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3 years ago

1 PHYSICS QUESTION - DEPENDENT AND INDEPENDENT VARIABLES. WILL GIVE BRAINLIEST!!!

Physics

1 answer:

nadya68 [22]3 years ago

3 0

Answer:

The dependant variable is obvrioulsy going to be the temperature of the watch, and the independan variable is going to be the amount of water ebing poured into the cups.

Explanation:

The temperature is the dependant variable by how it is the thing being observed, or recorded. Your independant variables could be a few things from wha information I have but it could be either, the amount of water being poured into the cups, the temperature of the water being poured, or the amount of time between each new temperature of wather bing poured into the cups.

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A .025 kg bullet is moving at 350 m/s when it hits and then passes through a 7 kg block of wood. After the bullet passes through

slamgirl [31]

Answer:

The final velocity of the wooden block is equal to $1.035 m/s$

Explanation:

Given that mass of bullet = $m_{b} = 0.025kg$

Mass of wood = $m_{w} = 7kg$

Initial velocity of bullet = $u_{b} = 350m/s$

Final velocity of bullet = $v_{b} = 60m/s$

Initial velocity of wood = o

Final velocity of wood = $v_{w]$

Here momentum is conserved so initial momentum = final momentum

$m_{b}u_{b} = m_{b} v_{b} + m_{w}v_{w}$ .

Upon substituting these values in above equation , we get

$v_{w} = 1.035m/s$ .

7 0

3 years ago

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

$\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL$

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0

3 years ago

What's saturns rotation

xeze [42]

The way it rotates is Counter-clockwise

4 0

3 years ago

calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s

zheka24 [161]

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

5 0

2 years ago

The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.31 m + (4.90 m/s2)t2 - (0.100 m

vovangra [49]

The equation of the car is given by the equation,

x(t) = 2.31 + 4.90t² - 0.10t⁶

If we are going to differentiate the equation in terms of x, we get the value for velocity.

dx/dt = 9.8t - 0.6t⁵

Calculate for the value of t when dx/dt = 0.

dx/dt = 0 = (9.8 - 0.6t⁴)(t)

The values of t from the equation is approximately equal to 0 and 2.

If we substitute these values to the equation for displacement,

(0) , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31

(2) , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51

Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters.

3 0

3 years ago