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3 years ago

A kangaroo jumps to a vertical height of 251 cm. how long (in seconds) was the kangaroo in the air?

Physics

1 answer:

julsineya [31]3 years ago

3 0

1.08 seconds i think. Is there no other option like how many cm they fall per second? That would help a lot.

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A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0

3 years ago

Pam has a mass of 42.4 kg and she is at rest on smooth, level, frictionless ice. Pam straps on a rocket pack. The rocket supplie

fredd [130]

Answer:

F = 2349.6 N

Explanation:

We can solve this exercise using the relationship of momentum and momentum

I = Δp

I = F t

As the woman accelerates at a distance of 29.1 m to go from rest to 56.8 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v / 2x

a = 56.8 2/2 29.1

a = 55.43 m / s²

Let's look for the time you need to get this speed

v = v₀ + a t

t = v / a

t = 56.8 / 55.43

t = 1,025 s

Let's clear the average force momentum from the momentum

F t = m v- m v₀

F = mv / t

F = 42.4 56.8 / 1.025

F = 2349.6 N

3 0

3 years ago

A car drives 40km West, and then 160km North. What is the displacement of the car?

Galina-37 [17]

Answer: 1

Explanation:

8 0

3 years ago

Read 2 more answers

Can someone check my answers please??

loris [4]

I checked them they are good.

3 0

4 years ago

Read 2 more answers

If dye x has a molar absorptivity, ε = 4.50 x 104 cm-1 m-1, at a certain wavelength, what concentration of the dye, in a 1.00 cm

NeTakaya

For this problem, the working equation should be used from the Beer's Law:

A = ∈lc,
where
A is the absorbance
∈ is the molar absorptivity
l is the length of path of the cuvette diameter
C is the concentration of the sample placed inside the cuvette

Substituting the values:

0.417 = (4.50×10⁴ cm⁻¹ M⁻¹<span>)(1 cm)(C)
Solving for C:
C = 9.27</span>×10⁻⁶ M

8 0

4 years ago