Ask question

Ask question

All categories

3 years ago

10

An outfielder with a mass of 86 Kilograms catches a baseball with a mass of 0.15 kilograms.The velocity of the ball just before

being caught is 40 meters per second.What is the impulse imparted by the ball onto the outfielder?
A. 6.00 Ns
B. 3,440 Ns
C. 2.00 Ns
D. 3,434 Ns

1 answer:

nordsb [41]3 years ago

5 0

Answer:

C

Explanation:

You might be interested in

You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

katrin2010 [14]

Answer:

b

c

e

h

Explanation:

Note that the swing direction was not giving in the question and direction could be sideways (in a turn) or in a track or both

The question show something in common ...acceleration

So let's look at the statements and pick the correct ones

a is false while b is correct as the train is accelerating

c is correct. The train is accelerating even thou the speed could not be ascertained

d is false and not feasible as the train is accelerating

e is true as the train maybe moving at a constant speed in a circle

f is false. This could be constant velocity in a circle. Same as g (false)

h is true. It's accelerating

7 0

3 years ago

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for

sveta [45]

Answer:

F = -4567.40 N

Explanation:

Given that,

The power developed by the engine, P = 196 hp

1 hp = 746 W

196 hp = 146157 W

Speed of the car, v = 32 m/s

Let F is the total friction force acting on the car. The product of force and velocity is called the power developed by the engine. It is given by :

$P=-F\times v$

$F=\dfrac{-P}{v}$

$F=\dfrac{-146157\ W}{32\ m/s}$

F = -4567.40 N

So, the total frictional force acting on the car is 4567.40 N. Hence, this is the required solution.

7 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

If you could travel at the speed of light, how long would it take to travel from our solar

Illusion [34]

Answer:

4.689 years

Explanation:

6 0

3 years ago

Using poiseulli assumption derive an expression for the rate of flow of a liquid through a horizontal tube in terms of a, l, p a

Marizza181 [45]

Answer:

im sorry im just trying this app please dont report

Explanation:

please im begging you maam/sir

7 0

3 years ago