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3 years ago

7

How many like does it take to get to the center of a tosipop

2 answers:

dmitriy555 [2]3 years ago

4 0

252-364 licks

I'm not sure if it's correct or not

Alja [10]3 years ago

4 0

364 licks to get to the center

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A car cruises at a constant rate of 50 miles per hour. How much time will it take to go 600 miles?

Len [333]

Speed is constant. 50 miles = 1 hour. 600/50 = 12. 1hr(12) = 12 hours.

7 0

3 years ago

on the surface of theearth ,the weight of a boy is 400N but on a mountainpeak his weight is 360N,Calculatethe value of ''g' on t

andreyandreev [35.5K]

The value of g at sea level is 9.81 ms^-2.

The boy's mass is constant wherever he is in the universe but his weight will depend on the strength gravity where he is.

By proportion its value on the mountain peak is (360 /400) * 9.81

= 0.9 * 9.81 = 8.83 ms^-2 to nearest hundredth, (answer).

7 0

3 years ago

The magnetic field at the center of a 1.40-cm-diameter loop is 2.50 mT . PART A) What is the current in the loop?

NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, $B=2.5\ mT=2.5\times 10^{-3}\ T$

(A) Magnetic field of a current loop is given by :

$B=\dfrac{\mu_oI}{2r}$

I is the current in the loop

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}$

I = 27.85 A

(B) Magnetic field at a distance r from a wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}$

r = 0.00222 m

$r=2.2\times 10^{-3}\ m$

Hence, this is the required solution.

3 0

3 years ago

What is the purpose of an experiment design?

Eddi Din [679]

An experimental design is used to assign variables for testing. In contrast to a control design where nothing is changed, the experimental design allows you to test various new inputs to see how they would vary from the original results.

5 0

3 years ago

A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T

sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

$\Delta U = Q-W$

$\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)$

Where

$\dot{m} =$ Mass flow

$h_1 =$ Specific Enthalpy (IN)

$h_2 =$ Specific Enthalpy (OUT)

$g =$ Gravity

$z_{1,2} =$ Heigth state (In, OUT)

$V_{1,2} =$ Velocity (In, Out)

Our values are given by,

$T_i = 10\°C$

$P_1 = 100kPa$

$\dot{m} = 5kg/s$

$z_2 = 20m$

For this problem we know that as pressure, temperature as velocity remains constant, then

$h_1 = h_2$

$V_1 = V_2$

Then we have that our equation now is,

$\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}$

$\dot{W} = \frac{(5)(9.81)(0-20)}{1000}$

$\dot{W} = -0.98kW$

8 0

3 years ago