Ask question

Ask question

All categories

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?

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

You might be interested in

According to Newton’s law of universal gravitation, which statements are true?

Arte-miy333 [17]

Answer: The statement first and the fourth statement are true.

Explanation:

According to Newton's gravitational law, every particle in the universe attracts every other particle with the force of attraction between the masses is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

As we move to higher altitude, the force of gravity on use decreases because the force of gravity is inversely proportional to the distance.

If the masses of the two objects are more then there will be greater force of gravity between them.

Therefore, the statement first and the fourth statement are true.

7 0

2 years ago

Read 2 more answers

How much force is required to pull a spring 3.0 cm from

avanturin [10]

Answer:

I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.

I hope it helps

plz let me know if it is wrong or right.

4 0

2 years ago

3. The expression 0.62 x10^3 is equivalent to...

Korolek [52]

$\\ \sf\longmapsto 0.62\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2+3}$

$\\ \sf\longmapsto 62\times 10^1$

$\\ \sf\longmapsto 62\times 10$

$\\ \sf\longmapsto 620$

5 0

1 year ago

Read 2 more answers

A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the

frozen [14]

Answer:

$8.0\cdot 10^5 N/C$

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

$E=k\frac{q}{(R+r)^2}$

where

$k=8.99\cdot 10^9 N m^2C^{-2}$ is the Coulomb's constant

$q=2.0 \mu C=2.0 \cdot 10^{-6}C$ is the charge on the sphere

$R=10 cm = 0.10 m$ is the radius of the sphere

$r=5.0 cm = 0.05 m$ is the distance from the surface of the sphere

Substituting, we find

$E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C$

3 0

2 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0

2 years ago