All categories

3 years ago

What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of?

1 answer:

7 0

We will apply the Newton's second Law so the we will be able to find the acceleration.
F (tot) = ma
a = F(tot) / m
a = 32.0 N / 65.0 kg = 0.492 m/s^2
Approximately 0.492 m/s^2 is her initial acceleration if she is initially stationary and wearing steel-bladed skates.

You might be interested in

Which stimulus increases blood glucose levels?

iVinArrow [24]

Well the options r not well described anyways, the answer goes:
1) Foods rich in carb, sugars, proteins...
2) Lack of activity/exercise
3) Not proper use of medicines
4) Imbalanced hormone
5) Dehydration
6) Stress (in some situations)
7) High intake of unhealthy beverages (i.e, juices, alcohol & alike)
And much more

Hope this was helpful :)

8 0

3 years ago

Which one of these contains two carbon atoms and six hydrogen atoms

kondor19780726 [428]

The second option
You can see it in the name that see represents carbon and H represents hydrogen on the periodic table. The numbers after it are the amount of items that are in the molecule

7 0

3 years ago

Read 2 more answers

A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV.

Mrrafil [7]

Answer:

The magnetic field required required for the beam not to be deflected is $B = 0.0036T$

Explanation:

From the question we are told that

The charge on the particle is $q = +2e$

The mass of the particle is $m = 6.64 *10^{-27} kg$

The potential difference is $V_a = 1.8 kV = 1.8 *10^{3} V$

The potential difference between the two parallel plate is $V_b = 120 V$

The separation between the plate is $d = 8 mm = \frac{8}{1000} = 8*10^{-3}m$

The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam after the region having a potential difference of 1.8kV

$KE_b = PE_b$

Generelly

$KE_b = \frac{1}{2} m v^2$

And $PE_b = q V_a$

Equating this two formulas

$\frac{1}{2} mv^2 = q V_a$

making v the subject

$v = \sqrt{\frac{q V_a}{2 m} }$

Substituting value

$v = \sqrt{\frac{ 2* 1.602 *10^{-19} * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }$

$v = 41.65*10^4 m/s$

Generally the electric field between the plates is mathematically represented as

$E = \frac{V_b}{d}$

Substituting value

$E = \frac{120}{8*10^{-3}}$

$E = 15 *10^3 NC^{-1}$

the magnetic field is mathematically evaluate

$B = \frac{E}{v}$

$B = \frac{15 *10^{3}}{41.65 *10^4}$

$B = 0.0036T$

6 0

3 years ago

Please select the word from the list that best fits the definition

d1i1m1o1n [39]

Answer:

inertia

Explanation:

the resistance of an object to moving or to stopping

a. gravity

b. inertia

c. gravitational force

7 0

2 years ago

I need help Mr or ms tutor

Ronch [10]

Explanation:

Height is the x-axis, and gravitational potential energy is the y-axis. As the height increases, the gravitational potential energy increases linearly.

7 0

3 years ago