If Chris drives 27 km N and then 40 km E , what is the distance and displacement

Hatshy [7]

The last one is correct (D)

8 0

3 years ago

A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubb

Svet_ta [14]

Answer:

The voltage will be 0.0125V

Explanation:

See the picture attached

4 0

3 years ago

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

5 0

3 years ago

A car starts from rest and goes down a slope with a constant

GREYUIT [131]

Answer:

25m/s

Explanation:

here we use 1st equation of motion

then we find this ans

3 0

3 years ago

A 1500 kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force

MakcuM [25]

The correct answer to the question is : 9375 N.

CALCULATION:

As per the question, the mass of the car m = 1500 Kg.

The diametre of the circular track D = 200 m.

Hence, the radius of the circular path R = $\frac{D}{2}$

= $\frac{200}{2}\ m$

= 100 m.

The velocity of the truck v = 25 m/s.

When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.

Hence, the force acting on the car is centripetal force.

The magnitude of the centripetal force is calculated as -

Force F = $\frac{mv^2}{R}$

= $\frac{1500\times (25)^2}{100}\ N$

= 9375 N. [ANS}

The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.

6 0

3 years ago