Ask question

Ask question

All categories

2 years ago

5

A 0.2-kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before

and just after impact with the floor is 10 m/s. Determine the magnitude of the impulse delivered to the floor by the steel ball.

2 answers:

nadya68 [22]2 years ago

7 0

Answer:

Explanation:

Given

mass of steel ball $m=0.2\ kg$

initial speed of ball $u=10\ m/s$

Final speed of ball $v=-10\ m/s$ (in upward direction)

Impulse imparted is given by change in the momentum of object

therefore impulse J is given by

$J=\Delta P$

$\Delta P=m(v-u)$

$\Delta P=0.2(-10-10)$

$\Delta =-4\ N-s$

so magnitude of Impulse =4 N-s

MAVERICK [17]2 years ago

4 0

Answer:

4 N s

Explanation:

mass, m = 0.2 kg

initial velocity, u = - 10 m/s (downward )

final velocity, v = + 10 m/s (upwards)

Impulse is defined a the change in momentum .

Impulse = m ( v - u)

Impulse = 0.2 ( 10 + 10)

Impulse = 4 N s

thus, the impulse is 4 N s .

You might be interested in

1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup

azamat

Answer:

$\dot{W} = 339.84 W$

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

$\dot{W} =\dot{m}\frac{V^{2}}{2}$

here $\dot{m}$ is mass flow rate and given as

$\dot{m} = \rho*Q$

$\dot{W} =\rho*Q\frac{V^{2}}{2}$

Putting all value to get minimum power

$\dot{W} =1.18*9*\frac{8^{2}}{2}$

$\dot{W} = 339.84 W$

7 0

2 years ago

Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec

AysviL [449]

The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
$E= \frac{\sigma}{2\epsilon_0}$
where
$\sigma$ is the charge density
$\epsilon_0$ is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
$E=30 N/C$

7 0

2 years ago

If a 3-kg rabbit's leg muscles act as imperfectly elastic springs, how much energy will they hold if the rabbit lands from a hei

DiKsa [7]

Answer;

- 15 J

Explanation;

-Potential energy is defined as mechanical energy, stored energy, or energy caused by its position.

-For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m /s² at the surface of the earth) and h is the height in meters.

Potential energy of the rabbit at the peak of its height is

PE = (3)(10)(0.5) = 15 J

(around 14.7 but because energy is lost, it is less than that)

3 0

2 years ago

Read 2 more answers

Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.

dimulka [17.4K]

Answer:

The distance of separation is $d = 0.092 \ m$

Explanation:

The mass of the each ball is $m= 10 g = 0.01 \ kg$

The negative charge on each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

$F = \frac{kq_1 * q_2}{d}$

=> $m* g = \frac{kq_1 * q_2}{d}$

here k the the coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

So

$0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}$

$d = 0.092 \ m$

5 0

2 years ago

An object on a number line moved from x = 15 cm to x = 165 cm and then

olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

$v_{avg}=\frac{x_{all}}{t}$

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

$v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}$

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

3 0

2 years ago

Read 2 more answers