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3 years ago

8

The velocity-time and acceleration-time graphs of a particle are given here. Its position-time graph may be given as: PLZ answer

fast!!!!! I will mark best as BRAINLEST!!!!!!! PLZ!!! TEST IN 2O MINS

1 answer:

artcher [175]3 years ago

8 0

Answer:

I would say the answer is B

Explanation:

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(answer only if you know the answer or I'll report) Help me please solve it w steps

Murrr4er [49]

X component of force Fx=Fcos45=50×cos×45

Y component of force Fy=Fsin45=50×sin×45

5 0

3 years ago

In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a 4.32 x 104-m length of wire

elena55 [62]

Answer:

Explanation:

Length, l = 4.32 x 10^4 m

speed, v = 7.07 x 10^3 m/s

magnetic field, B = 5.81 x 10^-5 T

The formula for the motion emf is given by

e = B x v x l

e = 5.81 x 10^-5 x 7.07 x 10^3 x 4.32 x 10^4

e = 17745.1 V

3 0

3 years ago

Evangelista Torricelli was the first person to realize that we live at the bottom of an ocean of air. He correctly surmised that

Radda [10]

Answer:

a) h = 8.02 10³ m b) yes

Explanation:

a) The pressure in a fluid is given by

P = ρ g h

The pressure in this case is the atmospheric pressure, 1.013 105 Pa, let's clear the height (h)

h = P / ρ g

h = 1.013 10⁵ / (1.29 9.8)

h = 8.02 10³ m

b) The height of Mount Everest is 8848 m

It is above this height, according to this model there would be no air to breathe

8 0

3 years ago

The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.

Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

$\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3$

$1\ kg=2.20462\ lb$

$1\ m=39.3701\ in$

$\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3$

So,

$10.1\ kN/m^3=0.03719\ lbs/in^3$

8 0

3 years ago

A mass of 10 g of oxygen fill a weighted pistonâ€“cylinder device at 20 kPa and 110Â°C. The device is now cooled until the tempe

mezya [45]

Answer:

The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

$\Delta V=V_{2}-V_{1}$

$\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})$

Put the value into the formula

$\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)$

$\Delta V=14.297\ L$

$\Delta V=14.3\times10^{-3}\ m^3$

Hence, The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

6 0

3 years ago