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3 years ago

What is a volume of a liquid in a 50ml cup

1 answer:

4 0

Answer:The scale of a 50-mL buret is divided into 0.1 mL increments. Therefore, when the liquid level in a buret is read, it is read and recorded to the nearest 0.01 mL.

Explanation:

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When operated on a household 110.0-V line, typical hair dryers draw about 1650 W of power. We can model the current as a long st

Dimas [21]

Answer:

Current = 15 A

Resistance = 7.33 ohm

Magnetic field = 1.62 x 10^-4 Tesla

Explanation:

V = 110 V, P = 1650 W, r = 1.85 cm,

(a) Let i be the current

P = V x i

i = 1650 / 110 = 15 A

(b) Let R be the resistance

V = i R

R = 110 / 15 = 7.33 Ohm

B = μ0 / 4π x 2i / r

B = 10^-7 x 2 x 15 / 0.0185 = 1.62 x 10^-4 Tesla

5 0

3 years ago

Please help asap!!!!

Lady bird [3.3K]

Answer:

4 is the answer

Explanation:

hope this helps

7 0

3 years ago

n electric motor rotating a workshop grinding wheel at 1.10 102 rev/min is switched off. Assume the wheel has a constant negativ

Schach [20]

Answer:

Time taken to stop = 6 sec

Explanation:

Given:

Rotation of wheel = 1.10 × 10² rev/min

Final velocity (v) = 0

Angular acceleration (a) = - 1.92 rad/s²

Find:

Time taken to stop

Computation:

Initial velocity (u) = (1.10 × 10² × 2π rad) / 60 sec

Initial velocity (u) = 11.52 rad / sec

We know that,

V = U +at

t = (v-u)a

t = (0 - 11.52) / (-1.92)

Time taken to stop = 6 sec

3 0

3 years ago

A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance betw

mario62 [17]

Answer:

$0.2$ mm

Explanation:

As we know that

$Y = \frac{m\lambda * D}{d}$

where

m represents the order of minimum

y represents the distance on the screen of the minimum from central axis

λ is the wavelength of the light

D is the distance between screen-to-slit

d represents the width of the slit

For first minima

$y_1 = \frac{1 * 633 * 10^{-9}*1.5}{d}$

For second minima

$y_2 = \frac{2 * 633 * 10^{-9}*1.5}{d}$

$Y_2 - Y_1 = 0.00475m\\\frac{633 * 10^{-9} * 1.5 }{d} = 0.00475\\d = 0.0002 m\\d = 0.2 mm$

3 0

3 years ago

Two wheels with fixed hubs, each having a mass of 1kg , start from rest, and forces are applied: F1= 1 N to the first wheel and

irina [24]

Answer:

Explanation:

$m = m_1 = m_2 = 1 kg$

The torques generated by tangential forces on the 2 wheels are

$T_1 = F_1R_1 = 1*0.5 = 0.5 Nm$

$T_2 = F_2R_2 = F_2*1 = F_2 Nm$

According to Newton's 2nd law, the angular accelerations generated by these torque would be

$\alpha_1 = \frac{T_1}{I_1} = \frac{0.5}{mR_1^2} = \frac{0.5}{1*0.5^2} = 2 rad/s^2$

$\alpha_2 = \frac{T_2}{I_2} = \frac{F_2}{mR_2^2} = \frac{F_2}{1*1^2} = F_2 rad/s^2$

For the 2nd wheel to have the same angular acceleration, its force must be

$\alpha_1 = \alpha_2$

$2 = F_2$

$F_2 = 2N$

6 0

3 years ago