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sergeinik [125]
3 years ago
14

HELP ME PLEASE *50 points*

Physics
2 answers:
aev [14]3 years ago
8 0
Their is no document for us to look at , can you add it so i can help you
kkurt [141]3 years ago
6 0

660 newtons

830 newtons

no

additional force would be 350 newtons

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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
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A 4.6-kg block of ice originally at 263 K is placed in thermal contact with a 15.7-kg block of silver (cAg = 233 J/kg-K) which i
Viktor [21]

Answer:

temperature at  326.44 K system achieve equilibrium

Explanation:

given data

mass of block of ice = 4.6 kg

temperature = 263 K

thermal contact =  15.7-kg

specific heat of silver  cAg = 233 J/kg-K

initially temperature = 1052 K

to find out

what temperature will the system achieve equilibrium

solution

first we consider final temperature of the system  is T

we know that specific heat of water (C w) = 4186 J(kg K)

and

specific heat of ice ( C i )  = 2030 J/(kg K)

and

latent heat of fusion of ice ( Lf ) = 3.33 × 10^{5} J/kg

and we know that system is insulated

so  heat lost by silver = heat generated by ice    .................1

so we can say

mass of silver × specific heat of silver  × ( initial temp - final temp ) = mass of  ice × specific heat of ice  × ( ice temp ) + mass of  ice × latent heat of fusion of ice + mass of  ice × specific heat of water  × (final temperature )  

put here value we get

mass of silver × specific heat of silver  × ( initial temp - final temp ) = mass of  ice × specific heat of ice  × ( ice temp ) + mass of  ice × latent heat of fusion of ice + mass of  ice × specific heat of water  × (final temperature )

15.7  × 233 × ( 1052 - T ) = 4.6 × 2030 × 10 + 4.6 × 3.33 × 10^{5} + 4.6 × 4286 × ( T - 273 )

solve we get

T =  326.44 K

so temperature at  326.44 K system achieve equilibrium

7 0
3 years ago
Find the velocity of a water wave in meters per second with frequency
Yuri [45]

Answer:

Velocity, v = 0.164 m/s

Explanation:

We have,

Frequency of wave, f = 0.04 Hz

Wavelength, \lambda=4.1\ m

It is required to find the velocity of a water wave. The speed of any wave is given in terms of wavelength and frequency. Its formula is :

v=f\lambda\\\\v=0.04\times 4.1\\\\v=0.164\ m/s

So, the velocity of a water wave is 0.164 m/s.

7 0
3 years ago
CAN SOMEONE TELL ME THE answer for this ?
Paladinen [302]

Answer:

b. light from earth is reflected by the moon surface

4 0
3 years ago
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An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10
Alexeev081 [22]

Answer:

a) c=1822.3214\ J.kg^{-1}.K^{-1}

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

  • mass of aluminium, m_a=0.1\ kg
  • mass of water, m_w=0.25\ kg
  • initial temperature of the system, T_i=10^{\circ}C
  • mass of copper block, m_c=0.1\ kg
  • temperature of copper block, T_c=50^{\circ}C
  • mass of the other block, m=0.07\ kg
  • temperature of the other block, T=100^{\circ}C
  • final equilibrium temperature, T_f=20^{\circ}C

We have,

specific heat of aluminium, c_a=910\ J.kg^{-1}.K^{-1}

specific heat of copper, c_c=390\ J.kg^{-1}.K^{-1}

specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)

Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)

Q_{in}=11375\ J

The heat released by the blocks when dipped into water:

Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)

where

c= specific heat of the unknown material

For the conservation of energy : Q_{in}=Q_{out}

so,

11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)

c=1822.3214\ J.kg^{-1}.K^{-1}

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0
3 years ago
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