before:
2m
What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?
(Change in momentum - momentum after-
after:
2m
momentum before) p - mv

Answers

Answer 1

Answer:

22

Explanation:

before:

2m

What is the change in momentum if the v before = 1m/s and the velocity after = 1m/s?

(Change in momentum - momentum after-

after:

2m

momentum before) p - mv


Related Questions

77. A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
a).004
s2
b).0056 m/s2 c).0079"
d).01 m/s2
M
m

Answers

Answer:

a

Explanation:

i just took the test

Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?

Answers

Answer:

The answer is below

Explanation:

Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.

After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:

(a - v) / 5.68 = 3.87

a - v = 21.9816

v = a - 21.9816

For motorcycle B:

(b - v) / 5.68 = 18.2

b - v = 103.376

v = b - 103.376

Therefore:

a - 21.9816 = b - 103.376

b - a = -21.9816 + 103.376

b - a = 81.3944

a) The difference between their speeds at the beginning was 81.3944 m/s

b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.

Therefore motorcycle B was moving faster

Which of the following is not a true statement?
A
B
C
D

Answers

Answer:

I think A

Explanation:

because I don't think a unknowned number can be a division problem

Is think is D but I’m not sure

Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws a rock straight upwards with an initial speed of 15 m/s, while the other person throws their rock straight downwards with an initial speed of 15 m/s. If both of the rocks miss the building and continue down to the street,
a. How long will it take for the rock thrown upwards to reach the ground?
b. How long will it take for the rock thrown downwards to reach the ground?
c. How fast will the upward thrown rock be travelling just before it hits the ground? The downward thrown rock?

Answers

Answer:

C

Explanation:

Hope this helps!!!!

A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 35 km/h at the 67-m mark. He then maintains this speed for the next 88 meters before uniformly slowing to a final speed of 32 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur

Answers

Answer:

0.705 m/s²

Explanation:

a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.

Using newton's law of motion:

v² = u² + 2as

v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h,  s = distance = 67 m

9.72² = 0² + 2a(67)

134a = 94.484

a = 0.705 m/s²

b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:

v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m

v² = u² + 2as

9.72² = 9.72² + 2a(88)

176a = 9.72² - 9.72²

a = 0

c) During the last distance, the speed slows down from 35 km/h to 32 km/h.

u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m

v² = u² + 2as

8.89² = 9.72² + 2a(45)

90a = 8.89² - 9.72²

90a = -15.4463

a = -0.1716 m/s²

The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.


What is the main cause of tides on Earth? *
1. Sun's gravity
2. Moon's gravity

Answers

2. Moons gravity

Explanation: water is connected to the moon the ocean rivers all that is connected to the mood:)

A positively charged oil drop of mass is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced 16 cm apart. If the mass of the drop is 8.0 10-15 kg and it remains stationary when the potential difference between the plates is 2.44 kV, what is the magnitude of the charge on the drop

Answers

Answer:

the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸  C

Explanation:

Given that;

mass of the drop m = 8.0×10⁻¹⁵ kg

distance d = 16 cm = 0.16 m

potential difference between the plates V = 2.44 kV = 2440 v

acceleration of gravity g = 9.81 m/s²

the magnitude of the charge on the drop = ?

weight is balanced by the electrostatic force

weight = mg = 8.0×10⁻¹⁵ kg × 9.81 m/s² = 7.848 × 10⁻¹⁴

we know that; V = Ed

E = V/d =  2440 / 0.16 = 15200 v/m

Electrostatic force = qE

so weight = qE

q = weight / E

q = 7.848 × 10⁻¹⁴ / 15200

q = 5.163 × 10⁻¹⁸  C

Therefore, the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸  C

Most new jobs in the United States will be in the _____.

Answers

in the service producing sector

anteaters are animals with long tongues that they used to wear a ant they sometimes free by poking their tongues into a bed and licking the ants up if an anteater is born with shortened tongue this would be an example of a

Answers

Answer: an ant eater who can’t eat

Explanation:

A hockey puck with mass 0.30 kg is sliding along the ice with initial speed of 12.68 m/s. A hockey player is heading toward the puck with his stick in hand. After the player strikes the puck, the puck reverses its direction and is traveling at double its speed before the strike. If the collision occurs in 0.05 s, what is the magnitude of the force the hockey player's stick applied to the puck

Answers

Answer:

F = 228.24 N

Explanation:

According Newton's 2nd Law, the impulse on one object is equal to the change in momentum of that object.I = F*Δt = Δp = pf - po (1)

       where pf = final momentum = m*vf

                  p₀ = initial momentum = m*v₀

Since after the strike, the puck reverses its direction and travels at double its speed before the strike, that means that vf = -2*v₀.Replacing in the right side of (1), we have:

       [tex]m*v_{f} - m*v_{o} = -2*v_{o} -m*v_{o} = -3*m*v_{o} = -3*0.3kg*12.68m/s = -11.41m/s (2)[/tex]

Replacing Δt = 0.05s, and solving for F in (1):

       [tex]F_{net} = \frac{-11.41m/s}{0.05s} = -228.24 N (3)[/tex]

which means that the force is applied in a direction opposite to the initial velocity of the puck.The magnitude of the force is just 228.24 N.

calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.

Answers

So you would multiply 4 x 16 to an equation of 12

You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Object 3 is a sphere of copper.
a. the density of tin is 5.75g/cm2. What is the mass of object 1 in kg if the rectangular block has a volume of 1.34L?
b. what is the volume in cubic inches of object 2 if the cube of aluminum 7.58 inches on a side?
c. what is the mass in kg of object 2? the density of aluminum is 2.70g/cm3
d. what is the volume in cm3 of object 3 if the sphere of copper has a diameter 8.62cm? the volume of the sphere is 4 {pi}^3/3
e. what is the mass in kg of object 3? Copper has a density of 8.96g/cm3

Answers

Answer: a. m = 7.7 kg

              b. V = 435.52 in³

              c. m = 1927 kg

              d. V = 335.37 cm³

              e. m = 3 kg

Explanation: Density is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.

The formula for density is

[tex]\rho=\frac{m}{V}[/tex]

Density's unit in SI is kg/m³, but it can assume lots of other units.

Some unit transformations necessary for the resolution of the question:

1 L = 1 dm³ = 1000 cm³

1 in³ = 16.3871 cm³

1 g = 0.001 kg

a. V = 1.34 L = 1340 cm³

[tex]\rho=\frac{m}{V}[/tex]

[tex]m=\rho.V[/tex]

m = 5.75 * 1340

m = 7705 g => 7.705 kg

Mass of object 1 with volume 1.34L is 7.7 kg.

b. A cube's volume is calculated as V = side³

V = 7.58³

V = 435.52 in³

Volume of object 2 is 435.52 in³.

c. Using 1 in³ = 16.3871 cm³ to change units:

V = 435.52 * 16.3871

V = 713689.4 cm³

Then, mass will be

[tex]m=\rho.V[/tex]

m = 2.7 * 713689.4

m = 1926961.4 g => 1927 kg

Mass of object 2 is 1927 kg.

d. Volume of a sphere is calculated as [tex]V=\frac{4}{3}.\pi.r^{3}[/tex]

Diameter is twice the radius, then r = 4.31 cm.

Volume is

[tex]V=\frac{4}{3}.\pi.(4.31)^{3}[/tex]

V = 335.37 cm³

Volume of object 3 is 335.37 cm³.

e. [tex]m=\rho.V[/tex]

m = 8.96 * 335.37

m = 3004.91 g => 3 kg

Mass of object 3 is 3 kg.

I NEED HELP WITH THIS QUESTION ASAP PLEASE!

Answers

Wood isn’t a medium. Pls give brainliest

NEED HELP ASAP
How does radiation help treat cancer?

(a) It cools down the cancer cells and weakens them.
(b) It cools down the non-cancerous cells and strengthens them.
(c) It heats up the cancer cells and weakens them.
(d) It heats up the non-cancerous cells and strengthens them.
THANK YOU

Answers

Answer:

c heats up cancer cells and weakens them.

Answer:

I think it is C

Explanation:

because

A wire has a length of 0.50 m and measures about 0.50 mm in its cross-sectional radius. At normal temperature, what is its resistance in Ohms, if Aluminum has a resistivity of 2.82x10^-8 Ohms*meter

Answers

Answer:

Explanation:

For resistance , the expression is as follows .

R = ρ L / S where ρ is specific resistance , L is length of wire and S is cross sectional area .

cross sectional area = π x ( .5 x 10⁻³ )²

S = .785 x 10⁻⁶ m²

Putting the values

R = 2.82 x 10⁻⁸ x .50 / .785 x 10⁻⁶

= 1.796 x 10⁻² ohm .

A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.

Answers

Answer:

Explanation:

The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .

The weight component acting on box parallel to incline plane

= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N

This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .

Force exerted by person = 98 N

distance travelled in 5 s

= velocity x time

= 2 x 5 = 10 m

Work done by person

= 98 x 10

= 980 J .

Solve this chemical equation: CH3CH2OH+__O2=CO2+__H2O

Answers

Answer:

some kind of chemical of which i do not know

Explanation:

A centrifuge is a device that rotates an object to produce an acceleration many times that of gravity Pilots are trained in such devices because they can simulate the same Accelerations that are experienced in certain types of flight. At which angular velocit would a space station of 30 m radius have to rotate to simulate Earth graivty?

Answers

Answer:

ω = 0.571 rad/s

Explanation:

given data

radius = 30 m

solution

we take here g = 9.8 m/s²

and g is express as

g = r × ω²     ....................1

put here value and we get

9.8 = 30 × ω²

solve it we get

ω = 0.571 rad/s

Why do people eat bo oty

Answers

Answer: I don't know my dude

Explanation:

What is the value of the angle of inclination of the slide?​

Answers

Answer:

63°

that's my answer

but then I am sorry if I'm wrong

Explanation:

90-27 = 63°

First to answer gets brainliest

Answers

Letter na that would be the answer

Astronaut 1 has a mass of 70 kgkg. Astronaut 2 has a mass of 80 kgkg. Astro 1 and 2 want to travel to separate planets, but they want to experience the same weight (in NN). Astro 1 visits a planet with gravitational acceleration 7.0 m/s2m/s2. What must be Astro 2 planet's agag to equal Astro 1's weight

Answers

Answer:

g₂ = 39 m / s²

Explanation:

Mass is the amount of matter that a body has, so it does not change, but the weight is the force with which the planet attracts this matter, this value does change

           W = m g

let's use subscript 1 for the first planet and subscript 2 for the second planet

           W₁ = m₁ g₁

           W₂ = m₂ g₂

They give us the values ​​of the masses and the value of g₁ = 7.0 m / s, let's find the value of the acceleration of gravity on Planet 2

             

They tell us that the weight of the two astronauts is the same therefore

          W₁ = W₂ = W

let's match the expressions

             m₁ g₁ =m₂ g₂

             g₂ = [tex]\frac{m_1}{m_2} \ g_1[/tex]

let's calculate

              g₂ = [tex]\frac{70}{80} \ 7.0[/tex]      

              g₂ = 39.2 m / s²

              g₂ = 39 m / s²

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water

Answers

Answer:

See explanation below

Explanation:

This is a typical exercise of free falling. In this case a rock thrown straight down from the bridge, and we are asked to determine the final velocity of the rock and it's displacement at those given times.

First, just for this problem, as the rock is going straight down, we'll say that the downward direction is positive, therefore, the following expressions to calculate velocity and speed will be:

V = V₀ + gt   (1)

X = V₀t + gt²/2   (2)

In this case, g = 9.8 m/s²

Now, let's see the displacement and velocity for each given time:

a) For  t = 0.5 s

V = 14 + (9.8)*0.5

V = 18.9 m/s

X = (14*0.5) + (9.8)(0.5)²/2

X = 7 + 1.225

X = 8.225 m

b) For t = 1.00 s

V = 14 + (9.8)*1

V = 23.8 m/s

X = (14*1) + (9.8)(1)²/2

X = 14 + 4.9

X = 18.9 m

c) For t = 1.5 s

V = 14 + (9.8)*1.5

V = 28.7 m/s

X = (14*1.5) + (9.8)(1.5)²/2

X = 21 + 11.025

X = 32.025 m

d) For t = 2 s

V = 14 + (9.8)*2

V = 33.6 m/s

X = (14*2) + (9.8)(2)²/2

X = 28 + 19.6

X = 47.6 m

e) For t = 2.50 s

V = 14 + (9.8)*2.5

V = 38.5 m/s

X = (14*2.5) + (9.8)(2.5)²/2

X = 35 + 30.625

X = 65.625 m

Hope this helps

kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)

0.2 mph
4.8 mph
5.5 mph
144.1 mph

it is actually science i couldn't find the word science

Answers

Answer:

4.8mph

Explanation:

Speed= Distance/time

Speed= 26.2/5.5

= 4.76mph

( To the nearest tenth ) = 4.8mph

Answer:

38.7 mph

Explanation:

I just add all the numbers together then divided them by 4, which is the amount of numbers you gave.

Help me I don't know what I'm doing ​

Answers

Answer:

C the metal handle because it is a good conductor

Answer:

D.

Explanation:

Although the metal handle will last longer, if heated up enough it could burn her hand.

a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time​

Answers

HERE IS YOUR ANSWER!

When you jump, you push down on the earth and it pushes back up against you. The earth pushing up against you is what causes you to go into the air. Why doesn’t your push cause the earth to go down if your push on the earth is equal and opposite of the earth's push on you?

Answers

That's a great question !

The answer is: It does !

A push on an object causes the object to accelerate in the direction of the force.  

The less mass the object has, the more the force accelerates it.

Now, when you jump, the forces on you and on the Earth are equal forces.

The up force on you causes you to accelerate up by some amount.

The down force on the Earth causes the Earth to accelerate down by some amount.

The Earth's mass is something like 5,972,000,000,000,000,000,000,000 kg, while your mass is something like 50 kg.

The Earth has something like 119,400,000,000,000,000,000,000 times as much mass as you have.

So your acceleration is something like 119,400,000,000,000,000,000,000 times as great as the Earth's acceleration.

==> The Earth's downward acceleration, caused by your jump, is there.  It's just too small to notice.

BUT . . . That's the reason why seismometers (instruments to detect and measure the vibrations from distant earthquakes) have to be located as far as possible from cities and busy roads.

In places that are too close to cities and roads, the Earth's surface is always vibrating, wiggling, jiggling, heaving and weaving, in reaction to the forces of people walking around, cars and trucks driving around, even rain falling down.  And kids jumping up and down !  

In such places, these people-motions are louder and stronger than the vibrations coming from distant earthquakes.  Seismometers wouldn't work there.    

WILL MARK BRAINLIEST IF CORRECT. Worth Lots Of Points. NEEDS TO BE WITH EXPLANATION PLEASE. PLEASE HELP!

Answers

The answer is 1•72+782

Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were 2.95 × 104 N/C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.26 cm?

Answers

Answer:

(A)  the acceleration  experienced by the proton 2.821 x 10¹² m/s²

(B) the speed of the proton is 2.67 x 10⁵ m/s

Explanation:

Given;

electric field experienced by the proton, E = 2.95 x 10⁴ N/C

charge of proton, Q = 1.6 x 10⁻¹⁹ C

mass of proton, m = 1.673 x 10⁻²⁷ kg

distance moved by the proton, d = 1.26 cm = 0.0126 m

(a)

The force experienced by the proton is calculated as;

F = ma = EQ

where;

a is the acceleration  experienced by the proton

[tex]a = \frac{EQ}{m} \\\\a = \frac{2.95\times 10^4 \ \times \ 1.6\times 10^{-19}}{1.673 \times 10^{-27}} \\\\a = 2.821 \times 10^{12} \ m/s^2[/tex]

(b) the speed of the proton is calculated;

v² = u² + 2ad

v² = 0 + (2 x 2.821 x 10¹² x 0.0126)

v² = 7.109 x 10¹⁰

v = √7.109 x 10¹⁰

v = 2.67 x 10⁵ m/s

a bus initially at rest accelerated of 4m/S2 at the end of 10 second find average velocity​

Answers

Answer:

V = 20m/s

Explanation:

Given the following data;

Acceleration = 4m/s²

Time = 10 secs

Initial velocity = 0m/s (since it's at rest).

To find the average velocity, we would use the first equation of motion;

[tex] V = U + at[/tex]

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting into the equation, we have;

V = 0 + 4*10

V = 40m/s

To find the average velocity;

Average velocity = (U + V)/2

Average velocity = (0 + 40)/2

Average velocity = 40/2

Average velocity = 20m/s

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