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aalyn [17]
2 years ago
8

At 20∘C, the hole in an aluminum ring is 2.800 cm in diameter. You need to slip this ring over a steel shaft that has a room-tem

perature diameter of 2.804 cm . 1. To what common temperature should the ring and the shaft be heated so that the ring will just fit onto the shaft? Coefficients of linear thermal expansion of steel and aluminum are 12×10−6 K−1 and23×10−6 K−1 respectively.
Physics
1 answer:
PtichkaEL [24]2 years ago
8 0

Answer:

Explanation:

 To measure increase in length due to thermal expansion,  the formula is

L₂ =L₁ ( 1 + α t ) where L₁ length increases to L₁ due to rise in temperature by t.

L₂ in both the case is same so

for aluminium

L₂ = 2.8 ( 1 + 23 X 10⁻⁶ t )

for steel

L₂ = 2.804 ( 1 + 12 X 10⁻⁶ t )

2.8 ( 1 + .000023t) = 2.804 ( 1 + .000012 t )

Solving this equation

t = 122C

Final temperature of both = 122 +20 = 142 C

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A +5.0-μC point charge is placed at the 0 cm mark of a meter stick and a -4.0-μC charge is placed at the 50 cm mark. What is the
algol13

Answer:

1.4 *10^6 N/C

Explanation:

The electric field caused by a charge at a certain point is given by the equation:

E = k \frac{q}{r} \^r

where k is the Coulomb constant equal to 8.99 *10^9 Nm^2/C^2, q the charge of the particle in coulombs, r is the distance from the point to the charge in meters.

\^r is the unitary vector that goes from the charge to the point. This vector will give us the direction of the Electric Field vector.

The unitary vector of the +5.0-μC charge will go to the right (+i), as the point is to the right of the charge. Then, the electric field caused by the charge will be:

E_1 = k \frac{q}{r^2} \^r = 8.99*10^9 Nm^2/C^2 \frac{5.0 *10^{-6}C}{(0.3m - 0m)^2}(+\^i) =  +0.5*10^6 N/C

The unitary vector of the -4.0-μC charge will go to the left (-i), as the point is to the left of the charge. Then, the electric field caused by the charge will be:

E_2 = k \frac{q}{r^2} \^r = 8.99*10^9 Nm^2/C^2 \frac{-4.0 *10^{-6}C}{(0.5m - 0.3m)^2}(-\^i) =  +0.9*10^6 N/C

The electric field at the 30 cm mark will be the addition of both electric field:

E_{total} = E_1 +E_2 = 0.5 *10^6 N/C + 0.9*10^6 N/C = 1.4 *10^6 N/C

3 0
2 years ago
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 4.15 m/s . Her husband Bruce suddenly reali
aleksandr82 [10.1K]

Answer:

a 15.22 m/s

b 45.65 m

Explanation:

Using the same formula,

x = vt, where

x is now 45.65, and

t is 3 s, then

45.65 = 3v

v = 45.65/3

v = 15.22 m/s

See the attachment for the part b. We used the distance gotten in part B, to find question A

5 0
2 years ago
A puck of mass m = 0.085 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless str
Rina8888 [55]

Answer:

T = 11.93 N

Explanation:

Newton's second law to the puck in the circular path

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in  radial direccion (N)

m : puck mass  (kg)

a : radial acceleration of the puck (m/s²)

Data:

m = 0.085 kg

L = 0.72 m = R : radium of the circular path (m)

θ=  one revolution = 2Π rad

t= 0.45 s

Angular speed of the puck

ω = θ/t

ω = 1 rev/0.45 s = (2π/0.45) rad/s

ω = 13.96 rad /s

Radial acceleration or centripetal

a = ω²*R

a = (13.96) ²* (0.72)

a = 140.3 m/s²

Magnitude of the tension in the string (T)

We apply the Formula (1)

∑F = m*a

T =  (0.085 kg )*  (140.3 m/s² )

T = 11.93 N

4 0
2 years ago
A thin copper rod 1.0 m long has a mass of 0.050 kg and is in a magnetic field of 0.10 t. What minimum current in the rod is nee
slamgirl [31]

Answer:

i = 4.9 A

Explanation:

Force on a current carrying rod due to magnetic field is given as

F = iLB

here we know that

i =current in the rod

B = 0.10 T

L = 1.0 m

now magnetic force is balanced by the weight of the rod

so we will have

iLB = mg

i(1.0)(0.10) = 0.05 \times 9.8

i = 4.9 A

8 0
2 years ago
You throw a baseball straight up into the air with a speed of 24.5 m/s. How long does it take the baseball to reach its highest
galben [10]
Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
Speed=24.5 m/s
Time=24.5/9.8=2.5 s
3 0
2 years ago
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