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Hatshy [7]
3 years ago
14

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

initially at rest, the carts locked together. What is the speed of the combined carts after the collision
Physics
1 answer:
mixer [17]3 years ago
8 0

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

as we know that

m_1 = 5 kg

m_2 = 2 kg

v_{1i} = 2.8 m/s

v_{2i} = 0 m/s

now from above equation we have

5(2.8) + 2(0) = m_1v+ m_2v

14 = (5+ 2) v

v = 2 m/s

so the speed of combined system is 2 m/s

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A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
4 years ago
8a. What is the equivalent resistance of the following circuit?
ollegr [7]

Answer: Take your pick

Explanation:

if they are all in parallel 1 /(1/100 + 1/300 + 1/50) = 30 Ω

if 50 is in parallel with 2 in series 1 / (1/(100 + 300) + 1/50) = 44.444...Ω

if 100 is in parallel with 2 in series 1 / (1/(50 + 300) + 1/100) = 77.777...Ω

if 300 is in parallel with 2 in series 1 / (1/(100 + 50) + 1/300) = 100 Ω

If 50 is in series with 2 in parallel 50 + 1/(1/100 + 1/300) = 125 Ω

If 100 is in series with 2 in parallel 100 + 1/(1/50 + 1/300) = 142.857...Ω

If 300 is in series with 2 in parallel 300 + 1/(1/50 + 1/100) = 333.333...Ω

If they are all in series 100 + 300 + 50 = 450 Ω

4 0
3 years ago
Which of the following are density labels? <br> a. Kg/L <br> b. g/m <br> c. g/mL <br> d. cm/g
bearhunter [10]
Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. It is a property that can be used to describe a substance.<span> </span><span>It has standard units of kg/m^3 or g/mL. So, the best answer is option C.</span>

4 0
3 years ago
Read 2 more answers
1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground
ladessa [460]

The acceleration of gravity on Earth is  9.8 m/s² .
The speed of a falling object keeps increasing smoothly,
in such a way that the speed is always 9.8 m/s faster than
it was one second earlier.

If you 'drop' the penny, then it starts out with zero speed. 
If you also start the clock at the same instant, then

         After  1.10 sec,  Speed = (1.10 x 9.8) = 10.78 meters/sec


         After  1.85 sec,  Speed = (1.85 x 9.8) = 18.13 meters/sec

But you want this second one given in a different unit of speed.
OK then:

     =  (18.13 meter/sec) x (3,600 sec/hr) x (1 mile/1609.344 meter)

     =    (18.13 x 3,600 / 1609.344)  (mile/hr)  =  40.56 mph  (rounded)

We did notice that in an apparent effort to make the question
sound more erudite and sophisticated, you decided to phrase
it in terms of 'velocity'.  We can answer it in those terms, if we
ASSUME that there is no wind, and the penny therefore doesn't
acquire any horizontal component of motion on its way down.

With that assumption in force, we are able to state unequivocally
and without fear of contradiction that each 'speed' described above ...
with the word 'downward' appended to it ... does become a 'velocity'.

3 0
3 years ago
At the north magnetic pole the earth’s magnetic field is vertical and has a strength of 0.62 gauss. The earth’s field at the sur
Anika [276]

Answer:

A) Dipole moment; m = 8.02 x 10^(22) J/T

B) I = 3.51 x 10^(9) A

Explanation:

The components of a magnetic field of a dipole are;

B_r = (μ_o•m/2πr³).cosθ

B_θ = (μ_o•m/4πr³).sin θ

B_Φ = 0

Let's make m the subject in the B_r equation ;

m = (2πr³•B_r)/(μ_o•cosθ)

Where;

B_r is magnetic field = 0.62 Gauss = 6.2 x 10^(-5) T

μ_o is the magnetic constant and has a value of 4π × 10^(−7) H/m

m is magnetic moment.

r is equal to radius of earth =6.371 x 10^(6)m

Thus, if we set θ = 0,we can solve for m as below;

m = (2π(6.371 x 10^(6))³•6.2 x 10^(-5) )/(4π × 10^(−7)•cos0)

Thus, m = 8.02 x 10^(22) J/T

B) Now, to find the current, let's use the expression for the magnetic field on the z-axis of the current ring.

B_z = (μ_o•Ib²/(2(z² + b²/2)^(3/2)))

So, let's set z = R and b = R/2

Thus, we now have;

B_z = (μ_o•I)/(5^(3/2)•R)

Making I the subject, we have;

I = [(5^(3/2)•R)•B_z]/μ_o

Plugging in the relevant values, we have;

I = [(5^(3/2) x 6.371 x 10^(6)) x 6.2 x 10^(-5)]/(4π × 10^(−7))

I = 3.51 x 10^(9) A

4 0
4 years ago
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