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solmaris [256]
3 years ago
10

A ball is tossed straight up from the surface of a small, sphericalasteroid with no atmosphere. The aball rises to a height equa

l tothe asteroid's raidus and then falls straight down toward thesurface of the asteroid.
1. What forces act on the ball while it is on the way up?

1) only a decreasing gravitational force that acts downward.
2) only an increasing gravitational forces that acts downward
3) only a constant gravitation force that acts downward
4) both a constant gravitational force that acts downward and adecreasing force that acts upward
5) no forces act on the ball

2. the acceleration of the ball at the top of its path is1) at its max vlue for the ball's flightb) Equal to the accleration at the surface of the asteroidc) equal to .5 the acceleration at the surface of the asteroidd) equal to 1/4 the acceleration at the surface of theasteroide) zero
Physics
1 answer:
Anastaziya [24]3 years ago
4 0

Answer:

1. 3) only a constant gravitation force that acts downward

2. b) Equal to the accleration at the surface of the asteroidc)

Explanation:

1.

  • In absence of atmosphere, the force that will act on the steroid is the force of gravity due to asteroid.
  • Gravitational force is a long range force acting always attractive in nature.
  • It is a contact-less field force which does not requires a medium.

Mathematically given as:

F=G.\frac{M_1.M_2}{R^2}

where:

G = gravitational constant

M_1\ \&\ M_2 are the mass of the two objects

R= radial distance between the two objects.

2.

The acceleration of the ball at the top height of the path is still under the influence of the gravity of the two masses so it will be equal to the acceleration due to gravity at the surface of the asteroid. Acceleration always remains constant.

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A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material
My name is Ann [436]

Answer:

a)   C = 4,012 10⁻¹⁴ F, b)  Q = 1.6 10⁻¹¹ C , c)   U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

       C = k e₀ A / d

Let's calculate

       C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

       C = 4,012 10⁻¹⁴ F

b) let's look  the charge

        C = Q / ΔV

         Q = C ΔV

         Q = 4,012 10⁻¹⁴ 400

         Q = 1.6 10⁻¹¹ C

c) The stored energy

        U = ½ C ΔV²

        U = ½ 4,012 10⁻¹⁴  400²

        U = 3.21 10⁻¹¹ J

4 0
3 years ago
The velocity of an object is the distance it travels per unit time. Suppose the velocity of a gilding bird is measured to be 52.
Elanso [62]

Answer:

d=7.115s

Explanation:

What problem says can be written mathematically as:

v=\frac{d}{t}

Where:

v=Velocity\\t=Time\\d=Distance

The problem itself it's really simple, we only need to replace the data provided in the previous equation, but first, let's convert the units of the velocity from cm/s to m/s because we have to work with the same units and working in meters is the most apropiate action, because is the base unit of length in the International System of Units:

52\frac{cm}{s} *\frac{1m}{100cm} =0.52\frac{m}{s}

Now, we can replace the data in the equation and find the time it will take the bird to travel 3.7 m:

0.52=\frac{3.7}{t}

Solving for t, multiplying by t both sides, and dividing by 0.52 both sides:

t=\frac{3.7}{0.52} =7.115384615s\approx7.115s

5 0
3 years ago
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
Hi i need help pls help me
r-ruslan [8.4K]
The correct answer would be the last one
4 0
3 years ago
Read 2 more answers
What is the main promblem with survey research?
zlopas [31]

Answer:

A major problem in all survey research is that respondents are almost always self-selected. Not everyone who receives a survey is likely to answer it, no matter how many times they are reminded or what incentives are offered.

Explanation:

<em><u>DISADVANTAGES</u></em>

Respondents may not feel encouraged to provide accurate, honest answers.

Respondents may not feel comfortable providing answers that present themselves in a unfavorable manner.

Respondents may not be fully aware of their reasons for any given answer because of lack of memory on the subject, or even boredom.

HAVE A GOOD DAY!

7 0
3 years ago
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