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dmitriy555 [2]
3 years ago
7

what specific heat of water explains why the climate doesn't vary much with season in places near large body water​

Physics
1 answer:
Crazy boy [7]3 years ago
6 0

Answer:

The specific heat of water is greater than that of dry soil, therefore water both absorbs and releases heat more slowly than land. ... This causes land areas to heat more rapidly and to higher temperatures and also cool more rapidly and to lower temperatures, compared to oceans.

Explanation:

You might be interested in
A horizontal force of 150 N is used to push a 40.0-kg packing crate a distance of 6.00 m on a rough horizontal surface. If the c
icang [17]

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

W = F x = \mu N x = \mu\ m\ g\ x

W = 150 \times 6

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

\mu = \frac {F}{m\timesg}

= \frac{150}{40\times 9.8}

= .383

We simply applied the above formulas so that each one part could calculate

3 0
3 years ago
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh
ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

  • Assuming no external forces acting during the collision, total momentum must be conserved.
  • Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
  • The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
  • So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

       p_{o1x} = 2.00 kg*m/s (1)

       p_{o1y} = 0 (2)

  • We can do the same for the particle moving along the positive y-axis:

        p_{o2x} = 0 (3)

        p_{o2y} = 4.00 kg*m/s (4)

  • Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
  • Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

       p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)

      p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s  (6)

  • Now, the total initial momentum, along these directions, must be equal to the total final momentum.
  • We can write the equation for the x- axis as follows:

       p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x}  (7)

  • We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

       p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)

  • Now, we can repeat exactly the same process for the y- axis, as follows:

       p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y}  (9)

  • We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

       p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)

  • Since we have the x- and y- components of the final momentum of  the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

       p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} }  = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)

  • We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)

  • The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

  • Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

       \frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)

  • So, the final kinetic energy has lost a 37% of the initial one.

6 0
2 years ago
 
Leto [7]

Answer:

w

Explanation:

..

6 0
3 years ago
Read 2 more answers
) The velocity function is v(t)=−t2+3t−2v(t)=−t2+3t−2 for a particle moving along a line. Find the displacement (net distance co
77julia77 [94]

Answer:

89.87m/s

Explanation:

Given the velocity function

v(t)=−t²+3t−2

In order to get the displacement function, we will integrate the velocity function as shown:

\int\limits^5_{-2} {v(t)} \, dt \\d(t)= \int\limits^5_{-2}{(-t^2+3t+2)} \, dt \\\\d(t)=[\frac{-t^3}{3}+\frac{3t^2}{2}+2t  ]^5_{-2}\\

at t = 5

d(5)=[\frac{-5^3}{3}+\frac{3(5)^2}{2}+2(5)  ]\\d(5)=[\frac{-125}{3}+\frac{75}{2}+10  ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s

at t = -2

d(-2)=[\frac{-(-2)^3}{3}+\frac{3(-2)^2}{2}+2(-2)  ]\\d(-2)=[\frac{-8}{3}+\frac{12}{2}+(-4)  ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s

Required displacement = d(5) - d(-2)

Required displacement = 89.2 - (-0.67)

Required displacement = 89.2 + 0.67

Required displacement = 89.87m/s

5 0
3 years ago
A sled is at rest at the top of a slope 2 m high. The sled has a mass of 45 kg. What is the sled's potential energy?
kap26 [50]

Answer:

PE = 882 J

Explanation:

Through the International System we know that the gravity of the earth is 9.8 m/s², so...

Data:

  • m = 45 kg
  • g = 9.8 m/s²
  • h = 2 m
  • PE = ?

Formula:

  • \boxed{\bold{PE=m*g*h}}

Replace and solve:

  • \boxed{\bold{PE=45\ kg*\ 9.8\frac{m}{s^{2}}*2\ m}}
  • \boxed{\boxed{\bold{PE=882\ J}}}

The potential energy of the sled is <u>882 Joules</u>.

Greetings.

8 0
3 years ago
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