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joja [24]
3 years ago
6

While visiting the beach, you enjoy the warm ocean water, but the sand burns your feet. That night you walk along the beach and

notice that the sand is colder than the ocean water. Why?
Group of answer choices

It takes a long time for sand to heat up, but it cools down very quickly. Water takes a short time to heat up and cool down.

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

Water is naturally colder than sand.

Sand is naturally colder than water.
Physics
1 answer:
garik1379 [7]3 years ago
8 0

The answer would be B..

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

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Answer:

Protons , neutrons and elections
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3 years ago
A jogger accelerates from rest to 4.86 m/s in 2.43 s. A car accelerates from 20.6 to 32.7 m/s also in 2.43 s. (a) Find the magni
Aleonysh [2.5K]

Explanation:

It is given that,

Initially, the jogger is at rest u₁ = 0

He accelerates from rest to 4.86 m, v₁ = 4.86 m

Time, t₁ = 2.43 s

A car accelerates from u₂ = 20.6 to v₂ = 32.7 m/s in t₂ = 2.43 s

(a) Acceleration of the jogger :

a=\dfrac{v-u}{t}

a=\dfrac{4.86\ m/s-0}{2.43\ s}

a₁ = 2 m/s²

(b) Acceleration of the car,

a=\dfrac{v-u}{t}

a=\dfrac{32.7\ m/s-20.6\ m/s}{2.43\ s}

a₂ = 4.97 m/s²

(c) Distance covered by the car,

d_1=u_1t_1+\dfrac{1}{2}a_1t_1^2

d_1=0+\dfrac{1}{2}\times 2\times (2.43)^2

d₁ = 5.904 m

Distance covered by the jogger,

d_2=u_2t_2+\dfrac{1}{2}a_2t_2^2

d_2=20.6\times 2.43+\dfrac{1}{2}\times 4.97\times (2.43)^2

d₂ = 64.73 m

The car further travel a distance of, d = 64.73 m - 5.904 m = 58.826 m

Hence, this is the required solution.

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2 years ago
Which of the following helps in designing an attractive 3-D craft?
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Answer:

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Explanation:

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The shuttles main engine provides 154,360 kg of thrust for 8 minutes. If the shuttle accelerated at 29m/s/s, and fires for at le
Vinil7 [7]

Answer:

The answer to the question is

3340800 m far

Explanation:

To solve the question, we note that acceleration = 29 m/s²

Time of acceleration = 8 minutes

Then if the shuttle starts from rest, we have

S = u·t+0.5·a·t² where u = 0 m/s = initial velocity

S = distance traveled, m

a = acceleration of the motion, m/s²

t = time of travel

S = 0.5·a·t² = 0.5×29×(8×60)² = 3340800 m far

3 0
3 years ago
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef
bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0
3 years ago
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