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3 years ago

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

Physics

2 answers:

almond37 [142]3 years ago

8 0

That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.

yawa3891 [41]3 years ago

6 0

The red at the bottom is just the holder for the beaker. It’s not apart of the density.

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A rope attached to a truck pulls a 180−kg motorcycle at 9.0m/s.The rope exerts a 400−N force on the motorcycle at an angle of 15

steposvetlana [31]

Answer:

115911.09915 J

Unchanged

Decreased

Explanation:

m = Mass of motorcycle = 180 kg

v = Velocity of motorcycle = 9 m/s

F = Force = 400 N

s = Displacement = 300 m

$\theta$ = Angle above the horizontal = 15°

Work done is given by

$W=Fscos\theta\\\Rightarrow W=400\times 300\times cos15\\\Rightarrow W=115911.09915\ J$

The work done is 115911.09915 J

It can be seen that the force applied is constant so it does not matter if the speed of the motorcycle changes.

Tension on the rope would be

$Fcos\theta-T=ma\\\Rightarrow T=Fcos\theta-ma$

Work done

$W=Ts\\\Rightarrow W=(Fcos\theta-ma)s$

So, the work done would be reduced.

7 0

3 years ago

Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what

hoa [83]

Answer:

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² = k*1.6²

$k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}$

The value of the spring constant is 187,500N/m

7 0

3 years ago

A small but measurable current of 1.2 10-10A exists in a copper wire whose diameter is 2.0 mm. Assume the current is uniform. (a

Rom4ik [11]

Q: A small but measurable current of 1.2×10⁻¹⁰A exists in a copper wire whose diameter is 2.0 mm. Assume the current is uniform. (a) Calculate the current density.

Answer:

3.82×10⁻⁵ A/m

Explanation:

Current density: This can be defined as the amount of charge passing through a conductor, per unit area, per unit time. The S.I unit of current density is A/m²

From the question above, the expression for current density is given as,

τ = I/A............... Equation 1

Where τ = current density of the copper wire, I = current flowing through the copper wire, A = cross sectional area of the copper wire

But,

A = πd²/4................. Equation 2

Substitute equation 2 into equation 1

τ = 4I/(πd²)............ Equation 3

Given: I = 1.2×10⁻¹⁰ A, d = 2 mm = 2×10⁻³ m, π = 3.14

Substitute into equation 3

τ = 4(1.2×10⁻¹⁰)/[3.14×(2×10⁻³)²]

τ = (4.8×10⁻¹⁰)/(1.256×10⁻⁵)

τ = 3.82×10⁻⁵ A/m