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3 years ago

When a car's velocity is positive and its acceleration is negative, what is happening to the car's motion?

Physics

1 answer:

Novosadov [1.4K]3 years ago

8 0

Answer:

Assuming rightward is positive, the velocity is positive whenever the car is moving to the right, and the velocity is negative whenever the car is moving to the left. The acceleration points in the same direction as the velocity if the car is speeding up, and in the opposite direction if the car is slowing down.

Explanation:

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A car traveling west at 70 miles per hour for 3 hours covers a distance of

gtnhenbr [62]

70x3=210 mph... its that easy

5 0

3 years ago

A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi

kati45 [8]

Explanation:

Drinking water requirement in town 2.0 m^3/s of water per second

nitrate in local well nitrate per 15 $\mathrm{m}^{3}$ of water

nitrate in distant reservoir $=5 \mathrm{~g} / \mathrm{m}^{3}$

Let the flow rate of well

flow rate of reservoir $=y m^{3} / s$

Drinking water requirement is $45 \mathrm{ppm}$ or $45 \mathrm{~g} / \mathrm{m}^{3}$

therefore, the total flow of drinking water

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3 years ago

A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .

kirill [66]

Answer:

<em>The distance the car traveled is 21.45 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

$\displaystyle a=\frac{6.6-0}{6.5}$

$a = 1.015\ m/s^2$

The distance is now calculated with [2]:

$\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}$

x = 21.45 m

The distance the car traveled is 21.45 m

6 0

3 years ago

At the equator earth rotates with a velocity of about 465 m/s.

Dafna1 [17]

The given velocity is 465 m/s.

Part a.
$465 \, \frac{m}{s} =(465 \times 10^{-3} \, \frac{km}{s})*( 3600 \, \frac{s}{h} ) = 1674 \, \frac{km}{h}$
Answer: 1674 km/h

Part b.
$1674 \frac{km}{h} = (1674 \, \frac{km}{h})*(24 \, \frac{h}{day} ) = 40176 \, \frac{km}{day}$
Answer: 40,176 km/day.

3 0

3 years ago

Read 2 more answers

A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr

Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0

3 years ago