An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?

Answers

Answer 1

Answer:

(a) The acceleration of the bird is $a = -1.02\ m/s$ . The negative sign indicated the opposite direction of motion. (b) The final speed is $v = 11.76\ m/s$ .

Given:

Initial speed, $u = 13\ m/s$

Final speed, $9.4\ m/s$

Time, $t = 3.5\ s$

The acceleration can be computed from the velocities and time. The standard unit of acceleration is a meter per second square.

(a)

The acceleration is computed as:

$a = v-u/t\na = 9.4-13/3.5\na = -1.02\ m/s$

Hence, the acceleration of the bird is $a = -1.02\ m/s$ . The negative sign indicated the opposite direction of motion.

(b)

The final speed as the given time can be computed from the first equation of motion. The first equation of motion gives the relation between final and initial speed, acceleration, and time.

The final speed at time 1.2 seconds is equal to:

$v = u+at\nv = 13+(-1.02)*1.20\nv = 11.76\ m/s$

Hence, the final speed is $v = 11.76\ m/s$ .

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Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

While leaning out a window that is 6.0 m above the ground, you drop a 0.60-kg basketball to a friend at ground level. Your friend catches the ball at a height of 1.6 m above the ground. Determine the following.(a) the amount of work done by the force of gravity on the ball.(b) the gravitational potential energy of the ball-earth system, relative to the ground when it is released.(c) the gravitational potential energy of the ball-earth system, relative to the ground when it is caught.

Answers

Answer:

a) W = 25.872 J

b) - 35.28 J

c) - 9.408

Explanation:

a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)

F = - mg (gravity is acting downwards)

F = - 0.6 × 9.8 = - 5.88 N

(H₂ - H₁) = (1.6 - 6) = - 4.4 m

W = (-5.88)(-4.4) = 25.872 J

b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J

c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J

An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

Answers

Given :

∅ = 60⁰

u = 4 m/s

g = 10m/s²

to find :

T = ?

Solution :

as per formula,

$t = (2u \: sin \theta)/(g)$

now put the value : $t \: = (2 * 4 * sin \: 60)/(10)$

as we know $sin60 \: = ( √(3) )/(2)$

therefore,

$t \: = (8 * ( √(3) )/(2) )/(10)$

as we solve this we get,

$t \: = ( 2√(3) )/(5)$

that's t = 0.69 sec

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0.8 seconds

Explanation:

time of flight = 2u/g

u=4m/s

g=10

= 8/10

= 0.8 sec

just a trial...not sure!!!

A truck’s suspension spring each have a spring constant of 769 N/m. If the potential energy of the right front spring is 250 J, how far is the spring compressed?

Answers

The energy stored in the body in a rest state is called potential energy.

There are two types of mechanical energy. The mechanical energy is consist of the following:-

Kinetic energy
Potential energy

According to the question, the solution is:-

The formula we used is $PE= (1)/(2)kx^(2)$

After putting the value the equation is stated as follows:-

$250 =(1)/(2) *769*x^(2)$

Hence the $x^(2)$ is equal to:-

$x^(2) = 651\nx= 0.81$ m

The spring compressed in 0.81m

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Answer:

x = 0.81 m

Explanation:

given,

spring constant, k = 769 N/m

Potential energy of the spring = 250 J

distance of spring compression = ?

using conservation of energy

potential energy will equal to the spring energy

$PE = (1)/(2)kx^2$

$250= (1)/(2)* 769* x^2$

x² = 0.650

x = 0.81 m

Hence, the spring is compressed to 0.81 m

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?

Answers

Answer:

$v_(su) = 19.44 m/s$

Explanation:

$m_(su)=5.68x10^(29)kg\nm_(sa)=5.68x10^(26)kg$

$T=9.29x10^8\nr_(o)=1.43x10^(12)$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_(su)*r_(o)=(m_(sa)+m_(su))*r_(1)$

solve to r1

$r_1=(m_(sa)*r_(o))/(m_(sa)+m_(su))=(5.68x10^(26)*1.43x10^(12))/(5.68x10^(26)+5.68x10^(26))$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = (m_(sa)*2*pi*r_1^2)/(T)$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_(sun)=(m_(sa)*2*\pi *( 2r_(o)*r_1 -r_1^2))/(T)$

Since

$L_(su)= m_(su)*v_(su)*r_1$

then

$v_(su)=(m_(sa)*2*pi*(2r_(o)*r_(1)-r_(1)^2))/(T*m_(sa)*r_1)$

$v_(su) = 19.44 m/s$

Final answer:

In a two-body system such as the Sun-Saturn system, both bodies orbit around their mutual center of mass, or barycenter. Given the Sun's significantly larger mass, this barycenter is near the center of the Sun, and hence the Sun's change in velocity relative to the center of mass of the system as Saturn completes half an orbit is effectively zero.

Explanation:

The problem here is asking for the change in velocity of the Sun relative to the center of mass of the Sun-Saturn system as Saturn completes half an orbit. This is a situation involving orbital physics and center of mass systems.

However, in an isolated two-body orbit system like this, the center of mass does not change velocity - it would remain constant, not unless acted upon by an outside force, which the problem instructs us to ignore.

Saturn and the Sun both orbit around their common center of mass (their barycenter). Given that the Sun is immensely more massive than Saturn, this center of mass is located very close to the center of the Sun.

So, while the Sun does indeed move a little due to Saturn's influence, the change in velocity of Sun relative to the center of mass of the system during the time Saturn completes half an orbit, for all intents and purposes, is zero.

This is especially true unless the problem specifically mentions that the Sun is initially at rest with respect to the center of mass. In any other case, the relative velocity remains constant and hence the change is zero.

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What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Answers

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?

Answers

Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
= (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
= (9.1×10∧-31 × 1,153,846.154∧2) ÷2
= (1.21154×10∧-18) ÷ 2
= 6.05769×10∧-19 J

Answer:

K = 6.02 × 10⁻¹⁹ J

Explanation:

The momentum (p) of an electron is its mass (m) times its speed (v).

p = m × v

v = p / m = (1.05 × 10⁻²⁴ kg.m/s) / 9.11 × 10⁻³¹ kg = 1.15 × 10⁶ m/s

We can find the kinetic energy (K) using the following expression.

K = 1/2 × m × v²

K = 1/2 × 9.11 × 10⁻³¹ kg × (1.15 × 10⁶ m/s)²

K = 6.02 × 10⁻¹⁹ J

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