All categories

4 years ago

Look at Figure 2. Describe the motion of the object in Graph A

Physics

1 answer:

dimaraw [331]4 years ago

3 0

The graphs show distance versus time data. For graph A, we can say that velocity is constant until it reaches time of 8 seconds. And a constant velocity is again exhibited starting at time 12 seconds. For graph B, the velocity is changing as time pass by since the slope is changing. At time of 2 seconds, graph A has greater velocity since it has a steeper slope.

You might be interested in

Tera performed multiple experiments to determine the best growing conditions for Azaleas. In one experiment, she tested the effe

Ira Lisetskai [31]

The answer should be D. Ur right

8 0

3 years ago

Read 2 more answers

A space probe is coasting through space at a steady speed of 100p feet per second. the booster rocket fires for 1/2 seconds so t

aleksandr82 [10.1K]

Answer:

Acceleration: $9800 ft/s^2$

Explanation:

The acceleration of an object is equal to the rate of change of velocity:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For the space probe in this problem, we have:

u = 100 ft/s (initial velocity)

v = 5000 ft/s (final velocity)

t = 0.5 s (time taken)

Therefore, the acceleration is

$a=\frac{5000-100}{0.5}=9800 ft/s^2$

6 0

3 years ago

What will create static electricity?

frosja888 [35]

Answer:

Friction between two objects causes a transfer of electrons from one object to the other.

Explanation:

3 0

3 years ago

DESPERATE WILL GIVE BRAINLIST

Leokris [45]

B, is older

Because if you look on the scale b is deeper

5 0

3 years ago

Read 2 more answers

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag fo

Sonbull [250]

Answer:

$2.267$

Explanation:

Drag force is given by

$F=\dfrac{1}{2}\rho Av^2C$

C = Drag coefficient is constant

A = Area is constant

$v_1$ = Velocity of the passenger jet = 1200 km/h = $\dfrac{1200}{3.6}\ \text{m/s}$

$v_2$ = Velocity of the prop plane = $\dfrac{1}{4}v_1$

$\rho_1$ = Density of the air where the jet was flying = $0.38\ \text{kg/m}^3$

$\rho_2$ = Density of the air where the prop plane was flying = $0.67\ \text{kg/m}^3$

$F\propto \rho v^2$

$\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267$

The ratio of the drag forces is $2.267$ .

5 0

3 years ago