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il63 [147K]
2 years ago
8

A student sects a leaf of length 7.2 cm to draw. Her drawing is 28.8 cm in length. What is the magnification of the drawing?

Physics
1 answer:
Alina [70]2 years ago
7 0

Answer:

A) x4

Explanation:

Magnification is equal to image size divided by the actual size, or M = I/A.

The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:

28.8 cm/7.2 cm = 4

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An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof
Pavel [41]

Answer:

T = 69^o C

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

Q_{ice} = Q_{coffee}

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2

here we have

m_1 = 15 gram

L = 333 kJ/kg = 333 J/g[/tex]

c_1 = c_2 = 4186 J/kg C = 4.186 J/g C

\Delta T_1 = T - 0

\Delta T_2 = 85 - T

now we have

15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)

4995 + 62.79T = 49813.4 - 586.04T

648.83 T = 44818.4

T = 69^o C

6 0
3 years ago
Which explains the information needed to calculate speed and velocity?
goblinko [34]

The second statement is the correct choice.  Don't make me type it out.

8 0
3 years ago
Read 2 more answers
Which conditions are low air pressure systems usually associated with?
Inga [223]

cloudy, wet weather                                          

4 0
3 years ago
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Which type of force absorbs shock in vehicles
Licemer1 [7]

Answer:

Vehicles typically employ both hydraulic shock absorbers and springs or torsion bars. In this combination, "shock absorber" refers specifically to the hydraulic piston that absorbs and dissipates vibration.

Explanation:

hope this helps

5 0
3 years ago
A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe
Liula [17]

Answer:

a) v₁fin = 3.7059 m/s   (→)

b) v₂fin = 1.0588 m/s     (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s   ⇒  Kin = 0 J

v₁fin = ?

h<em>in </em>= L = 0.7 m

h<em>fin </em>= 0 m   ⇒    U<em>fin</em> = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s   (→)

b)  Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s    (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒  v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒  v₂fin = 1.0588 m/s     (→)

7 0
3 years ago
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