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Tamiku [17]
3 years ago
12

The specific heat capacity of sea water is 4100 J/Kg°C and the boiling point of 100.6 °C. (i) Calculate the energy required to r

aise the temperature of 0.900 kg of this sea water from 10 °C up to its boiling point. Also mention the equation to be used. *
Physics
1 answer:
maw [93]3 years ago
6 0

Answer:

334.314 (kJ)

Explanation:

1) the formula for the required energy is: Q=c*m(Bp-t), where c - 4100 J/kg*C; m - 0.9 kg; Bp - 100.6 C; t - 10 C.

2) according to the formula above:

Q=4100*0.9*(100.6-10)=41*9*906=334314 (J).

You might be interested in
The airplane is flying with a constant velocity. Which force acting on the airplane below represents the friction from air resis
tankabanditka [31]

The correct answer to the question is  C i.e C represents the friction from air resistance.

EXPLANATION:

Before coming into any conclusion, first we have to understand friction.

The friction is the opposing force which acts tangentially between two bodies in contact when there is a relative motion between them.

The air resistance is that frictional force which is provided by the air to the moving body through it. Hence, the friction from air resistance will be directed opposite to the motion of the body.

In the given diagram, the airplane is going horizontally. The force A acts in forward direction while force C acts in backward direction. The forces B and D are acting vertically.  There is no motion in vertical direction. Hence, the net force of A and C will cause the airplane to move.

As the plane is moving along the direction of A, the frictional force must act along the direction of C.

8 0
3 years ago
Read 2 more answers
A 1200-kg ore cart is rolling at 10.8 m/s across a flat friction-free surface. a crane suddenly drops of ore vertically into the
Andre45 [30]
The value of the final speed depends on the mass of the ore.

Let's call m the mass of the ore. We can solve the exercise by requiring the conservation of momentum, which must be the same before and after the ore is loaded.

Initially, there is only the cart, so the momentum is
p=Mv=(1200 kg)(10.8 m/s)=12960 kg m/s
After the ore is loaded, the new mass will be (1200 kg+m), and the new speed is v_f. The momentum p is conserved, so it is still 12960 kg m/s. Therefore, we have
p=12960 kg m/s =(1200 kg+m)v_f
and so the final speed is
v_f =  \frac{12960 kg m/s}{1200 kg +m}
5 0
4 years ago
A squirrel jumps into the air with a velocity of 4 m/s at an angel of 50 degrees. What is the maximum height reached by the squi
Debora [2.8K]

Answer:

Explanation:

Assuming the squirrel is jumping off the ground, here's what we know but don't really know...

v₀ = 4.0 at 50.0°

So that's not really the velocity we are looking for. We are dealing with a max height problem, which is a y-dimension thing. Therefore, we need the squirrel's upward velocity, which is NOT 4.0 m/s. We find it in the following way:

v_{0y}=4.0sin(50.0) which gives us that the upward velocity is

v₀ = 3.1 m/s

Moving on here's what we also know:

a = -9.8 m/s/s and

v = 0

Remember that at the very top of the parabolic path, the final velocity is 0. In order to find the max height of the squirrel, we need to know how long it took him to get there. We are using 2 of our 3 one-dimensional equations in this problem. To find time:

v = v₀ + at and filling in:

0 = 3.1 - 9.8t and

-3.1 = -9.8t so

t = .32 seconds.

Now that we know how long it took him to get to the max height, we use that in our next one-dimensional equation:

Δx = v_0t+\frac{1}{2}at^2 and filling in:

Δx = 3.1(.32)+\frac{1}{2}(-9.8)(.32)^2 and using the rules for adding and subtracting sig fig's correctly, we can begin to simplify this:

Δx = .99 - .50 so

Δx = .49 meters

7 0
3 years ago
A 60.0 kg soccer player kicks a 0.4000 kg stationary soccer ball with 6.25 N of force. How fast does the soccer ball accelerate,
frozen [14]
F = ma
6.25 N = 0.4 kg · a
a = (6.25/0.4) m/s²      since N=kg·m/s²
a = 15.625 m/s² 

The answer is c) 15.6 m/s²
(Note that the mass of the soccer player is irrelevant.)
8 0
3 years ago
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
3 years ago
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