A car travels around an oval racetrack at constant speed. The car is accelerating:_______

A lightning bolt transfers 6.0 coulombs of charge from a cloud to the ground in 2.0 x 10-3 second. what is the average current d

AlladinOne [14]

The current is defined as the amount of charge transferred through a certain point in a certain time interval:
$I= \frac{Q}{\Delta t}$
where
I is the current
Q is the charge
$\Delta t$ is the time interval

For the lightning bolt in our problem, Q=6.0 C and $\Delta t= 2.0 \cdot 10^{-3}s$ , so the average current during the event is
$I= \frac{Q}{\Delta t} = \frac{6.0 C}{2.0 \cdot 10^{-3} s}=3000 A$

4 0

2 years ago

What is Yusef most likely finding?

Ratling [72]

Explanation:

yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .

4 0

1 year ago

A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x

mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

5 0

1 year ago

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

wolverine [178]

To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

$\omega =$ Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

$1000J > \frac{1}{2}mv^2$

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2$

$1000J = \frac{1}{2}mv^2$

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

6 0

1 year ago

A 2600-m-high mountain is located on the equator. how much faster does a climber on top of the mountain move than a surfer at a

puteri [66]

The climber move 0.19 m/s faster than surfer on the nearby beach.

Since both the person are on the earth, and moves with the constant angular velocity of earth, however there linear velocity is different.

Number of seconds in a day, t=24*60*60=86400 sec

The linear speed on the beach is calculated as

V1= $\frac{2πr}{t}$