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kupik [55]
2 years ago
12

An impala is an African antelope capable of a remarkable vertical leaf. In one recorded leap, a 45 kg impala went into a deep cr

ouch, pushed straight up for 0.21 s, and reached a height of 2.5 m above the ground. To achieve this vertical leap, with what force did the impala push down on the ground? What is the ratio of this force to the antelope's weight? What is the ratio of this force to the antelope's weight?
Physics
1 answer:
STatiana [176]2 years ago
8 0

Answer:

F =  1500 N

F/W = 500:147

Explanation:

Using the equation of motion, to get the initial velocity

v² = u²+2gs ............. Equation 1

Where v =final velocity, u = initial velocity, a = acceleration, s = distance.

make a the subject of the equation

Given: v = 0 m/s ( at the maximum height), s = 2.5 m, g = -9.8 m/s²(against gravity)

substitute into equation 1

0² = u² +2×2.5×(-9.8)

u² = 49

u = 7 m/s.

a = u/t

Where t = time = 0.21 s

a = 7/0.21

a = 33.33 m/s²

Recall that,

F = ma ........... Equation 2

Where F = force, m = mass of the impala.

Given: m = 45 kg and a =33.33 m/s²

Substitute into equation 2

F = 45(33.33)

F =1500 N.

Hence the force =1500 N.

Weight of the antelope = mg

W = mg

Where m = 45 kg, g = 9.8

W = 441 N.

F/W =1500/441

F/W = 500:147

Hence the ratio of force to antelope weight = 500:147

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Nady [450]

6

Explanation:

The mechanical advantage is a factor that measures how input force increases using a machine.

A lever is a simple machine with the fulcrum at the center.

To calculate the mechanical advantage M.A of levers we use the expression below;

M. A = \frac{F_{b} }{F_{a} } = \frac{a}{b}

F_{a} = input force

F_{b} = output force

a is the distance of the input force from the fulcrum

b is the distance of the output force from the fulcrum

Given

a = 36cm

b = 6cm

M.A = \frac{36cm}{6cm} = 6

learn more:

Torque brainly.com/question/5352966

#learnwithBrainly

6 0
2 years ago
Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m
OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

E_m = \frac{1}{2} m\omega_1^2A_1^2

The energy of the system having mass 2m is,

E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2

For the two expressions mentioned above remember that the variables mean

m = mass

\omega =Angular velocity

A = Amplitude

The energies of the two system are same then,

E_m = E_{2m}

\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2

\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}

Remember that

k = m\omega^2 \rightarrow \omega^2 = k/m

Replacing this value we have then

\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}

But the value of the mass was previously given, then

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}

\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}

\frac{A_1}{A_2} = 1

Therefore the ratio of the oscillation amplitudes it is the same.

5 0
2 years ago
The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls
kifflom [539]
Impulse = Integral of F(t) dt from 0.012s to 0.062 s

Given that you do not know the function F(t) you have to make an approximation.

The integral is the area under the curve.

The problem suggest you to approximate the area to a triangle.

In this triangle the base is the time: 0.062 s - 0.012 s = 0.050 s

The height is the peak force: 35 N.

Then, the area is [1/2] (0.05s) (35N) = 0.875 N*s

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2 years ago
A child’s toy rake is held so that its resistance length is 0.85 meters. If the mechanical advantage is 0.43, what is the effort
mart [117]

Answer:

1.28

Explanation:

7 0
1 year ago
a 10.0 kg block on a smooth horizontal surface is acted upon by two forces: a horizontal force of 30 N acting to the right and a
Eddi Din [679]

Answer:

a=-3\ m/s^2

Explanation:

<u>Second Newton's Law</u>

It allows to compute the acceleration of an object of mass m subject to a net force Fn. The relation is given by

F_n=m.a

The net force is the sum of all vector forces applied to the object. The block has two horizontal forces applied (in absence of friction): The 30 N force acting to the right and the 60 N force to the left. The positive horizontal direction is assumed to the right, so the net force is

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\boxed{\displaystyle a=-3\ m/s^2}

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7 0
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