A balloon-powered car rolls across the floor at a speed of 0.711 m/s. How long does it take to cover 8.25 m?

Which part of the following graph shows the object going at the fastest speed?

Alex73 [517]

Answer:

more point plzzzzz... eeeeeeeee

4 0

3 years ago

A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude

kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

$v_0=115 m/s$ (toward east)

So, decomposing along the x- and y- directions:

$v_{x0} = 115 m/s\\v_{y0} = 0$

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

$a=2.88 m/s^2$ (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

$a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2$

So the two components of the velocity after a time t = 25.0 s will be

$v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s$

So, the magnitude of the velocity of the plane will be

$v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s$

6 0

3 years ago

How does the mass of an object change the gravitational pull on two objects in a system? How does the distance between objects i

LuckyWell [14K]

Answer:The greater the mass, there greater then gravitational pull on their objects.

There greater the distance,the lower the gravitational pull between the object

Explanation:

Gravitational pull(force) is directly proportional to the products Of The masses. therefore if the mass increase,the gravitational pull(force) also increases.

Gravitational pull(force) is inversely proportional to distance.if the distance between the objects increases,the gravitational pull(force) decreases .

4 0

3 years ago

How does unequal solar heating lead to the Gulf Stream?

adoni [48]

Http://earthguide.ucsd.edu/virtualmuseum/virtualmuseum/OriginofGulfStream.shtml this website might help u find ur answer

8 0

3 years ago

Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2

dybincka [34]

Complete question is;

Block 1 is resting on the floor with block 2 at rest on top of it. Block 3, at rest on a smooth table with negligible friction, is attached to block 2 by a string that passes over a pulley, as shown in the attachment below. The string and pulley have negligible mass.

Block 1 is removed without disturbing block 2.

Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2, and physical constants as appropriate.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Looking at the attached image, if we consider the free body diagram for block 3, by using Newton's first law of motion, we will arrive at the formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is mass of block 3

Meanwhile doing the same with Block 2, the free body diagram would give us the formula; (m2)g - T = (m2)a

Making T the subject gives us;

T = (m2)g - (m2)a - - - (eq 2)

where;

g is acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is mass of block 2

To solve for the acceleration, we will just substitute (m3)a for T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0

3 years ago