What is the code to this?

A leaf floating down from a tree is an example of an object in free fall.

Nuetrik [128]

Answer:

Oi, mate its false

Explanation:

because if an leaf floats down from a tree it is not considered an object for a free-fall

7 0

4 years ago

For each situation, identify when sound would travel faster and why?

Rudiy27

A. Outside on a summer day, there are less particles that the sound bounces off of (snow and wind absorb the sound)

B. Water, sound travels faster through denser objects. Water has a higher density than air.

C. High air pressure, sound moves faster through matter that is closer together. This can be said because sound can't travel through space. (There is no atmosphere and consistent particles in space for sound to go through)

D. A piece of steel, It's denser than wood and water.

E. 90% nitrogen and 10% helium, nitrogen is denser than helium so it will move faster

8 0

3 years ago

A car, initially traveling at 81.8 mi/h, slows to rest in 7.1 s. What is the car's acceleration?

maksim [4K]

Answer:

a = -16.9 ft/s²

Explanation:

81.8 mi/hr(5280 ft/mi / 3600s/hr) = 120 ft/s

a = Δv/t = (0 - 120) / 7.1 = -16.9 ft/s²

if the initial direction is taken as the positive direction.

5 0

3 years ago

Cylindrical beaker of height 0.100 mm and negligible weight is filled to the brim with a fluid of density rhorhorho = 890 kg/m3k

Nesterboy [21]

Incomplete part of the question

A ball of density ρb = 5000 kg/m3 and volume V = 60.0 cm3 is then submerged in the fluid, so that some of the fluid spills over the side of the beaker. The ball is held in place by a stiff rod of negligible volume and weight. Throughout the problem, assume the acceleration due to gravity is g = 9.81 m/s2 . What is the weight Wb of the ball? Express your answer numerically in newtons.

What is the reading W2 of the scale when the ball is held in this submerged position? Assume that none of the water that spills over stays on the scale. Calculate your answer from the quantities given in the problem and express it numerically in newtons.

What is the force Fr applied to the ball by the rod? Take upward forces to be positive (e.g., if the force on the ball is downward, your answer should be negative). Express your answer numerically in newtons.

The rod is now shortened and attached to the bottom of the beaker. The beaker is again filled with fluid, the ball is submerged and attached to the rod, and the beaker with fluid and submerged ball is placed on the scale.

What weight W3 does the scale now show?

Answer:

(a) 2.94 N

(b) 1 N

(c) 2.42 N

(d) 3.42 N

Explanation:

(a)

From the definition of density, it's mass per unit volume hence mass is a product of density and volume. To get weight, we multiply mass by acceleration due to gravity

The weight of the ball is $W=\rho g V$

Where $\rho$ is the density, V is volume and g is acceleration due to gravity

Substituting density for 5000 Kg/m3 and g for 9.8 m/s2 and v for 0.00006 m3 then

$W= 5000 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3}=2.94 N$

(b)

Because the ball is being held up mostly by the rod, the fluid pressure on the bottom of the cylinder is just the same as before.

The scale does not "know" the ball is there at all.

That's why it still reads 1 N.

Therefore, the reading is 1 N

(c)

The buoyant force of the fluid on the ball is equal to the weight of the displaced fluid, namely,

$890 kg/m^{3} * 9.8 m/s^{2} * 0.00006 m^{3} = 0.52 N$

so the force needed for the rod to hold up the ball is 2.94 N - 0.52 N = 2.42 N.

(d)

Now the scale "feels" the weight of the ball,

so the scale reads the weight of the ball

PLUS the weight of the original fluid

MINUS the weight of the fluid that was displaced

= 2.94 N + 1.00 N - 0.52 N = 3.42 N

6 0

3 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$