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Dovator [93]
3 years ago
5

If an antelope is running at a speed of 10 mis, and can maintain that horizontal velocity when it jumps, how high must it jump i

n order to clear a horizqntal distance of 20m ? .)
Physics
1 answer:
pochemuha3 years ago
7 0

Answer:

Explanation:

Given

velocity of antelope v=10 m/s

antelope needs to cover a horizontal distance of 20 m

considering horizontal velocity antelope needs to cover 20 m in t sec which is given by

t=\frac{20}{10}=2 s

i.e. in 2 sec antelope needs to jump and return to ground for zero vertical displacement

h=ut+\frac{1}{2}at^2

h=vertical displacement

a=acceleration

t=time

for,  h_{max}, t=1 s

h_{max}=0+\frac{1}{2}g(1)^2

h_{max}=\frac{g}{2}

h_{max}=4.9 m

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6 0
3 years ago
Read 2 more answers
Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; H_0 = 51km/s/Mly

Age of the universe; t = \ ?

We know that, the reciprocal of the Hubble's constant ( H_0 ) gives an estimate of the age of the universe ( t ). It is expressed as:

Age\ of\ Universe; t = \frac{1}{H_0}

Now,

Hubble's constant; H_0 = 51km/s/Mly

We know that;

1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
2 years ago
TRUE OR FLASE
BaLLatris [955]

Answer:

true for first and false for second

Explanation:

5 0
2 years ago
Two technicians are explaining the information on a repair order. Technician A says that the cause written on the repair order i
bezimeni [28]
Technician A and B are correct . Because according to technician A, the cause written on the repair order is a diagnosis. Here, by diagnosis, he means that the problem is identified after examining the device and hence the judgement is made.
And according to B, you have to write the cause of the problems in the device that have been identified and the concern measures, which is also kind of diagnosis.

So, option D is correct.
4 0
3 years ago
A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on
Juliette [100K]

Answer:

k = 5\times 10^{4}\ N/m

b = 0.707\times 10^{3}

t = 7.1\times 10^{- 5}\ s

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

v^{2} = u^{2} + 2gh

where

u = initial velocity = 0

g = 10m/s^{2}

Thus

v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s

Force on the mass is given by:

F = mg = 1000\times 10 = 10000 N = 10\ kN

Also, we know that the spring force is given by:

F = - kx

Thus

k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m

Now, to find the damping constant b, we know that:

F = - bv

b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s

4 0
3 years ago
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