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Amanda [17]
3 years ago
12

The pressure in car tires is often measured in pounds per square inch (lb/in.2lb/in.2), with the recommended pressure being in t

he range of 25 to 45 lb/in.2lb/in.2. Suppose a tire has a pressure of 25.5 lb/in.2lb/in.2 . Convert 25.5 lb/in.2lb/in.2 to its equivalent in atmospheres. Express the pressure numerically in atmospheres.
Physics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

25.5 pounds per square inch are equivalent to 1.735 atmospheres.

Explanation:

An atmosphere equals 14.695 pounds per square inch. We find the equivalent of given pressure in atmospheres by means of simple rule of three:

x = 25.5\,\frac{lb}{in^{2}} \times \frac{1\,atm}{14.695\,\frac{lb}{in^{2}} }

x = 1.735\,atm

25.5 pounds per square inch are equivalent to 1.735 atmospheres.

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A boy drops a ball from an observation tower. The ball hits the ground in 5.0 s. What is the ball's velocity at the time of impa
Orlov [11]
The answer is <span>C. 49 m/s

The kinetic equation is:
v2 = v1 + a * t

v1 - initial velocity
v2 - final velocity
a - gravitational acceleration
t - time

We know:
v2 = ?
v1 = 0 (in free fall
a = 9.8 m/s
t = 5

</span>v2 = v1 + a * t
v2 = 0 + 9.8 * 5
v2 = 0 + 49
v2 = 49 m/s
7 0
3 years ago
A 0.10 g honeybee acquires a charge of +23 pC while flying.
kari74 [83]

Answer:

a) \frac{F}{w} =2.347\times 10^{-6}\ N

b) E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, m=10^{-4}\ kg

charge acquired by the bee, q_2=23\times 10^{-12}\ C

a.

Electrical field near the earth surface, E=100\ N.C^{-1}

Now the electric force on the bee:

we know:

F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}

F=E.q_2

F=100\times 23\times 10^{-12}

F=23\times 10^{-10}\ N

The weight of the bee:

w=m.g

w=10^{-4}\times 9.8

w=9.8\times10^{-4}\ N

Therefore the ratio :

\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}

\frac{F}{w} =2.347\times 10^{-6}\ N

b.

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

F=9.8\times10^{-4}\ N

E.q_2=9.8\times10^{-4}\ N

E\times 23\times 10^{-12}=9.8\times10^{-4}\ N

E=4.2609\times 10^7\ N.C^{-1} parallel to the earth surface.

  • In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.
3 0
2 years ago
A permanent magnet has a magnetic dipole moment of 0.160 A · m^2. The magnet is in the presence of an external uniform magnetic
Elena L [17]

Answer:

the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

Explanation:

The torque is given by :

\bar {N} = \bar {m} * \bar {B}

where ;

m = 0.160 A.m²

B = 0.0800 T

θ = 35°

So the magnitude of the torque N = mBsinθ

N = (0.160)(0.0800)(sin 35°)

N = 0.007341

N = 7.34×10⁻³ Nm

Hence, the magnitude of the torque  on the permanent magnet = 7.34×10⁻³ Nm

b) The potential energy \bar{U} = \bar{-m} * \bar{B}

U = -mBcosθ

U = (- 0.160)(0.0800)(cos 45)

U = -0.010485

U = -1.0485 ×10⁻² J

Thus, the potential energy (in J) of the system consisting of the permanent magnet and the magnetic field provided by the coils =  -1.0485 ×10⁻² J

6 0
2 years ago
A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf
kirza4 [7]

Answer:v=\sqrt{\frac{FL}{m}}

Explanation:

Given

Ball of mass m

maximum Bearable Tension in string is F

Let length of the cord be L m and moving at a speed of v m/s

Here Tension will Provide Centripetal Force

T=Centripetal Force

F=T=\frac{mv^2}{L}

v=\sqrt{\frac{FL}{m}}

8 0
3 years ago
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lys-0071 [83]

Answer:

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Explanation:

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