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3 years ago

10

The normal force which the path exerts on a particle is always perpendicular to the _________________ tangent to the path. trans

verse direction radial line. normal direction.

1 answer:

Marianna [84]3 years ago

7 0

Tangent to the pathh.

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Does anyone know this?!

Valentin [98]

Answer:

2 is the numerical answer.

Explanation:

Hello there!

In this case, according to the given information and formula, it is possible for us to remember that equation for the calculation of the average kinetic energy of a gas is:

$KE=\frac{3}{2} \frac{R}{N_A} T$

Whereas R is the universal gas constant, NA the Avogadro's number and T the temperature.

Which means that for the given ratio, we can obtain the value as follows:

$=\frac{\frac{3}{2} \frac{R}{N_A} T_1}{\frac{3}{2} \frac{R}{N_A} T_2} \\\\=\frac{T_1}{T_2} \\\\=\frac{500K}{250K} \\\\=2$

Regards!

8 0

3 years ago

When is the pressure of the man on the ground more – lying position or standing

Wittaler [7]

Hi,

<u>The man on the ground in standing position has more pressure</u>. This is because when he stands, only his legs are in contact with the ground. While lying, his body is more in contact with the ground, therefore, he exerts less pressure.

To the point, a man standing position on the ground had more pressure.

More is the area of contact, less is the pressure efforted.

Thank you...

7 0

2 years ago

A car that increase its speed from 20 km/h to 100 km/h undergoes -------acceleration,

stepladder [879]

negative acceleration- deceleration

8 0

3 years ago

What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a

Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

$A = W + \dfrac{W}{2}$

$A =\dfrac{3W}{2}$

$A =\dfrac{3mg}{2}$

When the car is at the bottom, the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

$N = m g + \dfrac{mv^2}{r}$

$\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}$

$\dfrac{mg}{2} = \dfrac{mv^2}{r}$

$v = \sqrt{\dfrac{rg}{2}}$

$v = \sqrt{\dfrac{40\times 9.8}{2}}$

v = 14 m/s

8 0

2 years ago

A sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density? <br>

Hoochie [10]

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

$Ro = m/V$

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

$V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]$

And the mass m = 4 [gramm] = 0.004 [kg]

$Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]$

3 0

2 years ago