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denis23 [38]
3 years ago
11

A hungry rat is placed in a maze. It walks the following path to find a piece of cheese. 4.0m N, 7.5 m E, 6.8 m S, 3.7 m E, 3.6

m S, 5.3 m W, 3.7 m N, 5.6 m W, 4.4 m S, 4.9 m W. Calculate the distance the rat walked. Find the rat’s displacement and direction.
Physics
2 answers:
storchak [24]3 years ago
6 0

Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

6.8 m S = - 6.8 j

3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

5.3 m W = - 5.3 i

3.7 m N, = 3.7 j

5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = -  4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

= -4.6 i -7.1 j

magnitude of displacement = \sqrt{(4.6^2+7.1^2)}

= 8.46 m

Direction

Tanθ = 7.1/ 4.6

θ = 57⁰ south of west .

distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9

= 49.5 m

ankoles [38]3 years ago
6 0

Answer:

equal to

Explanation:

i got it right hope it helped :)

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In a fission experiment observed by Hahn and Strassman, uranium-235 was bombarded by neutrons. The products of this ission react
wlad13 [49]

Answer:

helium-4 (90%) or tritium (7%).

Explanation:

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7 0
3 years ago
There is a species of bamboo that can grow 36 inches per day. If a plant grew at this rate and was measured at 40 inches initial
LekaFEV [45]

Answer:

It will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

Explanation:

From the question, the rate at which the species of the bamboo tree grows is 36 inches per day.

To determine how long it would take a plant 40 inches tall initially to grow at this rate (that is, 36 inches per day) to a height of 200 inches.

This means we will calculate the number of days it will take the plant to grow additional 160 inches ( 200 inches - 40 inches) at this rate.

Now,

If the plant grows 36 inches in 1 day

then it will grow 160 inches in x days

x = (160 inches × 1 day) / 36 inches

x = 160 / 36

x = 4\frac{4}{9} days or 4.44 days

Hence, it will take the plant 4\frac{4}{9} days or 4.44 days to grow to a height of 200 inches tall.

8 0
3 years ago
An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi
hoa [83]

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

4 0
2 years ago
Which element can form straight chains, branched chains, and rings
Mice21 [21]
Carbon atoms can form straight, and branched chains, and rings
6 0
3 years ago
Read 2 more answers
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

3 0
2 years ago
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