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2 years ago

Please help. All the information is in the image.

Physics

1 answer:

kondaur [170]2 years ago

5 0

The tiger's position at time t is given by

x (horizontal) = (4.5 m/s) t

y (vertical) = 7.5 m - 1/2 gt ²

Solve y = 0 for t to find the time it takes for the tiger to reach the ground :

0 = 7.5 m - 1/2 (9.8 m/s²) t ²

===> t = √(2 (7.5 m) / (9.8 m/s²)) ≈ 1.2 s

Evaluate x at this time :

x = (4.5 m/s) (1.2 s) ≈ 5.6 m

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Answer: find the attached files for the answer

Explanation:

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Please find the attached files for the solution

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3 years ago

The block in the drawing has dimensions L0×2L0×3L0, where L0 =0.5 m. The block has a thermal conductivity of 250 J/(s·m·C˚). In

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3 years ago

The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah

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3 years ago

A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses

stellarik [79]

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

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3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago