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Elden [556K]
3 years ago
11

A kangaroo can jump straight up to a height of 2.0 m. What is its takeoff speed

Physics
1 answer:
12345 [234]3 years ago
5 0
7.17m/s glad I could help
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What is the slit spacing of a diffraction necessary for a 600 nm light to have a first order principal maximum at 25.0°?
Sauron [17]

Explanation:

Given that,

Wavelength of light, \lambda=600\ nm=6\times 10^{-7}\ m

Angle, \theta=25^{\circ}

We need to find the slit spacing for diffraction. For a diffraction, the first order principal maximum is given by :

d\sin\theta=n\lambda

n is 1 here

d is slit spacing

d=\dfrac{\lambda}{\sin\theta}\\\\d=\dfrac{6\times 10^{-7}}{\sin(25)}\\\\d=1.41\times 10^{-6}\ m\\\\d=1.41\ \mu m

So, the slit spacing is 1.41\ \mu m.

6 0
3 years ago
Using the
xxMikexx [17]

Answer:

The principle of momentum conservation states that if there no external force the total momentum of the system before and after the collision is conserved.

Since momentum is a vector, we should investigate the directions and magnitudes of initial and final momentum.

\vec{P}_{initial} = \vec{P}_{final}\\\vec{P}_{initial} = m_1\vec{v}_1 + 0\\\vec{P}_{final} = m_1\vec{v}_1' + m_2 \vec{v}_2'

If the first ball hits the second ball with an angle, we should separate the x- and y-components of the momentum (or velocity), and apply conservation of momentum separately on x- and y-directions.

6 0
3 years ago
Which element below has chemical properties similar to the chemical properties of fluorine.
SVETLANKA909090 [29]
For number one the answer is Iodine because it is in the same group as fluorine. For number two the answer is Germanium for the same reason. For number three the answer is Aluminum for the same reason.
4 0
3 years ago
A yo-yo of mass M has an axle of radius b and a spool of radius R. Its moment of inertia can be taken to be MR2/2 and the thickn
kow [346]

Answer:

The tension in the cord is T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

Explanation:

Given:

M = mass

b = radius

R = spool of radius

The equation is:

bT=(\frac{MR^{2} }{2} )(\frac{a}{b} )\\T=\frac{MR^{2}a }{2b^{2} } (eq. 1)

The sum of forces in y:

∑Fy = Mg - T = Ma

Mg=(M+\frac{MR^{2} }{2b^{2} }  )a\\a=\frac{2b^{2}g }{2b^{2}+R^{2}  }

Replacing in eq. 1

T=\frac{MR^{2} }{2b^{2} } (\frac{2b^{2}g }{2b^{2} +R^{2} } )\\T=\frac{MR^{2}g }{2b^{2}+R^{2}  }

3 0
3 years ago
Read 2 more answers
A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori
olasank [31]

Explanation:

The given data is as follows.

        k = 130 N/m,       \Delta x = 17 cm = 0.17 m   (as 1 m = 100 cm)

     mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

             Energy = \frac{1}{2}kx^{2}

Now, spring energy gets converted into kinetic energy when the box is launched.

So,    \frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}

   \frac{1}{2} \times 130 \times (0.17)^{2} = \frac{1}{2} \times 2.8 \times v^{2}

          v^{2} = \frac{3.757}{2.8}

                     = 1.34

                v = 1.15 m/sec

Now,

           Frictional force = \mu \times mg

                                    = 0.15 \times 2.8 \times 9.8

                                    = 4.116 N

Also,  Kinetic energy = work done by friction

           \frac{1}{2}mv^{2} = F_{f} \times d

           \frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d

             1.8515 = 4.116 \times d

                 d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0
3 years ago
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