20 pts.

2 answers:

6 0

The cup is acted upon by an unbalanced force which is the cars acceleration, but before it was an object at rest that stayed at rest. This jet propels their body forward.

KIM [24]3 years ago

3 0

Answer:

The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.

Explanation:

Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."

Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.

You might be interested in

A person is filling a knee-high bucket with water using a garden hose and holding it such that water discharges from the hose at

Serggg [28]

Answer:

Yes i am agree with this suggestion

Explanation:

Given that we have to assume that there is no any frictional affects.

As we know that when height increases then the discharge level will decreases when discharge level decreases then the time of filling for the bucket will increase.So the bucket will fill faster if the hose lowered until knee level.

Yes i am agree with this suggestion

8 0

3 years ago

Can you solve it descriptively . thanks

Solnce55 [7]

Answer:

$|M_y| = 170.82 \ N.mm$

Explanation:

From the diagram affixed below completes the question

Now from the diagram; We need to resolve the force at point A into (3) components ; i.e x.y. & z directions which are equivalent to $F_x \ , F_y \ , F_z$

So;

$F_x$ = positive x axis

$F_y =$ Negative y axis

$F_z$ = positive z axis

Then;

$|M_x| = F_y *27-F_z*11 = 77 ----- equation(1) \\ \\ |M_z| = F_y*4 - F_x*11 = 81 ---- equation (2) \\ \\ |M_y| = F_x *27 - F_z *4 = ? ---- equation (3)$

From equation (1); Let's make $F_y$ the subject of the formula ; then :

$F_y = \frac{77+11F_z}{27}$

Substituting the value for $F_y$ into equation (2) ; we have:

$(\frac{77+11F_z}{27})4-F_x*11=81 \\ \\ 11(\frac{7+F_z}{27} ) 4- F_x -11 =81 \\ \\ 28+4 F_z - 27F_x = \frac{81*27}{11} \\ \\ 4F_z - 27F_x = 198.82 -28 \\ \\ 4F_z - 27F_x = 170.82 \\ \\ Since \ |M_y| = 4F_z-27F_x \\ \\ Then: \\ \\ \\ |M_y| = 170.82 \ N.mm$