4 years ago

How do defy gravity?

Physics

1 answer:

kaheart [24]4 years ago

6 0

Exert force upward.
Like when you pick something up from the floor, or walk up the stairs.

You might be interested in

If all of the forces acting on an object balance so that the net force is zero, then

nadya68 [22]

The answer is (A) the object must be at rest.

When all of the forces acting on an object balance, the net force is zero and hence the object will not move.

Yes, the direction might also change under some special circumstances.

6 0

4 years ago

a 2.35 water bucket is swung in a full cirlce of radius 0.824 m just fast enough so that the water doesn't fall out the top mean

tiny-mole [99]

Answer:

2.84 m/s

Explanation:

At the top position of the circular trajectory, the normal reaction is zero:

N = 0

So it means that the only force that is providing the centripetal force is the gravitational force (the weight of the bucket). Therefore we have:

$mg = m \frac{v^2}{r}$

where

m is the mass of the water bucket

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed of the bucket

r = 0.824 m is the radius of the circle

Solving for v,

$v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(0.824 m)}=2.84 m/s$

4 0

3 years ago

How to express whole numbers in scientific notation?

SCORPION-xisa [38]

Answer:

To write a number in scientific notation. First write a decimal point in the numbers so that there's only one digit to the left of the decimal point.

5 0

4 years ago

a school bus has stopped to allow children to get off the bus which graph shows the motion of the bus?

Usimov [2.4K]

Answer: what is it in

Explanation:

4 0

3 years ago

A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th

s344n2d4d5 [400]

Answer:

15 cm

Explanation:

$h_{o}$ = Diameter of the coin = 15 mm

$h_{i}$ = Diameter of the image of coin = 5 mm

$d_{o}$ = distance of the coin from mirror = 15 cm

$d_{i}$ = distance of the image of coin from mirror = ?

Using the equation

$\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}$

$\frac{d_{i}}{15} = \frac{- (5)}{15}$

$d_{i}$ = - 5 cm

$R$ = radius of curvature

Using the mirror equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}$

$\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}$

$R$ = - 15 cm

6 0

3 years ago