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3 years ago

Protons are the smallest particle of an element that retains the original Characteristics of the element. True or False

Physics

2 answers:

dlinn [17]3 years ago

5 0

Answer:

False; that’s an atom

Explanation:

Vedmedyk [2.9K]3 years ago

3 0

Answer:

false

Explanation:

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All of the following would be useful in trying to obtain procedural information to replicate an experiment previously published

Lera25 [3.4K]

Answer:

I am pretty sure that the answer would be a peer reviewed article.

Explanation:

I saw this because, an encyclopedia, published scientific journal, and a lab journal used in the original experiment, are all reliable sources of information.

7 0

3 years ago

Read 2 more answers

A student is preparing to take a bath when she realizes the hot water tap in the bathroom is not working the student go to the k

WITCHER [35]

Answer:

The final temperature of the bath water will be 55°C

Explanation:

A student is preparing to take a bath when she realizes that the hot water tap is not working.

The student goes to the kitchen and prepares 10 L of 100 degree Celsius water to mix with 10 L of 10 degree Celsius water. What will be the final temperature of the bath water?

Using the relation, Heat lost = Heat gained (assuming no heat is lost to the surroundings).

Heat lost by hot water = heat gained by cold water

Using the formula of heat, Q = mcΔT

where m = mass of water, c = specific heat capacity of water, ΔT = temperature difference

mass of water = density * volume (density of water = 1 kg/L)

mass of hot water, m₁ = 1 kg/L * 10 L = 10 kg;

mass of cold water, m₂ = 1 kg/L * 10 L = 10 kg;

specific heat capacity of water, c = 4200 j/kg

Let the final temperature of the mixture be T

-m₁cΔT₁ = m₂cΔT₂ ( since heat is lost by the hot water)

-m₁ΔT₁ = m₂ΔT₂

-10 * (T- 100) = 10 * (T - 10)

-10T + 1000 = 10T - 100

20T = 1100

T = 1100/20

T = 55°C

Therefore, the final temperature of the bath water will be 55°C

6 0

3 years ago

Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t

Bond [772]

Answer:

$\frac{D_{jet}}{D_{prop}}=2.865$

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

$D=\frac{1}{2}CpAv^2\\$

The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:

$\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865$

4 0

3 years ago

PLEASE HELP!! URGENTT!

oksano4ka [1.4K]

Answer:

120 Newton

Explanation:

Given the following data;

Mass = 12 kg

Angle = 4°

We know that acceleration due to gravity is equal to 10 m/s

To find the minimum force to stop the block from sliding;

Force = mgCos(d)

Where;

m is the mass of an object.

g is the acceleration due to gravity.

d is the angle of inclination (theta).

Substituting into the formula we have;

F = 12*10*Cos(4°)

F = 120 * 0.9976

F = 119.71 ≈ 120 Newton

3 0

3 years ago

A rocket is attached to a toy car that is confined to move in the x-direction ONLY. At time to = 0 s, the car is not moving but

laiz [17]

Answer:

What displacement must the physics professor give the car

= 12.91 METERS

Explanation:

Check the attached file for explanation

4 0

4 years ago