What does an electric current produce

A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou

koban [17]

Answer:

The value is $f = 0.05 \ Hz$

Explanation:

From the question we are told that

The mass of the car is $m = 615 \ kg$

The period of the circular motion is $T = 20 \ s$

The radius is $r = 80 \ m/s$

Generally the frequency of the circular motion is

$f = \frac{1}{T }$

=> $f = \frac{1}{ 20 }$

=> $f = 0.05 \ Hz$

3 0

3 years ago

You create a ramp using two text books and a 0.50m board. Using a timer you determine that a cart can roll down the ramp in 0.55

ahrayia [7]

Answer:

The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.

Explanation:

Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:

$v=v_0+at\\\\v^2=v_0^2+2ad$

Dividing the second equation by the first one, we obtain:

$v=\frac{v_0^2+2ad}{v_0+at}$

And, since $v_0=0$ , then:

$v=\frac{2ad}{at}\\\\v=\frac{2d}{t}\\\\v=\frac{2(0.50m)}{0.55s}\\\\v=1.81m/s$

It means that the velocity at the bottom of the ramp is 1.81m/s.

We could use this data, plus any of the two initial equations, to determine the acceleration:

$v=v_0+at\\\\\implies a=\frac{v}{t}\\\\a=\frac{1.81m/s}{0.55s}\\\\a=3.30m/s^2$

So the acceleration is 3.30m/s^2.

7 0

3 years ago

How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest<br> to 51 m/s?

Nesterboy [21]

Answer:

t = 16.5s

Explanation:

Given parameters:

Acceleration = 3.1m/s²

Initial velocity = 0m/s

Final velocity = 51m/s

Unknown:

Time taken = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{v - u }{t}$

v is the final velocity

u is the initial velocity

t is the time taken

So;

3.1 = $\frac{51 - 0}{t}$

3.1t = 51

t = 16.5s

8 0

3 years ago

What is the lowest possible temperature

HACTEHA [7]

The lowest possible temperature is absolute zero. However scientists have not reached this temperature, rather they have come very close to absolute zero.

6 0

3 years ago

Read 2 more answers

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = $4.5 \times 10^{-5} C$

Electric Field strength (E) = $2.0 \times 10^{4} N/C$

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

$U = q \times E \times d$

So now insert the values to find the answer

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

On further solving

$U = 0.027 J$

8 0

3 years ago