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Ivenika [448]
3 years ago
5

Starting at t = 0 s , a horizontal net force F⃗ =( 0.285 N/s )ti^+(-0.460 N/s2 )t2j^ is applied to a box that has an initial mom

entum p⃗ = ( -3.10 kg⋅m/s )i^+( 3.90 kg⋅m/s )j^ .
What is the momentum of the box at t = 1.90 s ?


Enter the x and y components of the momentum separated by a comma.
Physics
1 answer:
Irina-Kira [14]3 years ago
7 0

Answer:

Explanation:

We know that Impulse = force x time

impulse = change in momentum

change in momentum = force x time

Force F = .285 t -.46t²

Since force is variable

change in momentum = ∫ F dt  where F is force

= ∫ .285ti - .46t²j dt

= .285 t² / 2i - .46 t³ / 3 j

When t = 1.9

change in momentum = .285 x 1.9² /2 i  -  .46 x 1.9³ / 3 j

= .514i - 1.05 j

final momentum

= - 3.1 i + 3.9j +.514i - 1.05j

= - 2.586 i + 2.85j

x component = - 2.586

y component = 2.85

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Answer: Electromagnetic waves are generated by moving electrons. An electron generates an electric field which we can visualize as lines radiating from the electron Figure 10a. If the electron moves, say it vibrates back and forth, then this motion will be transferred to the field lines and they will become wavy Figure 10b.
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2 years ago
2. What is stroboscopic motion? -​
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Answer: The illusion of motion that occurs when a stationary object is first seen briefly in one location and, following a short interval, is seen in another location.

Explanation:

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3 years ago
An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a regi
son4ous [18]

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

\frac{1}{2}*m*v^{2}  =e*V

where :

m is the mass of electron

v is the velocity

V is the potential difference

v=\sqrt{\frac{2*e*V}{m} }    eq 1

Radius of electron moving in magnetic field is given by:

R=\frac{m*v}{q*B}       eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}

B=\sqrt{\frac{2*m*V}{e*R^{2} } }

B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3})  }{(1.60*10^{-19})*(0.170)^{2}  } }

B=9.1397*10^-4 Tesla

3 0
3 years ago
You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The
Sliva [168]

Answer:

Explanation:

Given

Height of ceiling is h=3.6\ m

Initial speed of Putty u=9.5\ m/s

Speed of Putty just before it strike the ceiling is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2-9.5^2=2\times (-9.8)\times 3.6

v^2=19.69

v=4.43\ m/s

time taken by putty to reach the ceiling

v=u+at

4.43=9.5-9.8\times t

t=\frac{5.07}{9.8}

t=0.517\ s

8 0
3 years ago
A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction
lara31 [8.8K]
<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed  = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

 

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

7 0
3 years ago
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