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Fudgin [204]
3 years ago
10

A ball experiences forces of 14 N [N] and 9.2 N [W]. A Free Body diagram is required. What is the acceleration of the ball if it

s mass is 225 g?
Physics
1 answer:
GrogVix [38]3 years ago
5 0

Answer:

First, we can define the North as the positive y-axis, and the East as the positive x-axis.

Here we have two forces:

F₁ = 14N to north

F₂ = 9.2 N to west.

Then the free body diagram will be the one that can be seen below:

Now we want to find the acceleration that the ball experiences if the mass is 225g

Knowing that 1k = 1000g, we can rewrite the mass as:

M = 0.225g

Here we also need to remember the second Newton's law:

F = M*a

Net force equals mass times acceleration.

First, let's find the net force:

F = F₁ + F₂ = 14N*(0, 1) + 9.2N*(-1, 0 )

F = (-9.2N, 14N)

The module of the force is then:

I F I = √( (-9.2N)^2 + (14N)^2) = 16.75 N

Then the acceleration is given by

16.75 N = 0.225kg*a

(16.75 N)/(0.225 kg) = a = 74 m/s^2

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If a nucleus decays by gamma decay to a daughter nucleus, which of the following statements about this decay are correct? (There
GarryVolchara [31]

Answer: Option (b) is the correct answer.

Explanation:

A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.

Symbol of a gamma particle is ^{0}_{0}\gamma. Hence, charge on a gamma particle is also 0.

For example, ^{234}_{91}Pa \rightarrow ^{234}_{91}Pa + ^{0}_{0}\gamma + Energy

So, when a nucleus decays by gamma decay to a daughter nucleus then there will occur no change in the number of protons and neutrons of the parent atom but there will be loss of energy as a nuclear reaction has occurred.

Thus, we can conclude that the statement daughter nucleus has the same number of nucleons as the original nucleus., is correct about if  a nucleus decays by gamma decay to a daughter nucleus.

5 0
3 years ago
A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?
Orlov [11]

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

V = u + at

15000 = 30000 + 60a

a = ( 15000-30000)/60

a = 9 × 10^5 m/s^2

4 0
3 years ago
In Niels Bohr's 1913 model of the hydrogen atom, the single electron is in a circular orbit of radius 5.29×10⁻¹¹m and its speed
Svet_ta [14]

The magnitude of the magnetic moment due to the electron's motion is 87.87 * 10^{-37}.

<h3>What is magnetic moment?</h3>

The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).

The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.

Calculations:

radius= 5.29 * 10^{-11} m\\

velocity=2.9* 10^{6} m/s

Working formula, M=N/A

I=\frac{charge flow }{time taken} =\frac{e}{time taken\\}

T= \frac{2xr}{v} =\frac{2xx * 5.29 * 10^{-11} }{2.9* 10^{6} }

   =15.16 * 10^{-5} s

I= \frac{1.6 * 10^{-19} }{15.16 * 10^{-5} }= 0.10 * 10^{-14}

                     =1 * 10^{-15} C

M=1x (1* 10^{-15} * (5.29 * 10^{-11} )^{2}

  =87.87 * 10^{-37}

To learn more about magnetic moment ,visit:

brainly.com/question/14298729

#SPJ4

4 0
1 year ago
According to recent typical test data, a Ford Focus travels 0.240 mi in 19.3 s, starting from rest. The same car, when braking f
Anit [1.1K]

Answer:

Explanation:

a )

While breaking initial velocity u = 62.5 mph

= 62.5 x 1760 x 3 / (60 x 60 )  ft /s

= 91.66 ft / s

distance trvelled s = 150 ft

v² = u² - 2as

0 = 91.66²  - 2 a x 150

a = - 28 ft / s²

b ) While accelerating initial velocity u = 0

distance travelled s = .24 mi

time = 19.3 s

s = ut + 1/2 at²

s is distance travelled in time t with acceleration a ,

.24 = 0 + 1/2 a x 19.3²

a = .001288 mi/s²

= 2.06 m /s²

c )

If distance travelled s = .25 mi

final velocity v = ? a = .001288 mi / s²

v² = u² + 2as

= 0 + 2 x .001288 x .25

= .000644

v = .025 mi / s

= .0025 x 60 x 60 mi / h

= 91.35 mph .

d ) initial velocity u = 59 mph

= 86.53 ft / s

final velocity = 0

acceleration = - 28 ft /s²

v = u - at

0 = 86.53 - 28 t

t = 3 sec approx .

4 0
3 years ago
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