The turning radius of a car is 36ft. What will be the distance in meters?

3. A rock is placed in a graduated cylinder containing water, and the cylinder is covered to avoid evaporation. The next morning

guajiro [1.7K]

Hi! Some rocks are porous material, like a sponge, with lots of air pockets inside, so they can absorbe water. In this case, the water level dropped 5.65m because tha volume of water went inside the rock filling the air pockets. The water replaced the air, that flowed to the outside water as bubbles and then to the air above the water.

7 0

3 years ago

Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri

Elanso [62]

First we can say that since there is no external force on this system so momentum is always conserved.

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}$

$0.48= 0.6*v_{1f} + 1.2*v_{2f}$

$0.8 = v_{1f} + 2v_{2f}$

now by the condition of elastic collision

$v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6$

$v_{2f} = 0.533 m/s$

also from above equation we have

$v_{1f} = -0.267 m/s$

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

8 0

4 years ago

If a train going 60 m/s hits the breaks, and it takes the train 85 seconds to stop, what is the train´s acceleration

Flauer [41]

Acceleration is change in velocity over time --> $a = \frac{Δv}{t}$

Δv or change in velocity equals the Final velocity minus the initial velocity

Our initial velocity here is 60m/s while our final velocity is 0 m/s

The time interval is 85 seconds so...

$a = \frac{0-60}{85}$

$a = \frac{-60}{85}$

$a = -0.706$

The negative is there to indicate that the train in slowing down or decelerating

4 0

4 years ago

At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 57.5 V/m. Fi

n200080 [17]

Answer:

$1.9167\times 10^{-7}\ T$

$2.9247498849\times 10^{-8}\ J/m^3$

$8.7742496547\ W/m^2$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

E = Electric field = 57.5 V/m

c = Speed of light = $3\times 10^8\ m/s$

Magnetic field is given by

$B=\dfrac{E}{c}\\\Rightarrow B=\dfrac{57.5}{3\times 10^8}\\\Rightarrow B=1.9167\times 10^{-7}\ T$

The magnetic field strength is $1.9167\times 10^{-7}\ T$

Energy density is given by

$u=\dfrac{1}{2}\epsilon_0E^2+\dfrac{1}{2\mu_0}B^2\\\Rightarrow u=\dfrac{1}{2}\times 8.85\times 10^{-12}\times 57.5^2+\dfrac{1}{2\times 4\pi \times 10^{-7}}(1.9167\times 10^{-7})^2\\\Rightarrow u=2.9247498849\times 10^{-8}\ J/m^3$