled(D)}{Time\, taken(t)}$

Therefore speed of motor boat while traveling upstream is

$v_{upstream}=\frac{92}{2}\frac{km}{h}=46\frac{km}{h}$

and speed of motor boat while traveling downstream is

$v_{downstream}=\frac{132}{2}\frac{km}{h}=66\frac{km}{h}$

Let speed of boat in still water be $v_b$ and rate of current be $v_w$

Therefore $v_{upstream}=v_b-v_w=46\frac{km}{h}$ ----(A)

and $v_{downstream}=v_b+v_w=66\frac{km}{h}$ ------(B)

Adding equation (A) and (B) we get

$2v_b= (46+66) \frac{km}{h}=112 \frac{km}{h}$

=> $v_b= 56 \frac{km}{h}$ ------(C)

Substituting the value of $v_b$ in equation (A) we get

$v_w= 10 \frac{km}{h}$

Thus the speed on boat in still water is $56 \frac{km}{h}$ and the rate of the current is $10 \frac{km}{h}$

5 0

4 years ago

PLEASE HELP A student climbs to the top of a press box where the cameras are. He wonders how many meters he is off of the ground

Feliz [49]

The height of the rail on top of the press box where the ball was dropped from is 11.025 m.

The given parameters:

Time of motion of the ball, t = 1.5 s
Let the height of the rail = h

<h3>Maximum height of fall;</h3>

The maximum height through which the ball was dropped from is calculated by applying second equation of motion;

$h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = \frac{1}{2} (9.8) (1.5)^2\\\\h = 11.025 \ m$

Thus, the height of the rail on top of the press box where the ball was dropped from is 11.025 m.

Learn more about height of projectiles here: brainly.com/question/10008919

3 0

2 years ago