What two types of motion are the same in which they both are under the influence of gravity?

Answers

Answer 1

Answer:

Orbit and Gravity

Explanation:

May I have brainliest please I would appreciate it! Have a great day!


Related Questions

In economics, _________ is the amount of a resource that firms and producers are willing and able to provide to the marketplace

Answers

sectors are the amount of resources

What are the SI units for distance and time?

Answers

Answer:

distance is meters, time is seconds

Answer:

distance is meters(m)

time is seconds (s)

Explanation:

Which property describes the number of waves that move past a point each second.
A) Speed
B) Frequency
C) Period
D) Wavelength

Answers

Answer:B) Frequency

Explanation:

Answer:

B) Frequency

Explanation:

field lines point out from all sides on a object. Some of the lines point into a second object nearby. Which statement best describes the objects? A. both objects are permanent magnets B. both objects are electromagnets C. the first object is negatively charged, and the second object is positively charged D. the first object is positively charged, and the second object is negatively charged

Answers

Answer:

The first object is positively charged, and the second object is negatively charged. (D)

Explanation:

a tennis ball is thrown straight up with an initial velocity of 22.5 m/s how much total time is the ball in the air

Answers

Answer:

4.6s

Explanation:

v=u+at

0=22.5+(-9.8)t

-22.5=-9.8t

t=-22.5/-9.8

t=2.295 s

The total time will double

2.295×2=4.59s

=4.6s

Calculate the TOTAL mechanical energy of pendulum is it swings from his highest point to its lowest point. Pendulum mass is 4 kg. Use your equations for gravitational potential energy and kinetic energy to determine these values based on the data given below. Total energy is the sum of gravitational potential energy and kinetic energy. In this problem, round gravity to: g = 10 m/s^2.

Answers

Answer:

its should be 2.0 and 4.5 on it

7. A car is moving at 50.0 mph when the driver applies brakes. Determine the distance it
covers before coming to a halt. Coefficient of static friction between the tires and surface of
the road is 0.514. Mass of the car is 1000 kg.

Answers

Answer:

The distance to come to a halt is approximately 49.53 meters

Explanation:

Thee given parameters are;

The speed of the car, v = 50 mph = 22.35 m/s

The mass of the car, m = 1000 kg

The coefficient of friction,  = 0.514

The force of friction of the brake = Mass × Gravity × Friction = 1000 × 9.81 × 0.514 = 5042.34 N

The initial kinetic energy of the car = 1/2×m×v² = 1/2 × 1000 × 22.35² = 249761.25 J

The work done by the brake = Force of the brake × Distance, d, to come to halt

By conservation of energy, we have;

The work done by the brake = The initial kinetic energy of the car

∴ The initial kinetic energy of the car = Force of the brake × Distance, d, to come to halt

The initial kinetic energy of the car = 249761.25 J = 5042.34 N × Distance, d, to come to halt

∴The distance to come to a halt = 249761.25 J /(5042.34 N) ≈ 49.53 meters

The distance to come to a halt ≈ 49.53 meters.

The car will cover 49.65 m distance before stopping due to application of brake.

Given data:

The mass of car is, M = 1000 kg.

The initial speed of car is, u = 50.0 mph = (50)(0.447 m/s) = 22.35 m/s.

The coefficient of static friction is, [tex]\mu = 0.514[/tex].

The given problem can be solved using the third kinematic equation of motion. Which is,

[tex]v^{2}=u^{2}+2as[/tex] ....................................................(1)

Here, v is the final speed, v = 0 (Because car is finally stopping)

s is the distance covered before stopping and a is the magnitude of acceleration.

Now, since frictional force opposes the motion. Then,

[tex]F = f\\\\ma = \mu \times m \times g\\\\a = 0.514 \times 9.8\\\\a =5.03 \;\rm m/s^{2}[/tex]

Substituting the values in equation (1) as,

[tex]0^{2}=23.35^{2}+2(-5.03)s\\\\s = \dfrac{2500}{2 \times 5.03}\\\\ s=49.65 \;\rm m[/tex]

Thus, the car will cover 49.65 m distance before stopping due to application of brake.

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A soccer player applies a force of 48.4 N to a soccer ball while kicking it. If the ball has
a mass of 0.44 kg, what is the acceleration of the soccer ball?
A. 27.3 m/s2
B. 21.3 m/s2
C. 110 m/s2
D. 104 m/s2

Answers

Answer:

C. 110 m/s2

Explanation:

Force = Mass x Acceleration

Since we have the force and the mass, we can rearrange this equation to solve for acceleration by dividing both sides by mass:

Force/Mass = (Mass x Acceleration)/Mass

Acceleration = Force/Mass

Now we just have to plug in our values and calculate!

Acceleration = 48.4/0.44

Acceleration = 110m/s/s

It is option C. 110 m/s2

Hope this helped!

Yea C is the answer the first answer is right

To get employees to work longer hours, employers often offer __________ in the form of extra pay.
A.
costs
B.
benefits
C.
incentives
D.
consequences

Answers

C- Incentives it’s like a reward ~!

Answer:

C

Explanation:

I took the test

the solubility of sugar is 30 degree Celsius at 220 what does it mean​

Answers

Answer:

So, that if you put 100 mL of 30 degree water in a beaker, you could add up to 220 grams of sugar and it would eventually dissolve. In other words, that mass of sugar should dissolve in 100 mL of water to make a solution that is just saturated.

The solubility of sugar is 30° C at 220. This means 220g of sugar is dissolved in 100g of solvent to make the saturated solution at 30° C. Solutions.

What is the solution?

Most frequently, the solubility is represented in terms of mass per volume of water. Solubilities are frequently stated as g/100 mL, g/100 g, or g/L of water.

The ratio you provided is probably 1 g of sugar to 100 mL of water. In other words, if you put 100 mL of water at 30 degrees in a beaker, you could add up to 220 grams of sugar, and it would finally dissolve.

To generate a solution that is just saturated, the mass of sugar should dissolve in 100 mL of water.

Therefore, at 220 degrees, sugar dissolves at 30 degrees. This indicates that to create the saturated solution at 30° C, 220g of sugar must be dissolved in 100g of solvent.

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matter makes up all ? and ?

Answers

and All lives matter

Yodelin has fifty quarters and dimes. Their total value is $9.80. Which of these systems of equations can be used to find the number of quarters (q) and
dimes (d) Yodelin has?

Answers

Answer:

q + d = 50

25q + 10d = 980

Explanation:

The equation that can be employed to determine the number of quarters(q), as well as, dimes(d) that Yodelin has would be:

q + d = 50...(1)

25q + 10d = 980...(2)

Multiplying (1) by 25

25q + 25d = 1250 ...(3)

25q + 10d = 980...(4)

subtracting (4) from (3)

15d = 270

d = 18

q = 50 - 18

    = 32

The system of the equations could be

q + d = 50

25q + 10d = 980

Calculation of the equation and the number of quarters and dimes:

Since Yodelin has fifty quarters and dimes. Their total value is $9.80.

So here the equations be

q + d = 50...(1)

25q + 10d = 980...(2)

Now we have to Multiplying (1) by 25

So,

25q + 25d = 1250 ...(3)

25q + 10d = 980...(4)

Now

subtracting (4) from (3)

15d = 270

d = 18

So,

q = 50 - 18

   = 32

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Determine the weight of a 10.0-kg person who is running with a speed of 5.0 m/s. Enter a
numerical answer.

Answers

Answer:

weight=mass*gravity

weight=10.0*10

so weight=100N

The shortest path between two points is:

1) displacement

2) breadth

Answers

Answer:

the shortest path between two points is displacement..

Explanation:

Hope this helps you✌️✌️

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The shorts path between two points is displacement.

1. Which of the following provides a correct answer for a problem that can be solved using the kinematic equations?
A A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
position is 32 m.
B A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
velocity is 16 m/s
C A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's travels a
distance of 16 m
A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
velocity is 24 m/s
E A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
position is 30 m.
1 of 14
2
3
5 6 7 8
9

Answers

Answer:

12

Explanation:

12

A wave having a wavelength of 1.5 meters and amplitude of 3.0 meters
travels a distance of 15 meters in 3.0 seconds. Determine the frequency
and the period of the wave. *
Your answer

Answers

Answer:

velocity, v = 15m / 3.0s

= 5 ms^-1

velocity = wavelength × frequency

5 ms^-1 = 1.5m × f

f = 5 ms^-1 / 1.5 m

f = 3.33 Hz

f = 1/T

T = 1/f

= 1/ 3.33 Hz

= 0.3 s

frequency, f = 3.33 Hz

period, T = 0.3 s

help do number 16 and 17​

Answers

Answer:

16 is d and i have no idea for 17

What is the name of the kind of stretch that involves stretching as far as you can and then holding for 10-30 seconds

Question 2 options:

PNF


ballistic


dynamic


static

Answers

Answer:

Static stretching.

Explanation:

It is static stretching because it is a form of stretching which u can do actively for a period of time and you hold position for about 30 to 60 seconds which allow the muscles and connective tissues to lengthen. It is done after work out with out movement in order to loosen up muscles so as to gain flexibility.

a car travels at a speed of 50 m/s for 13 seconds. what is the distance that the car travels?​

Answers

To find the distance, we should multiply the speed and time.

d = s x t

= 50m/s x 13 s

= 650 m

The answer is 650m. Let me know if I got wrong

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the decay constant  and half-life T1/2; (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30 hours after it is prepared?

Answers

Answer:

(a). The decay constant is [tex]1.55\times10^{-5}\ s^{-1}[/tex]

The half life is 11.3 hr.

(b). The value of N₀ is [tex]2.38\times10^{11}\ nuclei[/tex]

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity [tex]R_{0}=10\ mCi[/tex]

Time [tex]t_{1}=4\ hours[/tex]

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

[tex]R=R_{0}e^{-\lambda t}[/tex]

[tex]\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})[/tex]

[tex]\lambda=0.0000154\ s^{-1}[/tex]

[tex]\lambda=1.55\times10^{-5}\ s^{-1}[/tex]

We need to calculate the half life

Using formula of half life

[tex]T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}[/tex]

Put the value into the formula

[tex]T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}[/tex]

[tex]T_{\dfrac{1}{2}}=44.719\times10^{3}\ s[/tex]

[tex]T_{\dfrac{1}{2}}=11.3\ hr[/tex]

(b). We need to calculate the value of N₀

Using formula of [tex]N_{0}[/tex]

[tex]N_{0}=\dfrac{3.70\times10^{6}}{\lambda}[/tex]

Put the value into the formula

[tex]N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}[/tex]

[tex]N_{0}=2.38\times10^{11}\ nuclei[/tex]

(c). We need to calculate the sample's activity

Using formula of activity

[tex]R=R_{0}e^{-\lambda\times t}[/tex]

Put the value intyo the formula

[tex]R=10e^{-(1.55\times10^{-5}\times30\times3600)}[/tex]

[tex]R=1.87\ mCi[/tex]

Hence, (a). The decay constant is [tex]1.55\times10^{-5}\ s^{-1}[/tex]

The half life is 11.3 hr.

(b). The value of N₀ is [tex]2.38\times10^{11}\ nuclei[/tex]

(c). The sample's activity is 1.87 mCi.

Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force

Answers

Answer:

Resultant is 152 N at 28.5 degrees south to the 100 N force

Explanation:

I would appreciate the help

Two classmates, Aisha and Brandon, want to attend two school activities
over the coming weekend. They have one parking pass between them. The
probabilities that the classmates will attend each event are shown in the
table.
Alsha
Brandon
0.65
0.94
Probability of attending
the Saturday activity
Probability of attending
the Sunday activity
0.80
0.43
They decide to let the person more likely to attend both events have the
parking pass. Assuming that attendance at one activity is independent of
attendance at the other, who is more likely to attend both activities?
A. Brandon. He has a 0.40 probability of attending both activities
O B. Brandon. He has a 0.69 probability of attending both activities
C. Aisha. She has a 0.52 probability of attending both activities.

Answers

Answer:

The answer is D.

Explanation:

Which of the statements below is true?

An object’s weight and mass are equal.
An object’s weight is proportional to its mass.
An object’s weight is inversely proportional to its mass.
An object’s weight is always twice its mass.

Answers

Answer:

An object’s weight is proportional to its mass.

Where are streams located?

Answers

Answer:

Larger seasonal streams are more common in dry areas. Rain-dependent streams (ephemeral) flow only after precipitation. Runoff from rainfall is the primary source of water for these streams. Like seasonal streams, they can be found anywhere but are most prevalent in arid areas.

Explanation:

4. Which of the following statements is correct?
A Mass and weight are different names for the same thing
B The mass of an object is different if the object is taken to the Moon
C The weight of a cer is one of the forces acting on the car.
D The weight of a chocolate beris measured in kilograms

Answers

Answer:

Explanation:

A: wrong. Mass and weight are different.

B: Wrong. The mass here and the mass on the moon are the same. The weight, which is Mass * the acceleration  is equal to weight.

C: Correct.

D: Wrong. Weight is not measured in Kg. Mass is.

Consider a buggy being pulled by a horse.
Which is correct?
1. The horse can pull the buggy forward only
if the horse weighs more than the buggy.
2. The force on the buggy is as strong as the
force on the horse. The horse is joined to the
Earth by flat hoofs, while the buggy is free to
roll on its round wheels.
3. The horse pulls before the buggy has time
to react so they move forward.
4. The horse pulls forward slightly harder
than the buggy pulls backward on the horse,
so they move forward.

Answers

Answer:

Option 2

Explanation:

Answer:B

Explanation:because she started of with no money and now she made her money up

Hiro walks up a flight of stairs in his apartment building. Later in the day, he runs up the same flight of stairs.

Which statement about work and power describes Hiro’s actions?

He did more work running than walking.
He did more work walking than running.
He had more power running than walking.
He had more power walking than running.

Answers

Answer:

C) he had more power running than walking

Explanation:

saw it on a quizlet. not 100% sure tho

Hiro walks up a flight of stairs in his apartment building. Later in the day, he runs up the same flight of stairs so He had more power running than walking.

What is power?

In physics, power is the amount of energy transferred or converted per unit of time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.

Power is related to other quantities; for example, the power involved in moving a ground vehicle is the product of the traction force on the wheels and the velocity of the vehicle.

The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft

Since we know that Hiro will run on stairs to minimize the time of reaching his home so if time decreases then he has to increase his power as power is the ratio of work and time.

Hence Hiro walks up a flight of stairs in his apartment building. Later in the day, he runs up the same flight of stairs so He had more power running than walking.

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A student is on a skateboard facing a wall. The student and skateboard have a mass of 75 kilograms. The student pushes off of the wall with a force of 100 Newtons. What is the force of the wall pushing back on the student? What is the acceleration of the student as she moves away from the wall?

Answers

Answer:

F=-100N; a=1.3m/s^2

Explanation:

Force is being made by student, so wall counteracts that force by not moving so it is equally opposite.

The force of the wall pushing back on the student would be 100 Newtons, and the acceleration of the student as she moves away from the wall would be 1.33 m/s²

What is Newton's third law of motion?

Newton's third law of motion state that for every action force there exists a complementary reaction force that balances it.

As given in the problem a student is on a skateboard facing a wall. The student and skateboard have a mass of 75 kilograms. The student pushes off of the wall with a force of 100 Newtons,

The force of the wall pushing back the student = 100 Newtons

The acceleration of the student = 100/75

                                                     =1.33 m/s²

Thus, The student would be pushed back against the wall by a force of 100 Newtons, and she would accelerate away from the wall at a rate of 1.33 m/s².

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chemical formulawhat is the chemical formula ​

Answers

a set of chemical symbols showing the elements present in a compound and their relative proportions, and in some cases the structure of the compound.

That’s the definition

Answer: A chemical formula is a way of showing and representing information about a chemical proportion of atoms that constitute a particular chemical compound or molecule

Three toy boats with the same mass were in a lake. Two boats were moving and one was stopped. Each boat got bumped by another boat, but not in the same direction. All the boats changed speed as a result of being bumped. Use the information in the diagram to answer. Which toy boat exprienced the strongest force when it was bumped ? How do you know ?

Answers

The answer is " The orange and gray toy boats experienced the strongest force because both gained the same force as they sped up the blue boat lost force so it slow down

Explanation:

Trust me I did this one for an assignment

The boat that will experience the strongest force is the boat that has the highest speed.

According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

[tex]F = \frac{mv}{t}[/tex]

where;

F is the force experienced by the objectv is the velocity of the objectt is the time of motion

The force experienced by the boats is directly proportional to the velocity of their motion. Since the boats have the same mass, the force experienced by each boat will depend on the speed with which the boat moves.

Thus, the boat that will experience the strongest force is the boat that has the highest speed.

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