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2 years ago

Help me please (￣(ｴ)￣)

Physics

2 answers:

mihalych1998 [28]2 years ago

6 0

Answer:

<h3>Elastic Forces </h3><h2><em>Are</em><em> </em><em>not</em><em> </em><em>always</em><em> </em><em>conservative</em><em> </em></h2>

Explanation:

Elastic force is conservative only when Hooke's law is obeyed. After elastic limit Elastic force is not conservative as it dcesn't regain its original shape.

Len [333]2 years ago

3 0

Answer:

1) Are always conservative

Explanation:

Elastic forces are always conservative.

Hope it helps you.

please mark as the brainliest answer.

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A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.

olasank [31]

Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

T= r x F

T= r F sin∅

F= T/ (r * sin∅)

F= 5.2/ (0.91 * 1)

F= 5.71 N

5 0

3 years ago

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

kumpel [21]

The Answer is 11,121 Kg

6 0

3 years ago

Read 2 more answers

An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof

Blizzard [7]

240 meters that's it

7 0

3 years ago

Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various

Alexxx [7]

Answer:

$Q_{2}=1200cm^{3}/s$

Explanation:

Given data

Q₁=200cm³/s

We know that:

$F=n\frac{vA}{l}\\$

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

n₂=6.0n₁

Q=ΔP/R

$Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s$

3 0

2 years ago

Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has

a_sh-v [17]

Answer:

h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

Em₀ = U = m g h

final point. Lower, just before the crash

Em_f = K = ½ m $v_{b}^2$

energy is conserved

Em₀ = Em_f

m g h = ½ m v²

v_b = $\sqrt{2gh}$

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

p₀ = 2m 0 + m v_b

final instant. Right after the crash

p_f = (2m + m) v

the moment is preserved

p₀ = p_f

m v_b = 3m v

v = v_b / 3

v = ⅓ $\sqrt{2gh}$

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

Em₀ = K = ½ 3m v²

Final point. Higher

Em_f = U = (3m) g h'

Em₀ = Em_f

½ 3m v² = 3m g h’

we substitute

h’= $\frac{v^2}{2g}$

h’ = $\frac{1}{3^2} \ \frac{ 2gh}{2g}$

h’ = 1/9 h

6 0

2 years ago

Help me please (*￣(ｴ)￣*)​

Help me please (￣(ｴ)￣)