The slope of the line on a speed-time graph tells the speed.

If x2 = 60, what is the value of x? plus or minus square root 30 plus or minus square root 60 ±30 ±120

Zielflug [23.3K]

X2 = 60
/ 2 / 2
x = 30
Plus or minus square root 60

4 0

3 years ago

The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will __

Kipish [7]

Answer:

The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will decrease with time, the surface area between the tire and the road will increase with time, and the net force the tire exerts on the road will be constant with time.

Explanation:

when a wheel has an air leak, it means that the inside of the tire has less air, which means that there will be less air pushing the walls of the tire so that the air pressure decreases.

On the other hand, the tire begins to deform due to lack of air which increases the area of contact with the floor.

As the weight of the car remains constant and the air has a negligible mass the force towards the road will be the same

4 0

4 years ago

In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x

Andrei [34K]

Answer:

(a) $F_g=1.62*10^{-48}N$

(b) $F_e=3.68*10^{-9}N$

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

$F_g=-G\frac{m_1m_2}{r^2}$

Where G is the Cavendish gravitational constant, $m_1$ and $m_2$ are the masses of the electron and the proton respectively and r is the distance between them:

$F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N$

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

$F_g=1.62*10^{-48}N$

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

$F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N$

Its magnitude is:

$F_e=3.68*10^{-9}N$

6 0

3 years ago

When θ= 0 ̊, the assembly is held at rest, and the torsional spring is untwisted. if the assembly is released and falls downward

Georgia [21]

Rod is 450mm and disk has a radius of 75mm So there is a pin holding the assembly upwards which is when Θ=0 and at the pin there is a torsional spring with constant of k=20N m/rad. One end of the rod is attached to the pin and the other is attached to the disk.

7 0

4 years ago

Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1

Setler79 [48]

Answer:

The flow rate is $R =2.357 *10^{-4} \ m^3/s$