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Montano1993 [528]
2 years ago
13

A jogger runs 10.0 blocks do east, 5.0 blocks due South, and another two. Zero blocks do east. Assume all blocks are equal size,

what is the magnitude of the joggers net displacement?
Physics
1 answer:
Annette [7]2 years ago
6 0

Jogger moves in three displacements

d1 = 10 blocks East

d2 = 5 blocks South

d3 = 2 blocks East

now we can say

total displacement towards East direction will be

d_x = 10 + 2= 12 blocks

Total displacement towards South

d_y = 5 block

now to find the net displacement we can use vector addition

d = \sqrt{d_x^2 + d_y^2}

d = \sqrt{12^2 + 5^2}

d = 13 blocks

<em>so magnitude of net displacement will be equal to 13 blocks</em>

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A boy stands at a certain distance from a large building and blows a whistle. After 2.3s he hears the echo of the sound. He move
ddd [48]

Answer:

I) 57.5 m

Il) 50 m/s

Explanation:

Given that a boy stands at a certain distance from a large building and blows a whistle.

After 2.3s he hears the echo of the sound. The speed V of the sound will be:

V = 2X / T

Where X is the distance covered

V = 2X / 2.3

V = 0.87X ...... (1)

He moves 50m towards the building and blows his whistle again; this time the echo reaches him after 2.0s.

V = (2×50) / 2

V = 50 m/s

Substitutes the V into equation 1

50 = 0.87X

X = 50 / 0.87

57.5 m

Therefore, the Boys original distance from the building is 57.5m and the

Speed of sound in air is 50m/s

6 0
2 years ago
As new-forming stars grow by gravitationally attracting more material, they become brighter. If a star becomes too bright, the r
ratelena [41]

Answer:

Check the explanation

Explanation:

To tackle situations like the one above, the rate of gravity of that star must be equal to the rate of power output but we don’t have radius of that star. Also temp is not mentioned. And emissivity of star is also not mentioned. So the only possible way is like Einstein mass energy relationship E=mc^2=6.5380e40

power =E/Time so this energy is transferred per sec.

4 0
2 years ago
Read 2 more answers
A spherically symmetric charge distribution has a charge density given by ρ = a/r , where a is constant. Find the electric field
Airida [17]
Charge dQ on a shell thickness dr is given by 

dQ = (charge density) × (surface area) × dr 

dQ = ρ(r)4πr²dr 

∫ dQ = ∫ (a/r)4πr²dr 

∫ dQ = 4πa ∫ rdr 

Q(r) = 2πar² - 2πa0² 

Q = 2πar² (= total charge bound by a spherical surface of radius r) 

Gauss's Law states: 

(Flux out of surface) = (charge bound by surface)/ε۪ 

(Surface area of sphere) × E = Q/ε۪ 

4πr²E = 2πar²/ε۪ 

<span>E = a/2ε۪


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>
3 0
2 years ago
Read 2 more answers
Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is 7780 N, the
viva [34]

Answer:

V=19.08 m/s

Explanation:

Airplane gross weight w=7780 N

Airplane wing area S=16.6 m²

Air density p=1.2250 kg/m³

Maximum lift coefficient CL=2.1

To find

Stalling Speed

Solution

The equation to find stalling speed is given below

W=(1/2)S_{area}(V_{Stalling-speed} )^{2} (P_{Air-density} )(C_{Lmax} )\\ so\\V_{Stalling-speed}=\sqrt{\frac{2W}{S_{area}*(P_{Air-density} )(C_{Lmax} )} }\\V_{Stalling-speed}=\sqrt{\frac{2*7780}{16.6*2.1*1.225} }\\  V_{Stalling-speed}=19.08 m/s

5 0
2 years ago
Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2
dybincka [34]

Complete question is;

Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.

Block 1 is removed without disturbing block 2.

Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is mass of block 3

Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a

Making T the subject gives us;

T = (m2)g - (m2)a - - - (eq 2)

where;

g is acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is mass of block 2

To solve for the acceleration, we will just substitute (m3)a for T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0
2 years ago
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