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guajiro [1.7K]
2 years ago
9

You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the tr

ip are driven at 55 mi/h but then you get stuck in a standstill traffic jam for 20 minutes. In order to make up time, you then proceed at 75 mi/h for the rest of the trip. What is the magnitude of your average velocity for the whole trip?
Physics
1 answer:
Nitella [24]2 years ago
4 0

Answer:

51.2 mi/h

Explanation:

Total distance, d = 100 miles

First 60 miles with speed 55 mi/h

Next 40 miles with speed 75 mi/h

Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h

Time taken for 40 miles, t2 = 40 / 75 = 0.533 h

Time spent to get stuck, t3 = 20 min = 0.33 h

Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h

The average speed is defined as the ratio of total distance traveled to the total time taken.

Average speed = =\frac{100}{1.953}=51.2 mi/h

Thus, the average speed of the journey is 51.2 mi/h.

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An astronaut weighs 200 lb at sea level. The radius of the earth is 3960 miles. What force is exerted on the astronaut if he is
LiRa [457]

Answer:

85.31 N

Explanation:

Given,

Radius of Earth = 3960 miles = 6373 Km

Weight of Astronaut at Sea level = 200lb = 90.72 Kg

Altitude of Astronaut above Earth = 125 miles = 201.17 km

We know that,

F = m\times g --------------------------  (1)

where,

F = force on the object due to gravity also called the weight

m = mass of the object = 200 lb = 90.71 kg

g = acceleration due to gravity = 9.8 m/s²

Also,

F = \frac{GMm}{r^{2}} -------------------(2)

where,

F = force due to gravity

G = Gravitational constant = 6.67\times 10^{-11} Nm²/kg²

M = mass of the Earth = 5.97\times 10^{24} kg

r = distance between the two objects

here, r = (6373+201.17)km = 6574170 m

From equation (1),

m\times g = 90.71\\m=\frac{90.71}{g} \\m=\frac{90.71}{9.8}\\m=9.26 kg\\

Putting value of m in equation (2)

F = \frac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times 9.26}{6574170^{2}}\\

F=85.31N

6 0
2 years ago
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several
mamaluj [8]

Answer:

The  charge on the dust particle is  q_d  = 6.94 *10^{-13} \  C

Explanation:

From the question we are told that

    The length is  l = 2.0 \ m

    The width is  w = 4.0 \ m

   The charge is  q =  -10\mu C= -10*10^{-6} \ C

    The mass suspended in mid-air is m_a =  5.0 \mu g =  5.0 *10^{-6} \ g =  5.0 *10^{-9} \  kg

   

Generally the electric field on the carpet is mathematically represented as

           E =  \frac{q}{ 2 *  A  *  \epsilon _o}

Where \epsilon _o is the permittivity of free space with value \epsilon_o = 8.85*10^{-12}  \ \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

           E =  \frac{-10*10^{-6}}{ 2 *  (2 * 4 )  *  8.85*10^{-12}}

           E = -70621.5 \  N/C

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        F__{E}} =  F__{G}}

=>     q_d *  E  =  m * g

=>      q_d  =  \frac{m * g}{E}

=>      q_d  =  \frac{5.0 *10^{-9} * 9.8}{70621.5}

=>     q_d  = 6.94 *10^{-13} \  C

4 0
2 years ago
For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a
jonny [76]

Answer:

(a) 0.0178 Ω

(b) 3.4 A

(c) 6.4 x 10⁵ A/m²

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

R=\frac{L}{A\sigma }

R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

i = \frac{V}{R}

i = \frac{0.060}{0.0178}

i = 3.4 A

(c)

Current density is given as

J = \frac{i}{A}

J = \frac{3.4}{5.3\times10^{-6}}

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

E = \frac{J}{\sigma }

E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}

E = 9.01 x 10⁻³ V/m

5 0
2 years ago
A cubical shell with edges of length a is positioned so that two adjacent sides of one face are coincident with the +x and +y ax
Bingel [31]

Answer:

Q = ba⁴ * ε₀

Explanation:

From Gauss's Law, we know that

flux Φ = Q / ε₀

where ε₀ = 8.85e-12 C²/N·m²

and also,

Φ = EAcosθ

The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus

Φ = EAcos0

Φ = EA

E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then

E = ba²

Since A = a² for the cube face, we have

Q / ε₀ = E * A

Q / ε₀ = ba² * a²

so that

Q = ba⁴ * ε₀

5 0
2 years ago
There is an electromagnetic wave traveling in the -z direction in a standard right-handed coordinate system. What is the directi
wlad13 [49]

Answer: The direction of the electric field, E→, is pointed in the +y direction.

Explanation:

One can use the right hand rule to illustrate the direction of travel of an electromagnetic and thereby get the directions of the electric field, magnetic field and direction of travel of the wave.

The right hand rule states that the direction of the thumb indicate the direction of travel of the electromagnetic wave (<em>in this case the -z direction</em>) and the curling of the fingers point in the direction of the magnetic field  B→ (<em>in this case the +x direction</em>), therefore, the electric field direction E→ is in the direction of the fingers which would be pointed towards the +y direction.

6 0
2 years ago
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