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Greeley [361]
2 years ago
15

3. What exerts a greater force on the table of 2 kg book lying flat or a 2 kg book on its

Physics
1 answer:
Darina [25.2K]2 years ago
5 0

Answer:

A book on its side exerts a greater force.

Explanation:

Pressure = Force / Area

Assuming that 1kg = 10N

2kg = 20N

Area of book lying flat = 0.3m × 0.2m

                                     = 0.6m²

Pressure of book lying flat = 20N / 0.6m²

                                            = 30Pa (1 s.f.)

Area of book on its side = 0.2m × 0.05m

                                        = 0.01m²

Pressure of book on its side = 20N / 0.01m²

                                               = 2000Pa (1 s.f.)

Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.

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An electron and an antielectron (positron) each have a rest energy of 0.511 MeV , or approximately 8.2×10−14 J . When an electro
Mademuasel [1]

Answer:

Explanation:

Photon is also a particle . Hence when two particles like electron and positron annihilate to get completely changed to photons , a minimum of two photons of equal and opposite momentum and energy are produced flying in opposite direction  to conserve momentum and energy . Each photon will have same energy equal to 511 keV . It is so to conserve momentum and energy. Initially total momentum was zero so finally too total momentum should be zero.

8 0
3 years ago
Consider the numbers 23.68 and 4.12. The sum of these numbers has ____ significant figures, and the product of these numbers has
damaskus [11]

Answer:multiplying will give us 7 significant figures and addition will give us 3 significant figures

Explanation:

After multiplying the two numbers they resulting value will give a value in its 4 decimal places because both given values are in 2 decimal places. The 4 dp is gotten by the addition of the decimal places of both given numbers (2+2) and

The result of its addition will give us a value in its 1dp and 3 significant figures since the addition of 23.68 and 4.12 will give us 27.8

7 0
3 years ago
Calculate the pressure the fluid exerted on your diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is
erica [24]

Complete question:

A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.

Answer:

Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

Explanation:

Given;

density of water, ρ = 1000 kg/m³

diver's position below the surface of the water, h = 10 m

acceleration due to gravity, g = 9.8 m/s²

Let the atmospheric pressure, P₀ = 101325 Pa

The pressure 10 m below the surface of the water is calculated as;

P =  P₀  + ρgh

P = 101325 Pa  +  (1000 x 9.8 x 10)Pa

P = 199325 Pa

P = 1.99 x 10⁵ Pa.

Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa

5 0
2 years ago
A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f
Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram.  There are three forces acting on the car.  Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank.  If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0
2 years ago
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Answer:

The people with caculators will probably answer faster due to thier ablitiy to use a device of technology

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