A sailboat of mass m is moving with a momentum p. how would you represent its kinetic energy in terms of these two quantities?

A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr

Leokris [45]

The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

Rounding off to the nearest is 147N/m

The spring constant is 147N/m

Learn more about Hooke's law here

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7 0

2 years ago

An object of irregular shape has a characteristic length of L = 0.5 m and is maintained at a uniform surface temperature of Ts =

goblinko [34]

Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

Characteristic length = 0.5 m

Surface temperature = 400 K

Atmospheric temperature = 300 K

Velocity = 25 m/s

Air velocity = 5 m/s

Characteristic length of second object = 2.5 m

We have same shape and density of both objects so the reynold number will be same,

We need to calculate the value of the average convection coefficient

Using formula of reynold number for both objects

$R_{1}=R_{2}$

$\dfrac{u_{1}L_{1}}{\eta_{1}}=\dfrac{u_{2}L_{2}}{\eta_{2}}$

$\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}$

Here, $k_{1}=k_{2}$

$h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}$

$h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}$

Put the value into the formula

$h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}$

$h_{2}=20\ W/Km^2$

Hence, The value of the average convection coefficient is 20 W/Km².

7 0

3 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

3 years ago

What is the name

Molodets [167]

Answer:

<h3>It's called Resistance! </h3>

Explanation:

If the ratio is constant over a wide range of voltages, the material is said to "ohmic" material.

Hope it helps!

3 0

3 years ago

Convert mm3 into m3.

bixtya [17]

Answer:

$1 \times 10 { - }^{9}$

cubic metre or 1e-9

Explanation:

•By division. Number of cubic millimetre divided(/) by 1000000000, equal(=): Number of cubic metre.

•By multiplication. 83 mm3(s) * 1.0E-9 = 8.3E-8 m3(s)

4 0

3 years ago