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Lilit [14]
2 years ago
10

A cricket starts at the 5 meter mark on a ruler. It jumps to the 10 meter mark. Then, it jumps back to the 6 meter mark. Calcula

te the cricket’s displacement.
Physics
1 answer:
belka [17]2 years ago
4 0

We have that the cricket’s displacement is  

D_t=9meters

From the Question we are told that

Start position  5 meter mark

It jumps to the 10 meter mark.

it jumps back to the 6 meter mark.

Generally the equation for cricket’s displacement. is mathematically given as

D_t=Initial +final(displacement)\\\\D_t=(10-5)+(10-6)

D_t=9meters

For more information on this visit

brainly.com/question/24273210?referrer=searchResults

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If we could stop our eyes from quivering as we stared at a stationary object, the object would probably
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The correct answer is “vanish from sight.” I hope this helps you. Please mark as Brainiest
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How fast does sound travel? A. 1,115 feet per second B. 1,000 feet per second C. 2,000 feet per second D. 2,115 feet per second.
lys-0071 [83]

Answer:

A

Explanation:

6 0
3 years ago
I have four questions.
Inga [223]

Answer:

The potential energy of the ball is 784 joules. And the kinetic energy of it is 392 while falling halfway down.

Explanation:

PE = mass (2kg) * Gravitational acceleration (9.8 m/s^2)* height (40 meters)

KE = 1/2 mass (1 kg) * velocity^2 (19.8)

7 0
3 years ago
Read 2 more answers
A small plane flies 40.0 km in a direction 60° north of east and then flies 30.0 km in a direction 15° north of east. Use the an
lana66690 [7]

Answer:

d= 64.7 km

\theta = 40.9^{o}

displacement vector=r_xi + r_yj =  48.9i + 42.4j

Explanation:

total distance = 40 + 30 = 70 km

during 1st flight

r_1 x = 40*cos60

r_1 x = 20 km

r_1 y = 40*sin60

r_1 y = 34.64 km

during 2nd flight

r_2 x = 30*cos15

r_2 x = 28.9 km

r_2 y = 30*sin15

r_2 y = 7.76 km

the two component of r are:

r_x = r_1x + r_2x = 20 + 28.9 = 48.9 km

r_y = r_1y + r_2y = 34.64 + 7.76 = 42.4 km

Geographical Direction \theta = tan^{-1}\frac{r_y}{r_x} [tex]\theta = 40.9^{o}

Displacement d= \sqrt{r_x^2 + r_y^2}

                     d = \sqrt{48.9^2+42.4^2} = 64.7 km

d= 64.7 km

displacement vector=r_xi + r_yj =  48.9i + 42.4j

8 0
3 years ago
2. A Plate 0.02 mm distance from a fixed Plate moves at a velocity Of 0.6mls and requires a force of 1.962 N Per unit area to ma
Marizza181 [45]

Answer:

6.54 × 10⁻⁵ Pa-s

Explanation:

Since the shear force, F = μAu/y where μ = viscosity of fluid between plates, A = area of plates, u = velocity of fluid = 0.6 m/s and y = separation of plates = 0.02 mm = 2 × 10⁻⁵ m

Since F = μAu/y

F/A = μu/y where F/A = force per unit area

Since we are given force per unit area, F/A = 1.962 N per unit area = 1.962 N/m²

So,  μ = F/A ÷ u/y

substituting the values of the variables into the equation, we have

μ = F/A ÷ u/y

μ = 1.962 N/m² ÷ 0.6 m/s/2 × 10⁻⁵ m

μ = 1.962 N/m² ÷ 0.3 × 10⁵ /s

μ = 6.54 × 10⁻⁵ Ns/m²

μ = 6.54 × 10⁻⁵ Pa-s

5 0
2 years ago
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