Net force needed to accelerate a 1000-kg car at 0.5g

A wave with a height of 6 would have greater amplitude than a wave with a height of 5 true or false? HELP

BabaBlast [244]

Answer:

it could be either or because it doesnt just depend on the height but it also depends on the pressure but then again the question didnt ask anything about the pressure so the answer should be true

Explanation:

3 0

2 years ago

What is the answer I’m stuck on this one it’s a screenshot for the question

Dmitry_Shevchenko [17]

Answer:

deceleration is the opposite of acceleration

Explanation:

We know that acceleration is the increase of speed with respect to time. So deceleration must be represented on the graph as a decrease in speed over time.a

3 0

1 year ago

Find the position of the center of mass of the system of the sun and Jupiter? (Since Jupiter is more massive than the rest of th

8090 [49]

Answer:

$r_{cm} = 0.074 m$ from the position of the center of the Sun

Explanation:

As we know that mass of Sun and Jupiter is given as

$M_s = 1.98 \times 10^{30} kg$

$M_j = 1.89 \times 10^{27} kg$

distance between Sun and Jupiter is given as

$r = 7.78 \times 10^{11} m$

now let the position of Sun is origin and position of Jupiter is given at the position same as the distance between them

so we will have

$r_{cm} = \frac{M_s r_1 + M_j r_2}{M_s + M_j}$

$r_{cm} = \frac{1.98 \times 10^{30} (0) + (1.89 \times 10^{27})(7.78 \times 10^{11})}{1.98 \times 10^{30} + 1.89 \times 10^{27}}$

$r_{cm} = 0.074 m$ from the position of the center of the Sun

3 0

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

2 years ago

Two airplanes leave an airport at the same time. The velocity of the first airplane is 750 m/h at a heading of 51.3°. The veloci

abruzzese [7]

Refer to the diagram shown below.

In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.

The second airplane travels
b = 620*2.4 = 1488 mile

The angle between the two airplanes is
163° - 51.3° = 111.7°

Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles

Answer: 2335.4 miles

4 0

2 years ago

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