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Vsevolod [243]
3 years ago
9

Tafu is working with subatomic particles in the physics lab. a positron is traveling in a straight line down the particle accele

rator. tafu models its position with the function p(t)= t2 t â 1 where t>1. in the limit calculations below, if you need to use -â or â, enter -infinity or infinity. (a) calculate the limit
Physics
1 answer:
KatRina [158]3 years ago
4 0

Answer:

Infinity

Explanation:

When you have a limit approaching infinity one method you can use is to remember that the highest power dominants all of the other ones so when we have (t^2)/t-1 like in your problem, we see that t^2 has the highest power of 2 so if that's going to infinity and it's not being divided by any powers higher than it will just be inifinity.

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A square current-carrying loop is placed in a uniform magnetic field B with the plane of the loop parallel to the magnetic field
Radda [10]

Answer:

(A) a net torque but no net force on the loop.

Explanation:

The total force on the loop is zero because the forces on the opposite sides of the loop are equal but act in opposite directions and as a result they cancel each other out. The two forces on opposite sides to the axis of rotation each give rise to a torque about the axis of rotation. This torque is directed along the axis of rotation.

5 0
3 years ago
susan and joe push a table 18 meters down a hall by exerting a force of 200 newtons (n). what is the total amount of work done o
prisoha [69]

Answer:

3600joules

Explanation:

formula :W=FS

W=work done (J)

F=force (N)

S=displacement moved in the direction of force (m)

200N×18m

=3600J

3 0
2 years ago
A force of 3,200 kg x m/s^2 (Newton’s) acts on a truck giving it an acceleration of 2 m/s^2. What is the mass of the truck ?
Ivan

Answer:

According to newton's second law of motionF=ma Data:-F=3200kgm/sec² or N ,a=2m/sec² ,m=? solution :-F=ma here we have to find m so m=F/a ,m=3200/2=1600kg

3 0
3 years ago
bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc
USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

m_A = mass of car A;

v_{A1} = the initial velocity of car A;

v_{A2} = the final velocity of car A;

and

m_B = mass of car B;

v_{B1} = the initial velocity of car B;

v_{B2} = the final velocity of car B.

Then, the law of conservation of momentum demands that

m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}

And the conservation of kinetic energy says that

\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2

These two equations are solved for final velocities  v_{A2} and v_{B2} to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

\boxed{v_{A2} = -1.05m/s}

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

\boxed{v_{B2} = 3.49m/s}

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

4 0
3 years ago
Determine the accelerations that result when a 45 N net force is applied to 3kg object
svp [43]

Answer:

acceleration is 18

Explanation:

45N/3kg=18

4 0
3 years ago
Read 2 more answers
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