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bogdanovich [222]
3 years ago
12

a uniform resistance wire is stretched till its length becomes 4 times what happens to the resistance​

Physics
1 answer:
Rzqust [24]3 years ago
3 0
The answer is in the box


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Examine the nuclear reaction:
Stels [109]

Answer:

A change has occurred in a nucleus

7 0
4 years ago
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What is the mathematical expression for the electric ptential difference between points A and B when a charged particle is movin
ratelena [41]

Answer:

chang in V=\int\limits^B_A {E} \, ds

Explanation:

Since, as we know, the potential difference 'ΔV' is the difference of between the Potential energy per unit charge U/qo at one point 'B' to Potential energy per unit charge at other point 'A'. It so happens when a test charge 'qo' moves from point A to B, the potential difference becomes the change of potential energy of the system, i.e.

Potential Difference =\frac{change in U}{qo}=\int\limits^B_A {E} \, ds

4 0
3 years ago
If R = 20 Ω, what is the equivalent resistance between points A and B in the figure?​
poizon [28]

Answer:

c. 70 Ω

Explanation:

The R and R resistors are in parallel.  The 2R and 2R resistors are in parallel.  The 4R and 4R resistors are in parallel.  Each parallel combination is in series with each other.  Therefore, the equivalent resistance is:

Req = 1/(1/R + 1/R) + 1/(1/2R + 1/2R) + 1/(1/4R + 1/4R)

Req = R/2 + 2R/2 + 4R/2

Req = 3.5R

Req = 70Ω

6 0
3 years ago
. A projectile is fired with an initial velocity of 113 m słatan angle of 60.0
bonufazy [111]

Answer:

a) 9.99 s

b) 538 m

c) 20.5 s

d) 1160 m

Explanation:

Given:

x₀ = 0 m

y₀ = 49.0 m

v₀ = 113 m/s

θ = 60.0°

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

a) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find t.

vᵧ = aᵧ t + v₀ᵧ

(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)

t ≈ 9.99 s

b) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find y.

vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)

(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)

y ≈ 538 m

c) When the projectile lands, y = 0 m.  Find t.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²

You'll need to solve using quadratic formula:

t ≈ -0.489, 20.5

Since negative time doesn't apply here, t ≈ 20.5 s.

d) When the projectile lands, y = 0 m.  Find x. (Use answer from part c).

x = x₀ + v₀ₓ t + ½ aₓ t²

x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²

x ≈ 1160 m

4 0
3 years ago
Find the moment of force of 60 Newton about an axis of rotation at distance 20 cm from the force​
777dan777 [17]

Answer:

10 N.m

Explanation:

Force * Perpendicular distance.

= 20 N * 0.5 m

= 10 N.m

8 0
3 years ago
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