What is momentum please help meeeee


Answer 1
Answer: the quantity of motion of a moving body, measured as a product of its mass and velocity.

Answer 2
Answer: A golfer, putting on a green requires three strokes to “hole the ball.” During the first putt, the ball roles 5.0m due east. For the second putt, the ball travels 2.1m at an angle of 20 degrees north of east. The third putt is 0.50m due north. What displacement (magnitude

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5 letter word foramount of work done per second
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With a slope of 6.4%A car starts at 0mph, and has a top speed of 203
And has a base acceleration of 2.93ft/s2
How long would it take to reach 12.42 miles


Answer: 271.4 s


We are told the top speed (maximum speed)V_(max) the car has is:

V_(max)=203 mph=90.74 m/s taking into account 1 mile=1609.34 m

And the car's base acceleration (average acceleration)a_(ave) is:

a_(ave)=2.93 ft/s^(2)=0.89 m/s^(2)


a_(ave)=(V_(f)-V_(o))/(\Delta t)(1)


V_(f)=V_(max)=90.74 m/s is the car's final speed (top speed)

V_(o)=0 m/s because it starts from rest

\Delta t is the time it takes to reach the top speed

Finding this time:

\Delta t=(V_(f)-V_(o))/(a_(ave))(2)

\Delta t=(90.74 m/s - 0 m/s)/(0.89 m/s^(2))(3)

\Delta t=t_(1)=101.95 s(4)

Now we have to find the distance d the car traveled at this maximum speed with the following equation:

V_(f)^(2)=V_(o)^(2) + 2a_(ave) d(5)

Isolating d:


d=((90.74 m/s)^(2))/(2(0.89 m/s^(2)))(7)

d=4625.70 m(8)

On the other hand, we know the total distance D traveled by the car is:

D=12.42 miles = 19988.052 m

Hence the remaining distance is:

d_(remain)=D-d=19988.052 m - 4625.70 m(9)

d_(remain)=15362.35 m(10)

So, we can calculate the time t_(2) it took to this car to travel this remaining distance d_(remain) at its top speed V_(max), with the following equation:


Isolating t_(2):


t_(2)=(15362.35 m)/(90.74 m/s)(13)

t_(2)=169.45 s(14)

With this time t_(2) and the value of t_(1) calculated in (4) we can finally calculate the total time t_(TOTAL):

t_(TOTAL)=t_(1)+ t_(2) (15)

t_(TOTAL)=101.95 s + 169.45 s (16)

t_(TOTAL)=271.4 s s

Why is chemical energy a form of potential energy


Chemical Potential Energy is a form of Potential Energy related to the structral arrangment of Atoms or Molecules.

Give an example of an energy conversion that produces an unwanted form of energy


i think A light bulb that gives off heat. 

The magnitude of the gravitational field on the surface of a particular planet is 2g. The planet’s mass is half the mass of the Earth. What is the planet’s radius in terms of the radius Rg of Earth?


Final answer:

To find the planet's radius in terms of the radius Rg of Earth, use the equation g = GM/R^2 and substitute 2g for g. Solve for R to get R = sqrt(1/(2gMg)) * Rg.


To find the planet's radius in terms of the radius Rg of Earth, we need to understand the relationship between the gravitational field and the mass and radius of a planet. The magnitude of the gravitational field on the surface of a planet is given by g = GM/R2, where G is the gravitational constant, M is the mass of the planet, and R is its radius. For the planet in question, we are told that the magnitude of the gravitational field is 2g and its mass is half the mass of Earth. Since the gravitational field is 2g, we can substitute g with 2g in the equation and solve for R in terms of Rg:

2g = GM/R2 → 2gR2 = GM → 2gR2 = (GMg)/(2Rg) → R2/Rg = 1/(2gMg) → R = sqrt(1/(2gMg)) * Rg

Learn more about Gravitational field here:



Final answer:

To find the radius of a planet with a gravitational field twice that of Earth's and half the mass, the radius is calculated to be half of Earth's radius.


The magnitude of the gravitational field strength g on a planet is given by the equation g = G(M/R^2), where G is the universal gravitation constant, M is the planet's mass, and R is the planet's radius. Given that the gravitational field on the surface of the particular planet is 2g where g is Earth's gravitational field, and the planet's mass is half of Earth's mass, we can derive the planet's radius in terms of Earth's radius Rg. Setting up the proportion (G(1/2M_Earth)/(R^2)) / (G(M_Earth)/(Rg^2)) = 2, and simplifying, we find that R^2 = (1/4)Rg^2. Taking the square root of both sides gives us the final relation R = (1/2)Rg.

Learn more about gravitational field strength here:



Natural forces that can alter ecosystems include seasons and climate changes.

T or F


I’m not completely sure but I thought no that’s true. Sorry if I’m wrong.

A scientific study is repeated by four different groups. Only two groups get the same results. Are the data reproducible?


No, one more group must get the same result as the other group. This will make it more reliable and reproducible.