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2 years ago

11

What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road and (b) on an uphill road?

1 answer:

ElenaW [278]2 years ago

4 0

Answer:

a) zero b) zero

Explanation:

Newton's first law tells us that a body remains at rest or in uniform rectilinear motion, if a net force is not applied on it, that is, if there are no applied forces or If the sum of forces acting is zero. In this case there is a body that moves with uniform rectilinear motion which implies that there is no net force.

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A crude fermenter is set up in a shed in the backyard of a suburban house. Under anaerobic conditions with ammonia as the nitrog

Aleks04 [339]

Answer:

using calculations Heat losses will be 4512 J

5 0

3 years ago

Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.

Novosadov [1.4K]

Answer:

i) SF: $v(x) = \frac{(w_0* x )^2}{2L}$

ii) BM : $= \frac{(w_0*x)^3}{6L}$

Explanation:

Let's take,

$\frac{y}{w_0} = \frac{x}{L}$

Making y the subject of formula, we have :

$y = \frac{x}{L} * w_0$

For shear force (SF), we have:

This is the area of the diagram.

$v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0$

$= \frac{(w_0* x )^2}{2L}$

The shear force equation =

$v(x) = \frac{(w_0* x )^2}{2L}$

For bending moment (BM):

$BM = v(x) * \frac{x}{3}$

$= \frac{(w_0* x )^2}{2L} * \frac{x}{3}$

$= \frac{(w_0*x)^3}{6L}$

The bending moment equation =

$= \frac{(w_0*x)^3}{6L}$

5 0

3 years ago

ممكن الحل ............

Roman55 [17]

Answer:

i dont understand

Explanation:

4 0

3 years ago

Let be a real-valued signal for which when . Amplitude modulation is preformed to produce the signal . A proposed demodulation t

densk [106]

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t) be a real value signal for x ( jw ) = 0 , |w| > 200 $\pi$

g(t) = x ( t ) sin ( 2000 $\pi t )$

$x_{1} (t) = \frac{1}{2} x(t) sin ( 4000\pi t )$

next we apply Fourier transform

attached below is the remaining part of the solution

6 0

2 years ago

A reservoir is 1 km wide and 10 km long and has an average depth of 100m. Every hour, 0.1% of the reservoir's volume drops throu

Ksju [112]

Answer:

250.7mw

Explanation:

Volume of the reservoir = lwh

Length of reservoir = 10km

Width of reservoir = 1km

Height = 100m

Volume = 10x10³x10³x100

= 10⁹m³

Next we find the volume flow rate

= 0.1/100x10⁹x1/3600

= 277.78m³/s

To get the electrical power output developed by the turbine with 92 percent efficiency

= 0.92x1000x9.81x277.78x100

= 250.7MW

7 0

2 years ago