All categories

3 years ago

What is the difference between each distance traveled and displacement travled

1 answer:

8 0

Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.

While, Distance is a scalar magnitude that measures over the trajectory.

You might be interested in

A.) If its booster rockets accelerate the space shuttle at 15m/s2, how high will it be one minute after launch?

poizon [28]

Answer:

27,000 m

450 m/s

Explanation:

Assuming the initial velocity is 0 m/s:

v₀ = 0 m/s

a = 15 m/s²

t = 60 s

A) Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²

Δy = 27,000 m

B) Find: v_avg

v_avg = Δy / t

v_avg = 27,000 m / 60 s

v_avg = 450 m/s

5 0

3 years ago

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 1300 m. After

svetlana [45]

Explanation:

Below is an attachment containing the solution.

6 0

3 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

If you know the amount of air pressure exerted on a tabletop, how can you calculate the force exerted on the tabletop? Multiply

babymother [125]

Answer:

Multiply the air pressure by the area of the tabletop.

Explanation:

The relationship between pressure, force and area is given by:

$p=\frac{F}{A}$

where in this case, p is the air pressure, F is the force exerted and A the area of the tabletop. By re-arranging the equation, we can solve for F, the force exerted:

$p=\frac{F}{A}\\p\cdot A=\frac{F}{A}\cdot A\\pA=F$

So, the correct answer is:

The force exerted on the tabletop can be found by multiplying the air pressure by the area of the tabletop.

7 0

3 years ago

A racetrack curve has radius 70.0 m and is banked at an angle of 12.0 ∘. The coefficient of static friction between the tires an

zubka84 [21]

Answer:

See attachment below

Explanation:

7 0

3 years ago