Someone help me with this please

A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth

BaLLatris [955]

The answer would be 2.8m height on earth takes
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>

8 0

3 years ago

What part of a thunderstorm kills the most people each year?

alexandr1967 [171]

Here is the answer. The part of a thunderstorm that kills the <span> most people each year is the LIGHTNING. Thunder is only the sound created and will not hurt anyone, but it is the lightning that can kill anyone who will be struck by it. Hope this answers your question. Have a great day!</span>

5 0

3 years ago

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PLEASE HELP ME PLS PLS Imagine an object held at some height above the ground. It is released and falls toward the ground. Ignor

Stels [109]

Option e is true. The total energy is the sum of all the energies present in the system. The potential energy in a system is due to its position in the system.

<h3>What is the law of conservation of energy?</h3>

According to the Law of conservation of energy. Although energy cannot be generated or destroyed, it may be transferred from one form to another.

The following statements are true;

a)All of its energy must be potential energy before it falls.

b)At the conclusion of its fall, all of its energy must be transformed to kinetic energy.

c)During its fall, the sum of its kinetic and potential energy must match the initial quantity of potential energy.

d)Total energy = Kinetic Energy + Potential Energy.

Hence, option e is correct.

To learn more about the law of conservation of energy, refer ;

brainly.com/question/2137260

#SPJ2

8 0

2 years ago

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A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is

blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

4 0

3 years ago

A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu

lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

$F_{frk} = -\mu_{k} * F_{n} (1)$

where μk is the kinetic friction coefficient, and Fn is the normal force.

Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:

$F_{n} = F_{g} = m*g (2)$

We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

$F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)$

b)

According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
In this case, this net force is the friction force which we have already found in a).
Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
We can write the expression for a as follows:

$a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2 (4)$

c)

Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

$a = \frac{-v_{o} }{t} (5)$

Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

$t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)$

d)

From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
So, we can use the following kinematic equation in order to find the displacement before coming to rest:

$v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x (7)$

Since the puck comes to a stop, vf =0.
Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

$\Delta x = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m (8)$

e)

The total work done by the friction force on the object , can be obtained in several ways.
One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

$W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2} = -12.4 J (9)$

f)

By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
$P_{Avg} = \frac{\Delta E}{\Delta t} (10)$
If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

$P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)$

g)

The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

$P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)$

7 0

3 years ago