1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
elena-s [515]
3 years ago
13

A block of mass 2. 0 kg, starting from rest, is pushed with a constant force across a horizontal track. The position of the bloc

k as a function of time is recorded, and the data are shown in the table. What is the magnitude of the change in momentum of the block between zero and 4. 0 seconds?.
Physics
1 answer:
yan [13]3 years ago
5 0

The magnitude of the change in momentum of the block between zero and 4. 0 seconds is

dp=1.6m/s

<h3>What is the magnitude of the change in momentum of the block between zero and 4. 0 seconds?</h3>

Generally, the equation for the kinematics equation is mathematically given as

x=x+ut+0.5at^2

Therefore

1.6=0+0+0.5*a*1^2

a=0.2m/s^2

In conclusion,  change in momentum

dP=mv-mu

2.0*0.8-0

dp=1.6m/s

Read more about Force

brainly.com/question/13370981

You might be interested in
Is it illegal to attempt gerrymandering​
IRINA_888 [86]

Explanation:

The US Supreme Court has affirmed in Miller v. Johnson (1995) that racial gerrymandering is a violation of constitutional rights and upheld decisions against redistricting that is purposely devised based on race. However, the Supreme Court has struggled as to when partisan gerrymandering occurs (Vieth v.

8 0
3 years ago
A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is
Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

Speed of the car, v = 10 m/s

Radius of circular path, r = 30 m

Mass of the passenger, m = 60 kg

To find :

The normal force exerted by the seat of the car when the it is at the bottom of the depression.

Solution,

Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.

N=mg+\dfrac{mv^2}{r}

N=m(g+\dfrac{v^2}{r})

N=60\times (9.81+\dfrac{(10)^2}{30})

N = 788.6 Newton

N = 789 Newton

So, the normal force exerted by the seat of the car is 789 Newton.

6 0
3 years ago
Can someone please help me with science.
pav-90 [236]

On the dog's return trip (between <em>t</em> = 10 and <em>t</em> = 12.5 seconds), the slope of the position function is steeper than during the first 5 seconds, which means the dog ran home faster. The only option that captures this is D.

You can check to make sure that the dog indeed runs twice as fast on the return trip. The slope of the position function during the first 5 seconds is

(change in position) / (change in time) = (5 - 0) / (5 - 0) = 5/5 = 1

while during the return trip, it is

(0 - 5) / (12.5 - 10) = -5/2.5 = -2

Ignoring the sign (which only indicates the direction in which the dog was running), we see that the dog's speed on the return trip was indeed twice as high as during the first 5 s.

8 0
3 years ago
A car is traveling at a velocity of 22 m/s when the driver puts on the brakes
Brums [2.3K]

The car’s velocity at the end of this distance is <em>18.17 m/s.</em>

Given the following data:

  • Initial velocity, U = 22 m/s
  • Deceleration, d = 1.4 m/s^2
  • Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

V^2 = U^2 + 2dS

Substituting the values into the formula, we have;

V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}

<em>Final velocity, V = 18.17 m/s</em>

Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>

<em></em>

Read more: brainly.com/question/8898885

8 0
2 years ago
Explain why atomic radius decreases as you move to the right across a period for main-group elements but not for transition elem
Aleksandr-060686 [28]

Answer:

Explained.

Explanation:

Only the first question has been answered

In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus,  atomic radius decreases.

In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.

7 0
3 years ago
Other questions:
  • Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th
    8·1 answer
  • How long does it take oxygen to diffuse a distance of 1 mm in water at room temperature? How long does it take oxygen to diffuse
    6·1 answer
  • Which compound is an example of a binary ionic compound?
    6·2 answers
  • Name the force between the brakes and the wheels that does work to slow down the vehicle when the brakes are applied
    15·1 answer
  • A block slides to a stop as it goes 47 m across a level floor in a time of 6.35 s. a) What was the initial velocity? b) What is
    14·1 answer
  • What are 3 macro nutrients
    11·2 answers
  • A light spring with force constant 3.05 N/m is compressed by 7.80 cm as it is held between a 0.400-kg block on the left and a 0.
    7·1 answer
  • In the 1500's Galileo experimented and discovered many things. One of his famous experiments allowed him to discover that blank
    10·1 answer
  • An object that gives off electromagnetic waves based on its temperature
    7·1 answer
  • Which piece of furniture will have the most inertia and give the furniture movers the most difficulty in moving?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!