You hang a tv on your wall. What kind of energy does it have?

A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises

atroni [7]

Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

4 0

2 years ago

A single loop of wire with an area of 0.0780 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic

Katarina [22]

Answer:

The induced current is 26.7 mA

Explanation:

Given;

area of the loop, A = 0.078 m²

initial magnetic field, B₁ = 3.8 T

change in the magnetic field strength, dB/dt = 0.24 T/s

The induced emf is calculated as;

$emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V$

The resistance of the loop = 0.7 Ω

The induced current is calculated as;

$V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA$

4 0

3 years ago

If you separate vector B into its components. How many components will it have? Those components will be called?

Inessa [10]

The vector B will have two components and those components will be called resultant vectors.

<h3>What is a component vector?</h3>

A component vector is a unit vector that represents a given vector in a particular direction.

A vector can be represented in x - direction and y - direction.

x - component of the vector = Bcosθ
y - component of the vector = Bsinθ

Thus, the vector B will have two components and those components will be called resultant vectors.

Learn more about component vectors here: brainly.com/question/13416288

#SPJ12

3 0

2 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

What is the speed of sound for a noise that travels 2km in 5.8s?

Gnesinka [82]

Answer:

Explanation:

Speed is defined as the rate at which an object covers a particular distance. So the formula for determining speed is given as the ratio of distance to time taken for covering that distance.

Speed = Distance/Time

As here the distance is given in km units and time in s units, so the units of any one parameter should be changed. Since we know that speed of sound is always about 300 m/s. So it is better to convert the unit of distance from km to m.

Hence, now the distance traveled by the noise is 2000 m and time taken is 5.8 s.

So the speed of noise = Distance/Time = 2000/5.8=345 m/s.

Thus, the speed of noise is slightly greater than the speed of sound and it is found to be 345 m/s.

5 0

3 years ago