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3 years ago

The main difference between speed and velocity involves

Physics

2 answers:

mezya [45]3 years ago

4 0

The main difference between speed and velocity is A: direction.

vodomira [7]3 years ago

3 0

The Correct answer to this question for Penn Foster Students is: Direction

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3. The pressure at the bottom of the ocean is great enough to crush submarines with steel walls that are 10 centimeters thick. S

Elden [556K]

Answer:

just awnsered this one your awnser is the the second option

3 0

3 years ago

Read 2 more answers

A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik

shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle $=40^{\circ}$

$x =12\pm 0.27$

y=1.05

equation of trajectory of ball is given by

$y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }$

for x=12.27

$1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }$

u=10.69

for x=11.73

$1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }$

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0

3 years ago

An Ohmic material is one in which its resistance does not depend on the voltage across it or the current within it. In that case

Alexandra [31]

Voltage

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4 years ago

23. If a jogger runs 100 meters west and then turns around and runs 30 meters east. What

marshall27 [118]

Answer:

Explanation:

if he goes to the west, the east is opposite so 100-30

7 0

3 years ago

Read 2 more answers

To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You

drek231 [11]

Answer : The heat change of the cold water in Joules is, $1.6\times 10^3J$

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

$Density=\frac{Mass}{Volume}$

$Mass=Density\times Volume=1g/mL\times 45mL=45g$

Now we have to calculate the heat change of cold water.

Formula used :

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = $4.184J/g^oC$

$T_1$ = initial temperature of cold water = $24.7^oC$

$T_2$ = final temperature = $33.4^oC$

Now put all the given value in the above formula, we get:

$Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC$

$Q=1638.036J=1.6\times 10^3J$

Therefore, the heat change of cold water is $1.6\times 10^3J$

4 0

3 years ago