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2 years ago

Liquid nitrogen boils at 77.2 k. what is this temperature in degrees celsius

1 answer:

6 0

77.2 k is -195.95 C

To convert from Kelvin to Celsius, you simply subtract 273.15 from the temperature. So the expression becomes c = 77.2 - 273.15 = -195.95

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What do you know about water

Romashka-Z-Leto [24]

Answer:

It is vital for all known forms of life.
It provides no calories nor organic nutrients.
It forms precipitation in the form of rain and aerosols in the form of fog.
It's chemical symbol is H₂O

Explanation:

HOPE THAT HELPS

PLEASE MARK AS BRAINLIEST

7 0

2 years ago

A- 1000 m/s2 Xi-0m Xf-0.75m Vf-?

sleet_krkn [62]

Answer:

The final velocity of the object is, $v_{f}$ = 27 m/s

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, $x_{i}$ = 0 m

The final displacement of the object, $x_{f}$ = 0.75 m

The initial velocity of the object will be, $v_{i}$ = o m/s

The final velocity of the object, $v_{f}$ = ?

The average velocity of the object,

v = ( $x_{f}$ - $x_{i}$ )/ t

= 0.75 / t

The acceleration is given by the relation

a = v / t

1000 m/s² = 0.75 / t²

t² = 7.5 x 10⁻⁴

t = 0.027 s

Using the I equation of motion,

$v_{f}$ = u + at

Substituting the values

$v_{f}$ = 0 + 1000 x 0.027

= 27 m/s

Hence, the final velocity of the object is, $v_{f}$ = 27 m/s

8 0

2 years ago

Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k

NikAS [45]

Answer:

$Fr=12544N$

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation $Fr=usmg$ , where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

$Fr=0.8*1600kg*9.8\frac{m}{s^{2}}$

$Fr=12544N$

5 0

2 years ago

A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a

blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

$E=7*10^{9}N/C$

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field is mathematically given as

$E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\$

$E=7*10^{9}N/C$

For more information on this visit

brainly.com/question/21811998

4 0

2 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Then the electric field at the point of interest is estimated as:

$E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}$

6 0

3 years ago