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jeyben [28]
3 years ago
6

A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no

rmal force to the weight of the car when θ is equal to the following. (a) θ = 13°
Physics
1 answer:
padilas [110]3 years ago
5 0

Explanation:

The weight of the car is equal to, W_c=m\times g...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, F_N=mg\ cos\theta

or

F_N=mg\ cos(13).............(2)

The horizontal component of the force is, F_H=mg\ sin\theta

Taking ratio of equation (1) and (2) as :

\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}

\dfrac{F_N}{W_c}=cos(13)

\dfrac{F_N}{W_c}=0.97

or

\dfrac{F_N}{W_c}=\dfrac{97}{100}

Hence, this is the required solution.

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A block of wood has density 0.500 g/cm3 and mass 2 000 g. It floats in a container of oil (the oil's density is 0.750 g/cm3). Wh
Gennadij [26K]

Answer:

2,666.67cm^3

Explanation:

All we need to do in this problem is divide the mass of the wooden block to the oil's density.

2000/0.750 ≈ 2,666.67cm^3

Best of Luck!

6 0
2 years ago
A 1500-kg car accelerates from rest to 25 m/s in 7.0 s. what is the average power delivered by the engine? (1 hp = 746 w)
AVprozaik [17]
The solution for this is:
Work done = force * distance = m*a*d and power = energy/time 
The vo=0 and vf = 25 m/s and t=7 sec. This gives... 
3.6 m/s^2 as acceleration and d=87.5 meters and thus F=ma= 5400 N. 
Energy = 5400*87.5 = 4.7E5 Joules (2 sig. figs) and Power = 67,500 Watts or 90 HP (2 sig. figs again). 
5 0
3 years ago
Read 2 more answers
According to Newton's third law, two objects interacting under a force (such as gravity) both feel the same force. If the planet
asambeis [7]

Explanation:

That's because the Sun's acceleration is much smaller

8 0
2 years ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
A diver shines an underwater searchlight at the surface of a pond (n = 1.33). At what angle (relative to the surface) will the l
allsm [11]

Answer:

41.2°

Explanation:

Total internal reflection is the reflection of the incident ray at the interface between two media in which one of the media has a lower refractive index than the other. It occurs when the angle of incidence in the denser medium exceeds the critical angle.

The critical angle is the angle of incidence in the denser medium when the angle of incidence in the less dense medium is 90°.

Since

n= 1/sin C

C= sin^-(1/n)

C= sin^-(1/1.33)

C= 48.8°

Hence angle of incidence= 90-48.8 = 41.2°

4 0
3 years ago
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