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3 years ago

HELP: This is what I think can you correct me?

Physics

2 answers:

lora16 [44]3 years ago

7 0

It is indeed decreasing

Archy [21]3 years ago

4 0

Answer:

I think it is decreasing because it gets farther and the other 2 cars look smaller than the first so I also think it is decreasing

You might be interested in

Define Archemedics principle?

azamat

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.

6 0

4 years ago

A laboratory dish, 20 cm in diameter, is half filled with water. One at a time, 0.50/µL drops of oil from a micropipette are dro

aleksklad [387]

Answer:

n = 1.45

Explanation:

Given:

- Diameter of the dish d = 0.2 m

- Volume per drop V/drop = 0.5 uL

- wavelength of incident light λ = 600 nm

- minimum intensity @ 13 drops

- refractive index of oil : n

Find:

What is the index of refraction of the oil?

Solution:

- Destructive interference can be seen @ 13 drops. So we will use the relation of destructive interference:

n = m*λ / 2*t

- Compute the thickness of the film.

Volume of film = 0.5*13*10^-9 = 6.5*10^-9 m^3

Volume of cylinder = pi*d^2 / 4 * t

pi*d^2 / 4 * t = 6.5*10^-9 m^3

t = 4*(6.5*10^-9 )/ pi*0.2^2 = 2.069*10^-7 m

- Use the given relation @ m = 1:

n = λ / 2*t

n = (600*10^-9) / 2*2.069*10^-7

n = 1.45

3 0

4 years ago

A light ball is dropped straight down off a ledge and hits the ground. A heavy ball is thrown straight down from the ledge, rele

Anettt [7]

Answer:

(d) I and III

Explanation:

Motion of the 1st light ball

$s = gt_1^2/2$

$s = v_0t_2 + gt_2^2/2$

I. Since the heavier starts with an initial velocity v0, and both balls end up traveling the same distance s. From the 2 motion equation above we can conclude that their time must be different

II. Both balls are affected by the same gravitational acceleration g

III. Velocity before the impact of the light ball

$v_1^2 = 2gs$

And the heavy ball

$v_2^2 - v_0^2 = 2gs$

Since they have the same acceleration g and distance s. But the heavy ball has an initial push. It would end up with a larger speed.

7 0

3 years ago

1-) Two blocks of mass 3.SO kg and 8.00 kg are connected by a massless string that passes over a

Troyanec [42]

Answer:

The magnitude of the acceleration of each block is, a = 2.56 m/s²

The tension in the string is, T = 43.05 N

Explanation:

Given data,

The larger block of mass, M = 8.00 kg

The smaller block of mass, m = 3.50 kg

The formula for Atwood machine is,

Ma = Mg - T

ma = T - mg

Adding those equations,

a (M + m) = g ( M - m)

a = (M + m) / ( M - m)

Substituting the values,

a = (8 + 3.5) / (8 - 3.5)

= 2.56 m/s²

The magnitude of the acceleration of each block is, a = 2.56 m/s²

The tension in the string,

T = m(a + g)

= 3.5 ( 2.56 + 9.8)

= 43.05 N

The tension in the string is, T = 43.05 N

3 0

3 years ago

Use the table below to answer the following questions. Substance Specific Heat (J/g•°C) water 4.179 aluminum 0.900 copper 0.385

lbvjy [14]

1. $-8.78 \cdot 10^5 J$

The energy lost by the water is given by:

$Q=m C_s \Delta T$

where

m = 3.0 kg = 3000 g is the mass of water

Cs = 4.179 J/g•°C is the specific heat

$\Delta T=10.0C-80.0C=-70.0 C$ is the change in temperature

Substituting,

$Q=(3000 g)(4.179 J/gC)(-70.0 C)=-8.78 \cdot 10^5 J$

2. $3.24 \cdot 10^4 J$

The energy added to the aluminium is given by:

$Q=m C_s \Delta T$

where

m = 0.30 kg = 300 g is the mass of aluminium

Cs = 0.900 J/g•°C is the specific heat

$\Delta T=150.0 C-30.0C =120.0 C$ is the change in temperature

Substituting,

$Q=(300 g)(0.900 J/gC)(120.0 C)=3.24 \cdot 10^4 J$

3a. $-5.6^{\circ}C$

The temperature change of the water is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = -232 kJ=-2.32\cdot 10^5 J$ is the heat lost by the water

$m=10.0 kg=10000 g$ is the mass of water

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{-2.32\cdot 10^5 J}{(10000g)(4.179 J/gC)}=5.6^{\circ}C$

3b. $+10.2^{\circ}C$

The temperature change of the copper is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = 1.96 kJ=1960$ is the heat added to the copper

$m= 500 g$ is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{1960 J}{(500g)(0.385 J/gC)}=10.2^{\circ}C$

4. 42.9 g

The mass of the water sample is given by

$m=\frac{Q}{C_S \Delta T}$

where

$Q=4300 J$ is the heat added

$\Delta T=39 C-15 C=24C$ is the temperature change

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$m=\frac{4300 J}{(4.179 J/gC)(24 C)}=42.9 g$

5. 115.5 J

The heat used to heat the copper is given by:

$Q=m C_s \Delta T$

where

m = 5.0 g is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

$\Delta T=80.0 C-20.0C =60.0 C$ is the change in temperature

Substituting,

$Q=(5.0 g)(0.385 J/gC)(60.0 C)=115.5 J$

6. 0.185 J/g•°C

The specific heat of iron is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q = -47 J is the heat released by the iron

m = 10.0 g is the mass of iron

$\Delta T=25.0-50.4 C=-25.4 C$ is the change in temperature

Substituting,

$C_s = \frac{-47 J}{(10.0 g)(-25.4 C)}=0.185 J/gC$

8 0

4 years ago