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3 years ago

10

A good sleeping bag (among other equipment) is important for keeping warm when camping chegg

1 answer:

Naily [24]3 years ago

5 0

Answer:

For winter camping, you want a cold-weather sleeping bag that retains as much warmth as possible. Semi-rectangular and mummy-shaped bags are probably your best bet — because they are more form-fitting and have built-in hoods, they will maximize your heat retention during cold winter nights.

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Teams A and B are in a tug-of-war challenge. Team A wins the challenge. What can be said about Team A?

Harrizon [31]

Team a was physically stronger?

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3 years ago

Read 2 more answers

What type of metals are known as ferromagnetic metals

Sphinxa [80]

Answer:

those who are attracted to magnets. ex: iron, cobalt and nickel

4 0

3 years ago

a deer with a mass of 176 kg is running head-on towards you with a velocity of 19 m/s. you are going north. find the magnitude a

Igoryamba

Momentum = Mass × Velocity

According to this formula,

Momentum of deer = 176 × 19 = 3344 kg•m/s.

Since you are heading north and the deer is running towards you, the direction of the deer' s momentum is north as well.

6 0

4 years ago

Whats the net force

zavuch27 [327]

Answer:

13n pushing left

Explanation:

1563 - 1550

3 0

4 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago