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lapo4ka [179]
3 years ago
12

A 44.0 kg uniform rod 4.90 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a w

ire attached to its other end. The wire makes an angle of 30.0° with the horizontal, and is bolted to the wall directly above the hinge. If the wire can withstand a maximum tension of 1450N before breaking, how far from the wall can a 69.0kg person sit without breaking the wire?
Physics
1 answer:
pychu [463]3 years ago
4 0

Answer:

x ≤ 3.6913 m

Explanation:

Given

Mrod = 44.0 kg

L = 4.90 m

Tmax = 1450 N

Mman = 69 kg

A: left end of the rod

B: right end of the rod

x = distance from the left end to the man

If we take torques around the left end as follows

∑τ = 0   ⇒   - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0

⇒   - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0

⇒  -  (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0

⇒ x ≤ 3.6913 m

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Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

<u>h' = 603.08 m</u>

4 0
3 years ago
Which of the following is not a way to make exercising on a cold day safer?
IRISSAK [1]

Answer:

D.

Explanation:

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3 years ago
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Note that the scale bar under the photo is labeled 1 μm (micrometer). The scale bar works the same way as a scale on a map, wher
AnnZ [28]

Answer:

Hello your question has some missing parts attached below is the complete question

answer : 4 μm

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musickatia [10]
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3 years ago
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An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much
cupoosta [38]

Answer:

time will elapse before it return to  its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × 10^{7} m/s

uniform electric field E = 1.18 × 10^{4} N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × 10^{-31} kg ×a = 1.18 × 10^{4} × 1.602 × 10^{-19}

a = 20.75 × 10^{14} m/s²

so acceleration is 20.75 × 10^{14} m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × 10^{7} = 0 + 20.75 × 10^{14} (t)

t = 11.80 × 10^{-9} s

so time will elapse before it return to  its staring point is

time = 2t

time = 2 ×11.80 × 10^{-9}

time is 23.6 × 10^{-9} s

time will elapse before it return to  its staring point is 23.6 ns

7 0
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