Ask question

Ask question

All categories

3 years ago

11

A soccer ball is rolling with a constant acceleration of -1.5 m/s2 . At ti = 0 s, the ball has an instantaneous velocity of 14 m

/s. After 3s of rolling, what would be the velocity of the ball (in m/s)?

1 answer:

miskamm [114]3 years ago

6 0

Answer:

$v_{f} = 9.5\,\frac{m}{s}$

Explanation:

The final velocity is:

$v_{f} = v_{o} + a \cdot t$

$v_{f} = 14\,\frac{m}{s} + (-1.5\,\frac{m}{s^{2}} )\cdot (3\,s)$

$v_{f} = 9.5\,\frac{m}{s}$

You might be interested in

Why is Darwin criticized for his theory? and who were they

kondaur [170]

A few people criticized Darwin for his theory, such as the left-leaning biologists Stephen Jay Gould and Richard Lewontin, who fear the political implications of Darwinian theory. They fear that evolutionary theory, even when bolstered by modern genetics, and molecular biology, does not make reality probable enough.

4 0

2 years ago

Read 2 more answers

A ball is thrown upwards at an unknown speed. in a time of

alexandr402 [8]

Answer:

a) Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a) We have equation of motion $s=ut+\frac{1}{2} at^2$ , where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8 $m/s^2$

Substituting

$6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s$

Initial speed of the ball = 14.45 m/s

b) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

s = 6 m, u = 14.45 m/s, a = -9.8 $m/s^2$

Substituting

$v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s$

So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion $v^2=u^2+2as$ , where v is the final velocity

u = 14.45 m/s, v =0 , a = -9.8 $m/s^2$

Substituting

$0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m$

Maximum height reached = 10.65 m

8 0

3 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago

NEED HELP Two skaters stand facing each other. One skater’s mass is 60 kg, and the other’s mass is 72 kg. If the skaters push aw

stealth61 [152]

Answer:

the heavier skater has less momentum

hope it is helpful to you

3 0

2 years ago

Read 2 more answers

How fast would 40 Newtons of force accelerate a 2 kg object?

Digiron [165]

Answer:

$20 m/s^2$

Explanation:

We can solve this problem by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between its mass and its acceleration:

$F=ma$

where

F is the net force on the object

m is its mass

a is its acceleration

In this problem:

F = 40 N is the force on the object

m = 2 kg is its mass

Therefore, the acceleration of the object is

$a=\frac{F}{m}=\frac{40}{2}=20 m/s^2$

8 0

3 years ago