All categories

3 years ago

Examples of applied force

Physics

2 answers:

Vladimir [108]3 years ago

4 0

Answer:

Push - The most common form of force is a push through physical contact (like a lawnmower or shopping cart)

Pull - You can apply a force by directly pulling on an object (like pulling a wagon)

Explanation:

dem82 [27]3 years ago

3 0

-- you sit on a chair

-- gravity pulls something down

-- water pushes a log up to float

-- air holds the skin of a balloon out, in a ball-shape

-- my dog bites a bone

-- air holds the tires of my car out, in donut shapes

-- a book presses on a table

-- my shoes press down on the floor

-- my feet press down on the bathroom scale

-- the bathroom scale presses up on my feet

-- gravity pulls the Earth up toward me

You might be interested in

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$

6 0

3 years ago

The moon is 3.85 x 10 to the 8 m from earth and has a diameter of 3.48 x 10 to the 6 m. You have a pea (diameter = 0.50 cm) and

polet [3.4K]

Answer:

(a) dime

Explanation:

Convert all to metric unit:

0.5 cm = 0.005 m

1.8 cm = 0.018 m

71 cm = 0.71 m

In order to find out we would need to calculate the ratio R between the object diameter d and their distance s to our eyes:

$R_m = \frac{d_m}{s_m} = \frac{3.48*10^6}{3.85*10^8} \approx 0.009$

$R_p = \frac{d_p}{s_p} = \frac{0.005}{0.71} \approx 0.007$

$R_d = \frac{d_d}{s_d} = \frac{0.018}{0.71} \approx 0.0253$

Since the ratio of the dime is larger than the ratio of the moon, and the ratio of the pea is smaller than the ratio of the moon, only the (a) dime can cover your view of the moon.

3 0

3 years ago

8. An airplane is flying at 200 m/s when it touches the ground at the airport. It has a constant negative acceleration, and slow

Temka [501]

Answer:

Explanation:

Given

Initial velocity u = 200m/s

Final velocity = 4m/s

Distance S = 4000m

Required

Acceleration

Substitute the given parameters into the formula

v² = u²+2as

4² = 200²+2a(4000)

16 = 40000+8000a

8000a = 16-40000

8000a = -39,984

a = - 39,984/8000

a = -4.998m/s²

Hence the acceleration is -4.998m/s²

8 0

2 years ago

A box is pulled with a horizontal force of 500N and moves 5m what is work done

dalvyx [7]

The answer to the question is shown below:

We all know that formula for solving work done is the force multiplied by distance covered:
Work done = Force x distance
Distance = 5m
Force = 500 N
Work done = 500 N * 5m
Work done = 2500 J

4 0

3 years ago

A dart with mass md is launched toward a block of mass mb that is suspended from a string of length L. The dart is moving horizo

Yuki888 [10]

Answer:

A) Impulse is the same for both the objects

B) The higher is the speed, the greater will be the height.

Explanation:

Part a)

The time of interaction of the two bodies i.e the hanging mass and the stick is same. Thus, force caused by dart on the block = force caused by block on the dart. Hence, impulse is the same for both the objects.

Part B

The energy will be conserved in the entire reaction process

Hence, Kinetic energy = potential energy

0.5Mv^2 = gh(md+mb)

H is directly proportional to the square of speed.

Hence, the higher is the speed, the greater will be the height.

5 0

2 years ago