I need some help with this!

A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.

Savatey [412]

We are given the equation:

<span>x = 11t^2
</span>
We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,
x =53.24 meters

At t=2.95 sec,
x =95.73 meters

Velocity = (95.73 meters - 53.24<span> meters) / (2.95 s - 2.20 s ) = 56.65 m/s

</span>For (b)

At t=2.20 sec,
x =53.24 meters

At t=2.40 sec,
x =63.36 meters

Velocity = (63.36 meters - 53.24<span> meters) / (2.40 s - 2.20 s ) = 50.6 m/s</span>

4 0

3 years ago

The magnitude of the electric field between two parallel charged plates is 200. An electron moves to the negative plate 5. 0 cm

Mama L [17]

The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

<h3>What is an electric field?</h3>

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

The given data in the problem is given by;

E is the electric field = (200 N/C)

d is the distance = 5.0 cm.=0.05 m

Q is the charge of electrons= 1.602 x 10^-19 C

The formula for electric potential is given by;

$\rm V=Ed$

$\rm V=Ed \\\\ \rm V=200 \times 0.05 \\\\ \rm V= 10 \frac{Nm}{C} = 10 \frac{J}{C} = 10 V.$

The work is defined as the product of the potential difference and charge of an electron.

$\rm W= 10 \times 1.602 x 10^{-19} \\\\\ \rm W= 1.6 x 10^{-18 }J$

Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

To learn more about the electric field refer to the link;

brainly.com/question/15071884

8 0

2 years ago

The potential difference between A and B is 5.0 V. A proton starts from rest at A. When it reaches B what is its kinetic energy?

Aleksandr [31]

Answer:

total kinetic energy is 8 × $10^{-19}$ J

Explanation:

given data

potential difference = 5 V

e = 1.60 × $10^{-19}$ C

to find out

what is kinetic energy

solution

we will apply here conservation of energy that is

change in potential energy is equal to change in kinetic energy

so

change potential energy is e × potential difference

change potential energy = 1.60 × $10^{-19}$ × 5

change potential energy = 8 × $10^{-19}$ J

so change in kinetic energy = 8 × $10^{-19}$ J

and we know proton start from rest that mean ( kinetic energy is 0 ) so

change in KE is total KE

total kinetic energy is 8 × $10^{-19}$ J

3 0

3 years ago

What should a free-body diagram look like for a skydiver who has opened his parachute and is now slowing down as he falls?

Ksivusya [100]

Answer:

There is an arrow up for air resistance and an arrow down for gravity. The arrow up is longer than the arrow down.

Explanation:

The text of the problem says that the skydiver is slowing down: this means that he has an acceleration, which is directed opposite to the motion of the skydiver. Since the motion is downward, the acceleration must be upward.

There are two forces acting on the skydiver: the gravity (downward) and the air resistance (upward). According to Newton's second law:

F=ma

the acceleration has the same direction of the net force, so the net force must also be upward: therefore, the air resistance must be greater than the gravity, so the arrow up for air resistance is longer than the arrow down for gravity.

5 0

3 years ago

Read 2 more answers

A tube of mercury with resistivity 9.84 × 10 -7 Ω ∙ m has an electric field inside the column of mercury of magnitude 23 N/C tha

slava [35]

Answer:

The current through the tube is 73.39A.

Explanation:

The relationship between the resistivity $\rho$ , the electric field $E$ , and the current density $J$ is given by

$\rho = \dfrac{E}{J}$

This equation can be solved for $J$ to get:

$J = \dfrac{E}{\rho}$

Since the current is $I = J\cdot A$

$I= J\cdot A = \dfrac{E}{\rho} \cdot A$

Now, for the tube of mercury $\rho = 9.84*10^{-7}\: \Omega \cdot m$ , $E = 23N/C$ , and the area is $A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2$ ; therefore,

$I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2$

$\boxed{I = 73.39A.}$

Hence, the current through the mercury tube is 73.39A.

5 0

3 years ago