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Paladinen [302]
2 years ago
10

What is meant by specific latent heat of fussion​

Physics
1 answer:
Lunna [17]2 years ago
5 0

Answer:

Boiling point of water is 100 deg C. It takes about 540 cal to change 1 g of water at the boiling point to 1 g of steam at 100 deg C.

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Be sure to show all 4 steps for each of the problems below.
Kisachek [45]

Answer:

1=920

2=2300

Explanation:

from the question,

M¹=23

H¹=4

g=10(constant)

using the formula

P. E,= mgh

P. E= 23×10×4=920

P. E=23×10×10=2300

7 0
4 years ago
A wildlife researcher is tracking a flock of geese. The geese fly 4.0km due west, then turn toward the north by 40 degrees and f
BaLLatris [955]

Answer:

(a). The distance from the initial position is 6.68 km.

(b).  The magnitude of their displacement is 7.05 km.

Explanation:

Given that,

Geese fly 4.0 km due west, then turn to north by 40° and fly another 3.5 km.

(a). We need to calculate the distance from the initial position

Using formula of distance

D=AB+BC\cos\theta

Put the value into the formula

D=4.0+3.5\cos40

D=6.68\ km

(b). We need to calculate the magnitude of their displacement

Using formula of displacement

AC=\sqrt{CD^2+AD^2}

AC=\sqrt{(3.5\sin40)^2+(6.68)^2}

AC=7.05\ Km

Hence, (a). The distance from the initial position is 6.68 km.

(b).  The magnitude of their displacement is 7.05 km.

8 0
3 years ago
Read 2 more answers
How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr
Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

Q =mc\Delta T

where,

m = mass

c = specific heat

\Delta T = Change in temperature

Therefore the total heat exchange is given as

\Delta Q = Q_w+Q_v

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

Our values are given as,

Total mass is M_T = 200lb ,however the mass of solid vegetable and water is given as,

m_v= 0.4*200lb = 80lb

m_w=0.6*200lb=120lb

T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF

Replacing at our equation we have,

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)

\Delta Q = 22411.2Btu

Therefore the heat removed is 22411.2 Btu

6 0
3 years ago
Objects A and B both start at rest. They both accelerate at the same rate. However, object A accelerates for 3x the time as obje
mylen [45]

Answer:

v_Object A = 3v_Object B

OR

1/3v_Object A = v_Object B

Explanation:

Let's see what variables we have in this problem.

Since the objects both start at rest, we have an initial velocity of 0 m/s.

The objects accelerate, so we will have acceleration.

Object A accelerates for 3x the time as Object B, so we have time.

The problem wants us to compare their final speeds, so we have final velocity.

Check what constant acceleration kinematic equation has these variables:

  • v = v₀ + at  

Let's create values for the unknown variables and compare the final velocities.

  • v = ?
  • v₀ = 0 m/s (objects start at rest)
  • a = 5 m/s²
  • t = 10 s

Since Object A accelerates for 3x the time as Object B, we can use t = 30 s for Object A.

Let's write the two equations:

  1. v = (0) + (5)(10)
  2. v = (0) + (5)(30)

Simplify these equations.

  1. v = 50 m/s
  2. v = 150 m/s

Let's use another set of values to compare the final velocities to see if the velocities differ by 100 m/s or Object A has 3x the final velocity of Object B.

  • v = ?
  • v₀ = 0 m/s (objects start at rest)
  • a = 3 m/s²
  • t = 7 s

Write the two equations:

  1. v = (0) + (3)(7)
  2. v = (0) + (3)(21)  

Simplify these equations.

  1. v = 21 m/s
  2. v = 63 m/s

Now we can clearly see that the final velocities are differed by 3x. Object A has 3x the final speed compared to that of Object B.

3 0
3 years ago
A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
4 years ago
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