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3 years ago

5

Which equation expresses the conservation of mechanical energy in a system where the only forms of mechanical energy are kinetic

energy and gravitational potential energy? brainly?

1 answer:

scoundrel [369]3 years ago

7 0

HI
I think the answer is:
mgh×1/2 mv2=constant.
hope it helps.

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A man walks 80m to the East and then turns around and walks back (West) a distance of 20m. What is his total distance and displa

dedylja [7]

Answer:

Total Distance is 80+20=100m

Displacement is 80-20=60m to the west

Explanation:

use a calculator, this be easy level 1

5 0

3 years ago

Explain the difference between how intrusive and extrusive igneous rock form and how those formation processes affect the minera

rusak2 [61]

Extrusive molten rocks are formed from lava while the intrusive molten rocks are formed from magma.

<u>Explanation:</u>

Extrusive molten rocks originate from lava, framing at the outside of the Earth and cooling rapidly, which means they structure little precious stones. Intrusive volcanic rocks originate from magma, shaping profound underground and taking more time to cool, which means they structure bigger crystals.

Intrusive molten rocks are framed when magma cools gradually underneath the surface, while extrusive rocks are shaped when magma cools quickly at the surface.

6 0

4 years ago

The Achilles tendon is a fibrous tissue that connects your soleus to your calcaneus and can withstand 145 MPa of tensile stress.

Mnenie [13.5K]

Answer:

F = 1.2 10⁵ N

Explanation:

Force is defined by

P = F / A

F = P A

The area is

A = π r²

the radius is half the diameter

r = d / 2

A = π d²/4

we substitute

F = P π d²/4

let's calculate

F = 145 10⁶ π (3.25 10⁻²)²/4

F = 1.2 10⁵ N

5 0

3 years ago

Lions can run at speeds up to approximately 80.0 km / h. A hungry 109 kg lion running northward at top speed attacks and holds o

sukhopar [10]

Answer:

17.34 m/s

Explanation:

Given:

Mass of lion (m₁) = 109 kg

Initial speed of lion (v₁) = 80.0 km/h (Northward direction)

Mass of gazelle (m₂) = 39.0 kg

Initial speed of gazelle (v₂) = 78.5 km/h (Eastward direction)

Final velocity of both lion and gazelle is, $v_f=?$

First, let us convert the speeds from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

Therefore,

$v_1= 80.0\ km/h=80\times \frac{5}{18}=22.22\ m/s\\\\v_2=78.5\ km/h=78.5\times \frac{5}{18}=21.81\ m/s$

Now, the concept of conservation of total momentum is used here as this is a case of perfectly inelastic collision. In inelastic collision, the masses move together with same velocity after collision.

Here, as the lion and gazelle are moving in directions at right angles to each other, the vector sum of their momentums will give the net initial momentum of the system.

So, initial momentum is given as:

$P_i=\sqrt{P_1^2+P_2^2}\\\\Where,\\\\P_1\to initial\ momentum\ of\ lion\\P_2\to initial\ momentum\ of\ gazelle$

Now, we calculate P₁ and P₂.

$P_1=m_1v_1=(109\ kg)(22.22\ m/s) = 2421.98\ Ns\\\\P_2=m_2v_2=(39\ kg)(21.81\ m/s) = 850.59\ Ns$

Therefore, the net initial momentum of the system is given as:

$P_i=\sqrt{(2421.98)^2+(850.59)^2}=2567\ Ns$

The final momentum of the system is given as:

$P_f=(m_1+m_2)(v_f)\\\\P_f=(109+39)v_f\\\\P_f=148v_f$

From the law of conservation of momentum, the final momentum is equal to the initial momentum. So,

$P_f=P_i\\\\148v_f=2567\\\\v_f=\frac{2567}{148}=17.34\ m/s$

Therefore, the final speed of the lion-gazelle system is 17.34 m/s

3 0

3 years ago

A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration

JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

$v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}$

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

$v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m$

So, the maximum height reached by the rocket is:

$h=y_1+y_2\\h=160m+154.28m=314.28m$

(c) In the first part we have:

$v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s$

And in the second part:

$t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s$

So, the time it takes to reach the maximum height is:

$t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s$

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

$v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}$

$t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s$

So, the total time is:

$t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s$

7 0

4 years ago