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3 years ago

An object with a height of 1.0 cm is placed 1.4 cm from a convex lens that has a

Physics

1 answer:

Gennadij [26K]3 years ago

5 0

Answer:

A concave mirror has a focal length of. 10.0 cm. What is its radius of curvature? ... 20.0 cm. 62. An object located 18 cm from a convex mirror produces a virtual image 9 ... cm. 75 cm. 66. Find the image position and height for the object shown in ... 1 block 1.0 cm. Vertical scale: 2 blocks 1.0 cm. F. I1 hi. 1.0 cm di. 2.7 cm. O1.

Explanation:

Hope This Helps

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

$f=\frac{33.84}{0.935}$

$f=36.19\approx 36\ \textup{Hz}$

5 0

2 years ago

A piano emits frequencies the range from a low of about 28 Hz to a high of about 4200 Hz. Find the range of wavelengths in air a

Step2247 [10]

V=wave velocity , f= frequency, λ=wavelength 
Use it to find corresponding wavelengths for f=28 Hz 
λ= v/f= 337/28=12.036 m

for f=4200 Hz 
λ= v/f=337/4200= 0.08 m 
So max. wavelength is 12.036 m and 
Min Wavelength is 0.08 m 
So the range is between .08 m and 12.036 m
Hope this helps.

4 0

3 years ago

What was the effect of the Supreme Court case McCulloch v. Maryland?

AURORKA [14]

The effect was the decision that gave congress power under the necessary and proper clause act. States could also not impede on the valid constitutional excerpts powered by the Federal Government.

8 0

3 years ago

Read 2 more answers

A horizontal force of magnitude 46.3 n pushes a block of mass 4.14 kg across a floor where the coefficient of kinetic friction i

IrinaVladis [17]

A) Calling F the intensity of the horizontal force and d the displacement of the block across the floor, the work done by the horizontal force is equal to
$W=Fd = (46.3 N)(4.25 m)=196.8 J$

b) The work done by the frictional force against the motion of the block is equal to:
$W_f = -F_f d =- (\mu mg) d =-(0.609)(4.14 kg)(9.81 m/s^2)(4.25 m)=$
$=-105.1 J$
Part of these 105.1 Joules of work becomes increase of thermal energy of the block ( $\Delta E_B$ ), and part of it becomes increase of thermal energy of the floor ( $\Delta E_F$ ). We already know the increase in thermal energy of the block (38.2 J), so we can find the increase in thermal energy of the floor:
$\Delta E_F = 105.1 J - \Delta E_B = 105.1 J-38.2 J=66.9 J$

c) The net work done on the block is the work done by the horizontal force F minus the work done by the frictional force (the frictional force acts against the motion, so we must take it with a negative sign):
$W_{net}=W-W_f=196.8 J-105.1 J=91.7 J$
For the work-energy theorem, the work done on the block is equal to its increase of kinetic energy:
$W_{net} = \Delta K$
So, we have $\Delta K=+91.7 J$

5 0

3 years ago