Haley's swim team hosted a home meet and needed volunteers. How did she reveal to her coach that she is not a team player?

How do I find the velocity of a moving object?

Aliun [14]

If you know the distance and the time I travelled that distance.

You just have to divide the time from the distance to get velocity

V =d
_

t

8 0

3 years ago

DETERMINE THE LAUNCH ELEVATION AND MAXIMUM RANGE: The test vehicle should obtain a burnout velocity of 7200 m/s at 180 km altitu

fenix001 [56]

Answer:

Flight path angle= 15.12°, maximum range= 5.29× 10*6 km

Explanation:

u= 7200m/s, H= 180km= 180000m

Recall that

Maximum height, H= (u*2sin*2∆)/2g

180000= (7200×7200sin*2∆)/2×9.8

(18000×2×98)/7200×7200= sin*2∆

Sin∆= 0.2609

∆= 15.12°

Maximum range, R= u*2/g

(7200×7200)/9.8

= 5289795.92km

= 5.29× 10*6 km

8 0

3 years ago

If a force of 12 N is applied to an object with a mass of 2 kg, what will be its acceleration?

nadezda [96]

F = ma, Where F is force is in N, mass is in kg, = 2kg, a is acceleration in m/s²

12 = 2a

2a = 12

a = 12/2

a = 6

Acceleration = 6 m/s²

7 0

3 years ago

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

(C) at 33.23 m the speed will be 12 m/s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

(C) the position of the ball at 12 m/s

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago

A new spacecraft is built with an engine that is able to produce constant acceleration of 1g, where g = gravity’s acceleration o

QveST [7]

given that acceleration due to gravity is g = 10 m/s^2

speed of the rocket will reach to 0.9c

$v_f = 0.9* 3 * 10^8 = 2.7 * 10^8 m/s$

now by kinematics

$v_f = v_i + a*t$

$2.7 * 10^8 = 0 + 10*t$

$t = 2.7*10^7 s$

Part b)

distance traveled by it in above time

$d = v*t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}*10*(2.7*10^7)^2$

$d = 3.645 * 10^{15} m$

so this is the distance covered by the object

3 0

3 years ago