What is the force of a punch if it accelerates at 0.2m/s2 and has a mass of 2kg?

Air at 27oC and 1 atm flows over a flat plate 40 cm in length and 1 cm in width at a speed of 2 m/s. The plate is heated over it

irga5000 [103]

Answer:

Heat transferred = 22.9 watt

Explanation:

Given that:

$T_1$ = 27°C = (273 + 27) K = 300 K

$T_2$ = 600°C = (600 +273) K = 873 K

speed v = 2 m/s

length x = 40 cm = 0.4 cm

width = 1 cm = 0.001 m

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

From the tables of properties of air, the following values where obtained.

$k = 0.02476 \ W/m.k \\ \\ \rho = 1.225 \ kg/m^3 \\ \\ \mu = 18.6 \times 10^{-6} \ Pa.s \\ \\ c_p = 1.005 \ kJ/kg$

To start with the reynolds number; the formula for calculating the reynolds number can be expressed as:

reynolds number = $\dfrac{\rho \times v \times x }{\mu}$

reynolds number = $\dfrac{1.225 \times 2 \times 0.4}{18.6 \times 10^{-6}}$

reynolds number = $\dfrac{0.98}{18.6 \times 10^{-6}}$

reynolds number = 52688.11204

Prandtl number = $\dfrac{c_p \mu}{k}$

Prandtl number = $\dfrac{1.005 \times 18.6 \times 10^{-6} \times 10^3}{0.02476}$

Prandtl number = $\dfrac{0.018693}{0.02476}$

Prandtl number = 0.754963

The nusselt number for this turbulent flow over the flat plate can be computed as follows:

Nusselt no = $\dfrac{hx}{k} = 0.0296 (Re) ^{0.8} \times (Pr)^{1/3}$

$\dfrac{h \times 0.4}{0.02476} = 0.0296 (52688.11204) ^{0.8} \times (0.754968)^{1/3}$

$\dfrac{h \times 0.4}{0.02476} =161.4252008}$

$h =\dfrac{161.4252008 \times 0.02476}{ 0.4}$

h = 9.992 W/m.k

Recall that:

The heat transferred from the plate can be calculate by using the formula:

Heat transferred = h×A ×ΔT

Heat transferred = $h\times A \times (T_2-T_1)$

Heat transferred = 9.992 × (0.4 × 0.01) ×(873-300)

Heat transferred = 22.9 watt

6 0

3 years ago

A wire with radius 23 cm has a current of 7 A which is distributed uniformly through its cross sectional area. If you were to us

Rina8888 [55]

Answer:

The magnetic induction of the magnetic field is 0.0005293 mT

Explanation:

Data given

I = 7 A = the total current in the wire

r = 23 cm = the radius of the wire = 0.23 meter

r' = 2cm = the measurement point, which should be inside the wire = 0.02 meter

Let's consider the current density is constant in the wire, ⇒ the current enclosed is a function of the enclosed area

I(enclosed) = Jπ r ²

we can consider the current density as the total current over the whole area:

I(enclosed) = I / (πr ²) * πr' ²

I(enclosed) = (I* r'²)/ (r ²)

with I = total current in the wire = 7A

With r = the radius of wire = 0.23 meter

with r' = the distance of point from the center of wire 0.02 meter

We plug this into ampere's law:

∮ *B *dl =μ 0 * (I* r'²)/ (r ²)

with B = Magnetic flux density (in Tesla) or magnetic induction

with dl = an infinitesimal element (a differential) of the curve C

with µ0 = the magnectic constant = 4π*10^−7 H/m

We can simplify this, by using an Amperian loop can write this as:

B *( 2 π r') = μ 0 * (I* r'²)/ (r ²)

Because the circumference of a circle is 2 π r , when we integrate over length at a distance r ′ from the center of wire whose crossection is a circle we get 2 π r ′

When we isolate B, we get:

B = µo *(Ir'/2 π r ²)

B = 4π*10^−7 * ((7*0.02)/2*π*0.23²)

B =5.293 *10 ^-7 T = 0.0005293 mT

The magnetic induction of the magnetic field is 0.0005293 mT

6 0

3 years ago

Why does a spoon appear to be bent when it’s in a glass of water?

crimeas [40]

When light goes from one medium to another, it changes angle. This is why it will look bent.

5 0

3 years ago

An experiment is carried out to measure the extension of a rubber band for different loads.

PtichkaEL [24]

Complete question is;

An experiment is carried out to measure the extension of a rubber band for different loads.

The results are shown in the image attached.

What figure is missing from the table?

Answer:

17.3 cm

Explanation:

The image attached showed values for load, extension and initial length.

Now, the first length there is 15.2 cm and as such it's corresponding extension is 0 because it has no preceding measured length.

The second measured length is 16.2 cm. Since it's initial measured length is 15.2 cm, then the extension has a formula; final length - initial length.

This gives: 16.2 - 15.2 = 1 cm

This corresponds to what is given in the table.

For the next measured length, it is blank but we are given the extension to be 2.1 cm. Now, since the initial measured length is 15.2 cm.

Thus;

2.1 cm = Final length - 15.2 cm

Final length = 15.2 + 2.1

Final length = 17.3 cm

3 0

2 years ago

I NEED A 100% ACCURATE ANSWER FOR THIS QUESTION ASAP NO LINKS !!!

marysya [2.9K]

Answer:

3. Higher in some places and lower in other places

7 0

3 years ago

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What is the force of a punch if it accelerates at 0.2m/s2 and has a mass of 2kg?​

What is the force of a punch if it accelerates at 0.2m/s2 and has a mass of 2kg?