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Morgarella [4.7K]
4 years ago
11

An acorn with a mass of 0.0300 kilograms falls from a tree. What is its kinetic energy when it reaches a velocity of 5.00 m/s? 1

.47 J .075 J 0.375 J cannot be determined without knowing how far the acorn has fallen
Physics
2 answers:
Tju [1.3M]4 years ago
6 0
KE = 1/2 m v^2.

That's 0.375 J.
ozzi4 years ago
6 0

Answer

Kinetic energy of acorn is 0.375 J .

Explanation:

Formula

Kinetic energy is defined by

K.E = \frac{1}{2}mv^{2}

Where m is mass and v is velocity .

As given

An acorn with a mass of 0.0300 kilograms falls from a tree with a velocity of 5.00 m/s .

m = 0.0300 Kg

v = 5 m/s

Putting all the values in the formula

K.E = \frac{1}{2}\times 0.0300\times 5\times 5

K.E = \frac{1\times 0.0300\times 5\times 5}{2}

K.E = \frac{0.75}{2}

K.E = 0.375 J

Therefore the kinetic energy of acorn is 0.375 J .

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A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s
posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m

The spring constant of the net is 20130.76 N

From Hooke's Law

F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m

The net would strech 0.03167 m

If h = 35 m

From energy conservation

65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0

Solving the above equation we get

x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45

The compression of the net is 1.52 m

4 0
4 years ago
A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d
Ad libitum [116K]

Answer:

Option (D) is correct.

Explanation:

The balloon lands horizontally at a distance of 420 m from a point where it as released.

Velocity of air balloon along +X axis =10 m/s

velocity of ball=4 m/s along + X axis

the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s

time taken= 30 s

The distance traveled is given by d= v t

d= 14 (30)

d= 420 m

Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.

6 0
3 years ago
My older brother received a ticket for driving 80 miles per hour. This information describes my brother's
natima [27]
Your brother would be described as a speeder.
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3 years ago
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A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
4 years ago
If i = 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N ⋅ m , what are the lengt
Alik [6]

Answer:

Length of the sides of the square loop is given by

s = √[(τ)/(NIB sin θ)]

Explanation:

The torque, τ, experienced by a square loop of area, A, with N number of turns around the loop and current of I flowing in the wire, with a magnetic field presence, B, and the plane of the loop tilted at angle θ to the x-axis, is given by

τ = (N)(I)(A)(B) sin θ

If everything else is given, the length of a side of the square loop, s, can be obtained from its Area, A.

A = s²

τ = (N)(I)(A)(B) sin θ

A = (τ)/(NIB sin θ)

s² = (τ)/(NIB sin θ)

s = √[(τ)/(NIB sin θ)]

In this question, τ = 0.076 N.m, I = 1.70 A

But we still need the following to obtain a numerical value for the length of a side of the square loop.

N = number of turnsof wire around the loop

B = magnetic field strength

θ = angle to which the plane of the loop is tilted, measured with respect to the x-axis.

6 0
3 years ago
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