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Karo-lina-s [1.5K]
3 years ago
6

A dog walks 26m, East, 29m, West, and 7m, East. What is his displacement?

Physics
1 answer:
sergeinik [125]3 years ago
7 0

Answer:

4m East

Explanation:

26-29= -3 (so that means 3m west)

-3+7= 4 (so 4m East)

hope it helps

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Allison wants to determine the density of a bouncing ball. which metric measurements must she use?
lubasha [3.4K]
Density depends on mass and volume so option D is correct answer. Hope this helps!
3 0
3 years ago
a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f
harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for \DeltaDAC:

tan\theta = \frac{d}{x}

⇒ \theta = tan^{-1} \frac{d}{x}

Now for the \DeltaBAC:

tan\theta = \frac{d + h}{x}

⇒ \theta = tan^{-1} \frac{d + h}{x}

Now, differentiating w.r.t x:

\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}]

For maximum angle, \frac{d\theta }{dx} = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} -  tan^{-1} \frac{d}{x}][/tex]

0 = \frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}

\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}

After solving the above eqn, we get

x = \sqrt{\frac{d}{d + h}}

The observer should stand at a distance equal to x = \sqrt{\frac{d}{d + h}}

4 0
3 years ago
One gallon of paint (volume = 3.79 X 10-???? m3) covers an area of 25.0 m2. What i the thickne s of the fresh paint on the wall?
Drupady [299]

Answer:

Explanation:

Given

Volume of paint is V=3.79\times 10^{-3}\ m^3

Area of cover A=25\ m^2

Suppose paint to be a rectangular box with thickness t and volume V

therefore we can write as

V=A\times t

t=\frac{V}{A}

t=\frac{3.79\times 10^{-3}}{25}

t=1.516\times 10^{-3}\ m

t=1.516\ mm  

6 0
3 years ago
Which is capeble of housing astronaughts while they conduct reasearch
dusya [7]

Answer:

2

Explanation:

2

5 0
2 years ago
Read 2 more answers
A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal
Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction  

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

v_x = v_i cos \theta

where

v_i = 40 m/s is the initial velocity

\theta=20^{\circ} is the angle of projection

Substituting,

v_x = (40)(cos 20^{\circ})=37.6 m/s

And therefore, the range of the projectile is:

d=v_x t = (37.6)(5.0)=188 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0
3 years ago
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