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Alexeev081 [22]
2 years ago
12

6 verify it in the laboratory. State Hooke's law. Describe how you can A force of 40 N stretches a wire through 30 cm. What forc

e will stretch it through 5.00 and through what length will a force of 100N stretch it? What assumption have you made? State Hooke's law​

Physics
1 answer:
4vir4ik [10]2 years ago
8 0
If you assume a) that the elastic limit wont be met, and b) that is a linear elastic material, then
f=kx
40=k.03 or k= 1333N/m

5cm?
force=k*.05=1333*.05=67N

100=kx
x= 100/1333 = 7.5 cm
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When a garden hose with an output diameter of 20 mm is directed straight upward, the stream of water rises to a height of 0.13m
SpyIntel [72]

Answer: h = 0.52m

Explanation:

Using the equation of out flow;

A1 × V1 = A2 ×V2

Where A1 = area of the first nozzle

A2 = area of the second nozzle

V1= velocity of flow out from the first nozzle

V2 = velocity of flow out from 2nd nozzle

But AV= area of nozzle × velocity of water = volume of water per second(m³/s).

Now we can set A×V = Area of nozzle × height of rise.

Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)

D1 = 20mm= 0.02m; h1 = 0.13m

D2 = 10mm = 0.01m; h2= ?

h2 = π(D1/2)²× h1 /π(D2/2)²

h2 = (0.02/2)² × 0.13/(0.01/2)²

= (0.01)² ×0.13 /(0.005)²

= 1.3 × 10^-5/(5 × 10^-3)²

= 1.3 × 10^-5/25 × 10^-6

= (1.3/25) 10^-5 × 10^6

= 0.052× 10

= 0.52m

7 0
3 years ago
A billiard ball traveling at 3.4 m/s has an elastic head-on collision with a billiard ball of equal mass that is initially at re
Tpy6a [65]

Answer:

Explanation:

Given

Initial velocity of first billiard ball u_1=3.4\ m/s

Initial velocity of second billiard ball u_2=0\ m/s

After collision first ball comes to rest

suppose m is the mass of both the balls

Conserving momentum to get the speed of second ball after collision

Initial momentum P_i=mu_1+mu_2=m\cdot 3.4+0

Final momentum P_f=mv_1+mv_2

where v_1 and v_2 are the speed of first and second ball respectively

P_i=P_f

m\cdot 3.4+0=0+m\cdot v_2

v_2=3.4\ m/s

thus speed of second ball after collision is equal to speed of first ball before collision

3 0
3 years ago
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How is photon energy calculated? (choose 2)
raketka [301]
Using Placks’s constant and frequency
5 0
3 years ago
Puck 1 is moving 10 m/s to the left and puck 2 is moving 8 m/s to the right. They have the same mass, m.
Julli [10]

Answer:

(a) the total momentum of the system before the collision = -2m kg.m/s.

(b) the total momentum of the system after the collision = -2m kg.m/s.

(c) puck 1's velocity after the collision in component form = (5.44 i, 2.54 j)

Explanation:

Given;

mass of Puck 1 , = m

mass of Puck 2, = m (since they have the same mass m)

initial velocity of Puck 1, u₁ = 10 m/s to the left

initial velocity of Puck 2, u₂  = 8 m/s to the right

Let the rightward direction be positive direction

Let the leftward direction be negative direction

(a) the total momentum of the system before the collision;

P₁ = (initial momentum of Pluck 1) + (initial momentum of Pluck 2)

P₁ = (-mu₁) + mu₂

P₁ = mu₂ - mu₁

P₁ = m(u₂ - u₁)

P₁  = m(8 - 10)

P₁  = -2m kg.m/s

(b) the total momentum of the system after the collision;

Based on the principle of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision.

Thus, the total momentum of the system after the collision is -2m kg.m/s.

(c) puck 1's velocity after the collision in component form

v = (v_x, v_y)\\\\v = (vcos \theta , vsin \theta)\\\\v = (6cos 25^0 , 6sin25^0)\\\\v = (5.44i, 2.54j)m/s

8 0
3 years ago
In line 19 inertness most nearly means
neonofarm [45]

If you provided the aforementioned line 19, that would kinda be great.

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3 years ago
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