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Alchen [17]
3 years ago
12

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

oad (the rolling resistance). The density of air is 1.2 kg/m3.
(a) What are the drag forces in newtons at 77 km/h and 106 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.) at 77 km/h N at 106 km/h N
(b) What are the drag forces in newtons at 77 km/h and at 106 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.) at 77 km/h N at 106 km/h N Supporting Materials
Physics
1 answer:
sweet [91]3 years ago
6 0

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}

a).

ρ= 1.2 \frac{kg}{m^{3} }, A_{t}= 0.7 m^{2}, D_{t}= 0.28

F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N

b).

ρ= 1.2 \frac{kg}{m^{3} }, A_{h}= 2.44 m^{2}, D_{h}= 0.57

F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N

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we measure a voltage difference of 5.0 V between two points on the conducting paper between two parallel conducting electrodes.
Alex17521 [72]

Answer:

E=1824.81 V/m

Explanation:

Given that

Voltage difference = 5 V

Distance ,D= 3 mm

 θ = 24°

As we know that electric filed given as

E=\dfrac{V}{d}

Given that D is 24° with respect to the perpendicular to the electrodes.So we have to take cos component of D.

d= D cosθ

d= 3 cos24°

d = 2.74 mm

So

E=\dfrac{V}{d}

E=\dfrac{5}{2.74\times 10^{-3}}

E=1824.81 V/m

8 0
3 years ago
A 2.0 kg block is released from rest at the top of a curved incline in the shape of a quarter of a circle of radius R = 3.0 m. T
Zigmanuir [339]

The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :

<em>PE</em> = <em>m g h</em>

<em>PE</em> = (2.0 kg) (9.8 m/s²) (3.0 m)

<em>PE</em> = 58.8 J

Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :

<em>PE</em> = <em>KE</em>

58.8 J = 1/2 <em>m v</em> ²

117.6 J = (2.0 kg) <em>v</em> ²

<em>v</em> = √((117.6 J) / (2.0 kg))

<em>v</em> ≈ 7.668 m/s ≈ 7.7 m/s

3 0
3 years ago
James Bond is trying to escape his enemy on a speedboat but
SVEN [57.7K]

Answer:

100 m/s

Explanation:

Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁

Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is  completely stationary. Let v₁ is speed of bond's boat.

It is the concept of the conservation of momentum. It remains conserved. So,

m_1u_1+m_2u_2=m_1v_1+m_2v_2

Putting all the values, we get :

0+(2m_1)50=m_1v_1+(2m_2)(0)\\\\100m_1=m_1v_1\\\\v_1=100\ m/s

So, Bond's boat is moving with a speed of 100 m/s after the collision.

3 0
3 years ago
Two satellites, A and B are in different circular orbits
jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is \frac{1}{2}

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

v_{oA} = 2v_{oB}        (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

v_{o} = \farc{2\piR}{T}

where

R = Radius of the orbit

Now,

For satellite A:

v_{oA} = \farc{2\piR}{T_{a}}

Using eqn (1):

2v_{oB} = \farc{2\piR}{T_{a}}           (2)

For satellite B:

v_{oB} = \farc{2\piR}{T_{b}}              (3)

Now, comparing eqn (2) and eqn (3):

\frac{T_{a}}{T_{b}} = \farc{1}{2}

6 0
3 years ago
A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 280.5 meters from the wall and the sound of h
ValentinkaMS [17]

Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

d= distance from the wall = 790.5 m

t= time = the unknown

t= 2d/V

t= 2 × 790.5/ 329.2

t= 4.80 seconds

8 0
3 years ago
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