All categories

2 years ago

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

Physics

1 answer:

BaLLatris [955]2 years ago

8 0

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>

You might be interested in

An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going

KatRina [158]

3.negative acceleration....

5 0

2 years ago

Nitrogen (n2) gas within a piston–cylinder assembly undergoes a compression from p1 = 20 bar, v1 = 0.5 m3 to a state where v2 =

Bingel [31]

Part a)

As we know that

$P_1V_1^{1.35} = P_2V_2^{1.35}$

here we know that

P1 = 20 bar

V1 = 0.5 m^3

V2 = 2.75 m^3

from above equation

$20* 0.5^{1.35} = P * (2.75)^{1.35}$

$P = 2 bar$

so final state pressure will be 2 bar

Part b)

now in order to find the work done

$W = \int PdV$

$W = \int \frac{c}{V^{1.35}}dV$

$W = c\frac{V^{-0.35}}{-0.35}$

$W = \frac{P_1V_1 - P_2V_2}{0.35}$

$W = \frac{20* 0.5 - 2 * 2.75}{0.35}* 10^5 = 12.86 * 10^5 J$

3 0

2 years ago

Read 2 more answers

A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio

dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

$I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)$

So, τ = 3/2*m*r²*α (2)

When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

$I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)$

Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

τ = 3/4*m*r² * (2α) = 3/2*m*r²

same result than in (2), so the torque remains the same.

7 0

2 years ago

The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from

Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

$L_T=L_S+10log(n)$

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

$L_T=60+10 log(5)$

= 66.98 db

7 0

1 year ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

2 years ago

Read 2 more answers