1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
liq [111]
3 years ago
13

What to forces effects the momentum of an object

Physics
2 answers:
noname [10]3 years ago
7 0
<span>Any force can affect the momentum of an object</span>
faltersainse [42]3 years ago
4 0

Any force can affect the momentum of an object
if it changes the object's speed.

You might be interested in
An information signal consists of a 25 Hz and a 75 Hz sine waves summed together. It is sampled at a frequency of 500 Hz, the hi
denpristay [2]

Answer:

Explanation:

An information contains

25Hz and 75Hz sine wave

Sample frequency is 500Hz

The analogy signal are generally

y(t) = Asin(2πx/λ - wt), w=2πf

y1(t)=Asin(2πx/λ - wt)

y1(t)=Asin(2πx/λ - 2π•25t)

y1(t)=Asin(2πx/λ - 50πt)

Similarly

y2(t)=Asin(2πx/λ - 150πt)

Using Nyquist theorem

Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.

From sampling

f(nyquist)=f(sample)/2

f(nyquist)=500/2

f(nyquist)=250Hz

From signal

The highest frequency is 150Hz

F(nyquist) = 2×F(highest)

f(nyquist)= 2×150

f(nyquist)= 300Hz

Sample per frequency Ns is given as

Ns=F(sample)/F(highest signal)

Ns=500/150

Ns=3.33sample/period

This is above nyquist rate of 2sample/period

So signal below 300Hz reproduced without aliasing.

The highest resulting frequency is 300Hz

6 0
3 years ago
10 rations measure into appropriate quanitilies
zavuch27 [327]
Hold up let me add them
8 0
3 years ago
Which statement(s) correctly compare the masses of protons, neutrons, and electrons? Check all that apply
umka21 [38]

__Protons and neutrons have similar mass.

__Electrons are smaller than a proton or a neutron.

Explanation:

The three particles involved in this problem are:

  • Proton: it is positively charge, it is found in the nucleus of the atom, and its mass is m_p = 1.67 \cdot 10^{-27} kg
  • Neutron: it has no electric charge, it is also found in the nucleus of the atom, and its mass is approximately equal to that of the proton (just slightly larger)
  • Electron: it has negative electric charge, it orbit around the nucleus of the atom, and its mass is much smaller than that of the proton: m_e = 9.11 \cdot 10^{-31} kg

We can now analyze each of the given statement:

__Protons and neutrons have similar mass.  --> TRUE

__Protons and electrons have similar mass.  --> FALSE, the electron is much lighter

__Neutrons and electrons have similar mass.  --> FALSE, the neutron is much heavier

__Protons are smaller than a neutron or an electron.  --> FALSE, protons are similar to the neutrons

__Neutrons are smaller than a proton or an electron.  --> FALSE, neutrons are similar to the protons

__Electrons are smaller than a proton or a neutron. --> TRUE

Learn more about atoms:

brainly.com/question/2757829

#LearnwithBrainly

3 0
3 years ago
Read 2 more answers
Tutorial Exercise An unstable atomic nucleus of mass 1.83 10-26 kg initially at rest disintegrates into three particles. One of
kogti [31]

Answer:

A) v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) K_total = 373.08 × 10^(-15) J

Explanation:

We are given;

Mass of unstable atomic nucleus; M = 1.83 × 10^(-26) kg

Mass of first particle; m1 = 5.03 × 10^(-27) kg

Speed of first particle in y-direction; v1 = (6 × 10^(6) m/s) j^

Mass of second particle; m2 = 8.47 × 10^(-27) kg

Speed of second particle in x - direction; v2 = (4 × 10^(6) m/s) i^

Now, we don't have the mass of the third particle but since we are told the unstable atomic nucleus disintegrates into 3 particles, thus;

M = m1 + m2 + m3

1.83 × 10^(-26) = (5.03 × 10^(-27)) + (8.47 × 10^(-27)) + m3

m3 = (1.83 × 10^(-26)) - (13.5 × 10^(-27))

m3 = 4.8 × 10^(-27) kg

A) Applying law of conservation of momentum, we have;

MV = (m1 × v1) + (m2 × v2) + (m3 × v3)

Now, the unstable atomic nucleus was at rest before disintegration, thus V = 0 m/s.

Thus, we now have;

0 = (m1 × v1) + (m2 × v2) + (m3 × v3)

We want to find the velocity of the third particle v3. Let's make it the subject of the formula;

v3 = [(m1 × v1) + (m2 × v2)]/(-m3)

Plugging in the relevant values, we have;

v3 = [(5.03 × 10^(-27) × 6 × 10^(6))j^ + (8.47 × 10^(-27) × 4 × 10^(6))i^]/(-4.8 × 10^(-27))

v3 = [(30.18 × 10^(-21))j^ + (33.88 × 10^(-21))i^]/(-4.8 × 10^(-27))

v3 = -[6.29 × 10^(6)]j^ - [7.06 × 10^(6)]i^

B) Formula for kinetic energy is;

K = ½mv²

Now,total kinetic energy is;

K_total = K1 + K2 + K3

K1 = ½ × 5.03 × 10^(-27) × (6 × 10^(6))²

K1 = 90.54 × 10^(-15) J

K2 = ½ × 8.47 × 10^(-27) × (4 × 10^(6))²

K2 = 67.76 × 10^(-15)

To find K3, let's first find the magnitude of v3 because it's still in vector form.

Thus;

v3 = √[(-6.29 × 10^(6))² + (-7.06 × 10^(6))²]

v3 = 9.46 × 10^(6) m/s

K3 = ½ × 4.8 × 10^(-27) × (9.46 × 10^(6))²

K3 = 214.78 × 10^(-15) J

K_total = (90.54 × 10^(-15)) + (67.76 × 10^(-15)) + (214.78 × 10^(-15))

K_total = 373.08 × 10^(-15) J

7 0
3 years ago
A baseball thrown by a pitcher is hit by a batter. At the moment when the ball hits the bat, the force exerted on the bat by the
Slav-nsk [51]

Newton's Laws of motion describe the motion of an object based on the applied force

The correct option that gives the relationships between the forces is the the option;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>(Either option A or B without the minus symbol before </em>F_{bat}<em>, likely typographical error)</em>

<em />

Reason:

Force exerted on the bat by the ball = F_{ball}

Force exerted on the ball by the bat = F_{bat}

Given that the batter hits the ball with the bat, the force exerted by the bat on the ball, F_{bat}, is reacted to by the force of the ball acting on the bat, F_{ball}

According to Newton's third law of motion, action and reaction are equal and opposite. Therefore, at the moment when the ball hits the bat, we have;

\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

<em>Where the force of the bat is high, the ball is accelerated to travel at high speed</em>

Learn more Newton's Laws of motion here:

brainly.com/question/24522313

8 0
2 years ago
Other questions:
  • What term in physical science is defined as a change in position
    11·1 answer
  • Please help me on calculating Net Force, examples E and D in the attachment. Would be grateful if you'd explain. Thank you.
    14·1 answer
  • A solid disc with a radius of 5.00 m and a mass of 20.0 kg is initially at rests and lies on the plane of the paper. A smaller s
    13·1 answer
  • The relationship between voltage, E, current, I, and resistance, Z, is given by the equation E = IZ. If a circuit has a current
    5·2 answers
  • A total solar eclipse is visible from
    9·2 answers
  • After watching a news story about a fire in a high rise apartment building, you and friend decide to design an emergency escape
    14·1 answer
  • Which of the following is an accurate comparison of the weight of an astronaut on the moon and the Earth? The weight of the astr
    8·2 answers
  • A rectangular solid made of carbon has sides of lengths 1.0 cm, 2.0 cm, and 4.0 cm, lying along the x, y, and z axes, respective
    15·1 answer
  • wich statement is true about the law of conservation of energy for an object released from a certain height above the ground​
    10·1 answer
  • 100 degrees C equals how many degrees F?​
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!