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2 years ago

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A 145-g baseball is thrown so that it acquires a speed of 25 m/s. What was the net work done on the ball to make it reach this s

peed, if it started from rest?

1 answer:

inysia [295]2 years ago

6 0

When the ball has left your hand and is flying on its own, its kinetic energy is

KE = (1/2) (mass) (speed²)

KE = (1/2) (0.145 kg) (25 m/s)²

KE = (0.0725 kg) (625 m²/s²)

<em>KE = 45.3 Joules</em>

If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from: It MUST have come from the hand that threw the ball. The hand would have needed to do <em>45.3 J</em> of work on the ball before releasing it.

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A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con

Nataly [62]

Answer:R=1607556m

θ=180degrees

Explanation:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Using analytical method :

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

Note that θ is in the second quadrant, so add 180

θ=180-7.9256*10^6=180degrees

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2 years ago

"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati

Rasek [7]

Answer:

99.95%

Explanation:

A double pulsar system named PSR J0737-3039A/B in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.

A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

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2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

MArishka [77]

Complete Question:

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.

Answer:

m = 0.001 M

For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/

6 0

1 year ago

A compact, dense object with a mass of 2.90 kg is attached to a spring and is able to oscillate horizontally with negligible fri

enot [183]

(a) 80 N/m

The spring constant can be found by using Hooke's law:

$F=kx$

where

F is the force on the spring

k is the spring constant

x is the displacement of the spring relative to the equilibrium position

At the beginning, we have

F = 16.0 N is the force applied

x = 0.200 m is the displacement from the equilibrium position

Solving the formula for k, we find

$k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m$

(b) 0.84 Hz

The frequency of oscillation of the system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 80 N/m is the spring constant

m = 2.90 kg is the mass attached to the spring

Substituting the numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz$

(c) 1.05 m/s

The maximum speed of a spring-mass system is given by

$v=\omega A$

where

$\omega$ is the angular frequency

A is the amplitude of the motion

For this system, we have

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

$A=0.200 m$ (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)

Substituting into the formula, we find the maximum speed:

$v=(5.25 rad/s)(0.200 m)=1.05 m/s$

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.

(e) 5.51 m/s^2

In a simple harmonic motion, the maximum acceleration is given by

$a=\omega^2 A$

Using the numbers we calculated in part c):

$\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s$

$A=0.200 m$

we find immediately the maximum acceleration:

$a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2$

(f) At the position of maximum displacement: $x=\pm 0.200 m$

According to Newton's second law, the acceleration is directly proportional to the force on the mass:

$a=\frac{F}{m}$

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

$F=kx$

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

$E=K=\frac{1}{2}mv_{max}^2$

where

m = 2.90 kg is the mass

$v_{max}=1.05 m/s$ is the maximum speed

Solving for E, we find

$E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J$

(h) 0.99 m/s

When the position is equal to 1/3 of the maximum displacement, we have

$x=\frac{1}{3}(0.200 m)=0.0667 m$

so the elastic potential energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J$

and since the total energy E = 1.60 J is conserved, the kinetic energy is

$K=E-U=1.60 J-0.18 J=1.42 J$

And from the relationship between kinetic energy and speed, we can find the speed of the system:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s$

(i) 1.84 m/s^2

When the position is equal to 1/3 of the maximum displacement, we have

$x=\frac{1}{3}(0.200 m)=0.0667 m$

So the restoring force exerted by the spring on the mass is

$F=kx=(80 N/m)(0.0667 m)=5.34 N$

And so, we can calculate the acceleration by using Newton's second law:

$a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2$

8 0

1 year ago

The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and

mina [271]

Answer:

$\frac{dQ}{dt}= 4312 W$

Explanation:

As we know that base of the slab is given as

$A = 11 \times 8$

$A = 88 m^2$

now we know that rate of heat transfer is given as

$\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)$

here we know that

$k = 1.4 W/m k$

Also we have

$x =0.20$

$\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)$

$\frac{dQ}{dt}= 4312 W$

7 0

1 year ago