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
alina1380 [7]
3 years ago
8

Consider the relationship of the variables in Newton’s second law. In a drag car race, the force applied to the car is doubled b

y the driver stepping on the gas pedal. The acceleration of the car will . The mass of the car will . The velocity of the car will .
Physics
2 answers:
kkurt [141]3 years ago
5 0

When the driver mashes the gas, puts the pettle to the mettle, and doubles the engine force applied to the car . . .

-- the acceleration of the car doubles

-- the mass of the car doesn't change

-- the velocity of the car starts to increase in the direction in which the car is already moving.  If the car's speed was already increasing forward before the force adjustment, then the rate at which it increases due to the doubled force will double.

daser333 [38]3 years ago
4 0

Recall that the force on an object is related to the mass and acceleration of that object by the formula F = ma, where m is the mass of the object and a is its acceleration. What happens when we double F? Well, you might remember from algebra that, in order to keep our equality true, if we double one side, we must also double the other, so our equation becomes 2F = 2ma. Now, this means one of two things: either the mass has doubled, or the acceleration has doubled.

We can tell right away that it'd be absurd if a race car doubled in mass every time it hit the gas, so the quantity doubling must be the <em>acceleration. </em>If we call the car's current velocity v1, we'll be adding the doubled acceleration to get its new velocity. Mathematically, v = v1 + 2a.

We can now conclude that, by doubling the force:

  • The acceleration of the car will double,
  • The mass of the car will stay the same, and
  • The velocity of the car will increase by double the original acceleration
You might be interested in
Convert 3402kgm/s to 20000Newtons
oee [108]

The 3,402 has units of kg-m/s.  That's momentum.  The 20,000 has units of Newtons.  That's force.  Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely.  You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push.  The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it.  The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

<em>T = 0.1701 second</em>

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0
3 years ago
Two particles with charges of 2nC and 5nC are separated by a distance of 3m. The charge 2nC is placed on the left. Find the forc
Vanyuwa [196]

Answer:

1\cdot 10^{-8} N to the left

Explanation:

The magnitude of the electrostatic force between two charges is given by the following equation:

F=k\frac{q_1 q_2}{r^2}

where:

k=9\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the magnitude of the two charges

r is the distance between the two charges

Moreover, the force is:

- Attractive if the charges have opposite sign

- Repulsive if the charges have same sign

In this problem, we have:

q_1=2nC=2\cdot 10^{-9}C is the magnitude of charge 1

q_2=5nC =5\cdot 10^{-9}C is the magnitude of charge 2

r = 3 m is the distance between the two charges

Substituting, we find the force on both charges:

F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N

Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).

5 0
3 years ago
Which type of wave is classified as electromagnetic?
Nimfa-mama [501]

Answer: B) Light

Explanation: light waves are Electromagnetic waves. Visible light is one of the many types of electromagnetic waves. The others are mechanical waves.

8 0
3 years ago
Read 2 more answers
Solar cells convert what type of energy into electrical energy?
zysi [14]
I think the correct answer from the choices listed above is option D. Solar cells convert electromagnetic energy into electrical energy. This is because the photons from the rays of light are electromagnetic waves or particles. Hope this answers the question.
6 0
4 years ago
Read 2 more answers
A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon
Anettt [7]
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s and acceleration g=9.81 m/s^2:
S_x(t)=v_xt
S_y(t)=v_y t- \frac{1}{2} gt^2
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0
Therefore:
v_y t -  \frac{1}{2}gt^2=0
which has two solutions:
t=0 is the time of the beginning of the motion,
t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m
4 0
3 years ago
Other questions:
  • The motor in a refrigerator has a power output of 400 W. If the freezing compartment is at 273 K and the outside air is at 306 K
    15·1 answer
  • The car has a mass of 1400 kg and travels at a velocity of 25 m/s.Calculate the kinetic energy of the car.
    12·1 answer
  • Which of the following statements is true about hydrothermal vents?
    8·2 answers
  • Which correctly describes the Law of Conservation of Energy? Question 6 options:Energy can be created or destroyed, but cannot c
    9·1 answer
  • Which type of reaction releases the most energy? nuclear chemical physical
    8·2 answers
  • Runner A is initially 5.8 km west of a flagpole and is running with a constant velocity of 8.6 km/h due east. Runner B is initia
    9·1 answer
  • What factors does weight depend on?
    9·1 answer
  • Explain how momentum is determined and conserved.<br><br> ASAP!!
    11·1 answer
  • Falling objects drop with an average acceleration of -9.8 m/s2. If an object falls from rest from a tall building, how long will
    6·1 answer
  • The mass of earth 6*10^24kg its radius is 6.4*10^3km now calculate due to gravity on earth<br>​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!