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TEA [102]
3 years ago
14

The kangaroo stops at a lake for a drink of water, then starts hopping again to the south. After 5 seconds, the kangaroo is goin

g 2.5 m/s south. What is the kangaroo’s acceleration?
Physics
1 answer:
kiruha [24]3 years ago
6 0

Answer:

<u>A kangaroo hops 60 m to the east in 5 s. What is the kangaroo's average velocity? ... The kangaroo stops at a lake for a drink of water and then starts hopping again to the south. Each second, the kangaroo's velocity increases 2.5 m/s.</u>

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A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of th
Artemon [7]

Answer:

0.14 J

Explanation:

The maximum velocity is the amplitude times the angular frequency.

vmax = Aω

ω = vmax / A

ω = (3.2 m/s) / (0.06 m)

ω = 53.3 rad/s

For a spring-mass system:

ω = √(k / m)

ω² = k / m

k = ω²m

k = (53.3 rad/s)² (0.050 kg)

k = 142 N/m

The elastic potential energy is:

EE = ½ kx²

EE = ½ (142 N/m) (0.044 m)²

EE = 0.14 J

6 0
3 years ago
A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa
aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:        

The expression for the power radiated is as follows;

P=A\epsilon\sigma T^{4}

Here, A is the area, \sigma is the stefan's constant,\epsilon is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}

Here, A_{1} is the area of the sphere, A_{2} is the area of the cube,T_{1}  is the temperature of the sphere and T_{2}  is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

A_{1}T_{1}^{4}=A_{2}T_{2}^{4}

The area of the sphere is A_{1}=4\pi \times r^{2}.

Here, r is the radius of the sphere.

The area of the cube is A_{2}=6\times a^{2}.

Here, a is the edge of the cube.

Put A_{1}=4\pi \times r^{2} and A_{2}=6\times a^{2}.

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}  ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

V_{1}=V_{2}

Here,V_{1},V_{1} are the volumes of the sphere and the cube.

\frac{4}{3}\pi r^{3}=a^{3}

\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}  

Put this value in the equation (1).

T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}

Put T_{1}=500°c.

T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}

T_{2}=472.2^{\circ}c

Therefore, the value of the celsius temperature of the cube is 472.7°c.    

5 0
3 years ago
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