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Whitepunk [10]
3 years ago
14

65 POINTS! PLEASE ANSWER EVERY QUESTION! NEED HELP ASAP!

Physics
1 answer:
otez555 [7]3 years ago
3 0
Maybe you can split up the questions. I will try to answer your first question.

1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)

2. Momentum: p = m₁v₁ + m₂v₂

m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B

Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0

Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0

3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.

4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7

5.You figure out.



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A particle is moving along the x-axis so that its position at any time t is greater than and equal to 0 is given by x(t)=2te^-t?
erastovalidia [21]
For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left
6 0
3 years ago
d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s
Harrizon [31]

Answer:

a' = 0.35 m/s^2

Explanation:

Let say the empty wagon has mass "M"

now by newton's II Law we will have

F = Ma

now it is given that empty wagon is pulled with acceleration 1.4 m/s/s

now we will have

F = 1.4 M

now a child of mass three times the mass of wagon is sitting on the empty wagon

so here we have

F = (M + 3M) a

1.4 M = 4M a'

so we have

a' = 0.35 m/s^2

8 0
3 years ago
A 0.290 kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato
Sergeu [11.5K]

Answer:

A) The speed of the potato at the lowest point of its motion is 7.004 m/s

B) The tension on the string at this point is 8.5347 N

Explanation:

Here we have that the height from which the potato is allowed to swing  is 2.5 m

Therefore we have ω₂² = ω₁² + 2α(θ₂ - θ₁)

Where:

ω₂ = Final angular velocity

ω₁ = Initial angular velocity = 0 rad/s

α = Angular acceleration

θ₂ = Final angle position

θ₁ = Initial angle position

However, we have potential energy of the potato

= Mass m×Gravity g× Height h

= 0.29×9.81×2.5 = 7.1125 J

At he bottom of the swing, the potential energy will convert to kinetic energy as follows

K.E. = P.E. = 7.1125 J

1/2·m·v² = 7.1125 J

Therefore,

v² = 7.1125 J/(1/2×m) = 7.1125 J/(1/2×0.290) = 49.05

∴ v = √49.05 = 7.004 m/s

B) Here we have the tension given by

Tension T in the string = weight of potato + Radial force of motion

Weight of potato = mass of potato × gravity

Radial force of motion of potato = mass of potato × α,

where α = Angular acceleration = v²/r and r = length of the string

∴ Tension T in the string = m×g + m×v²/r = 0.290×(9.81 + 7.004²/2.5)

T = 8.5347 N

4 0
3 years ago
Read 2 more answers
The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the
Studentka2010 [4]

Answer:

V(t)= 240V* H(t-5)

Explanation:

The heaviside function is defined as:

H(t) =1 \quad t\geq 0\\H(t) =0 \quad t

so we see that the Heaviside function "switches on" whent=0, and remains switched on when t>0

If we want our heaviside function to switch on when t=5, we need the argument to the heaviside function to be 0 when t=5

Thus we define a function f:

f(t) = H(t-5)

The -5 term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (H(t-5) =1 when t\geq 5, so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

V(t)= 240V* H(t-5)

I have made a sketch for you, and added it as attachment.  

5 0
3 years ago
A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at
Vesna [10]

Answer:

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

1 x 32= k x 2

K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

my''+Cy'+Ky=F

Now by putting the values

1 y"+8 y'+ 16y=6 cos 4 t ----(1)

The general solution of equation Y=CF+IP

Lets assume that at steady state the equation of y will be

y(IP)=A cos 4t+ B sin 4t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -4A sin 4t+4B cos 4t

y"=-16A cos 4t-16B sin 4t

Now put the values of y" , y' and y in equation 1

1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

So by comparing the coefficient both sides

-16A+32B+16A=0  So B=0

-16 B-32 A+16B=6  So A=-3/16

y=-3/16 cos 4t

Now to find the CF  of differential equation 1

y"+8 y'+ 16y=6 cos 4 t

Homogeneous version of above equation

m^2+8m+16=0

So CF =(C_1+tC_2)e^{-2t}

So the general equation

Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t

Given that t=0 Y=0 So

C_1=\dfrac{3}{16}

t=0 Y'=0 So

C_2 =\dfrac{3}{8}

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

The above equation is the general equation for motion.

3 0
3 years ago
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