Answer:
A(predicts an organisms ability to adapt to its enviroment, it is not a fact that each organization can adapt)
Explanation:
If an object travels on a circular path is an acceleration? What is changing to cause an acceleration?
The cannon on a battleship can fire a shell a maximum distance of 33.0 km.
(a) Calculate the initial velocity of the shell.
Answer:
v = 804.23 m/s
Explanation:
Given that,
The maximum distance covered by a cannon, d = 33 km = 33000 m
We need to find the initial velocity of the shell. Let it is v. It can be calculated using the conservation of energy such that,
[tex]v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 33000} \\\\v=804.23\ m/s[/tex]
So, the initial velocity of the shell is 804.23 m/s.
b) Two skaters collide and grab on to each other on a frictionless ice. One of them, of mass 80 kg, is moving to the right at 5.0 m/s, while the other of mass 70 kg is moving to the left at 2.0 m/s. What are the magnitude and direction of the two skaters just after they collide
Answer:
The two skaters move with a speed of 1.73 m/s after the collision in the right direction.
Explanation:
Given that,
The mas of skater 1, m₁ = 80 kg
The speed of skater 1, u₁ = 5 m/s (right)
The mass of skater 2, m₂ = 70 kg
The speed of skater 2, u₂ = -2 m/s (left)
Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
Put all the values,
[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]
So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.
A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 and 1510 N, respectively. The acceleration of the cable is 0.620 m s 2 , upward. What is the tension in the cable (a) below the worker and ( b) above the worker
Answer:
Explanation:
a)
Below the worker , the tension in cable is pulling the crate . Let the tension be T₁ .
weight of crate is acting downwards .
Total weight 1510 N.
Net force acting on both = T₁ - 1510
Applying second law of Newton ,
T₁ - 1510 = 1510 / 9.8 x 0.62 [ 1510 / 9.8 = mass of crate ]
T₁ - 1510 = 95.5
T₁ = 1605.5 N.
b )
Above the worker , the tension in cable is pulling both the worker and the crate . Let the tension be T₂ .
weight of both worker and crate is acting downwards .
Total weight = 965 + 1510 = 2475 N.
Net force acting on both = T₂ - 2475
Applying second law of Newton ,
T₂ - 2475 = 2475 / 9.8 x 0.62 [ 2475 / 9.8 = mass of both worker and crate ]
T₂ - 2475 = 156.6
T₂ = 2631.6 N.
what will be the gravitational force between two heavenly bodies if the masses of both are tripled keeping the distance between them constant
Answer:
If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
crushing chalk into powder is and irreversible change. is this example a physical or chemical change?Why?
Answer:
It is a example of physical change
An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance
Answer:
0.857 cm
Explanation:
We are given that:
The focal length for a convex lens to be (f) = 1.5cm
The object distance (u) = - 2.0 cm
We are to determine the image distance (v) = ??? cm
By applying the lens formula:
[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]
By rearrangement and making (v) the subject of the above formula:
[tex]v = \dfrac{uf}{u-f}[/tex]
replacing the given values:
[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]
[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]
v = 0.857 cm
A coin and feather are dropped in a moon. what will fall earlier on ground.give reasons.if they are dropped in the earth,which one will fall faster
Answer:
on the moon, they will fall at the timeon earth, the coin will fall faster to the groundExplanation:
A coin and feather dropped in a moon experience the same acceleration due to gravity as small as 1.625 m/s², and because of the absence of air resistance both will fall at the same rate to the ground.
If the same coin and feather are dropped in the earth, they will experience the same acceleration due to gravity of 9.81 m/s² and because of the presence of air resistance, the heavier object (coin) will be pulled faster to the ground by gravity than the lighter object (feather).
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
A ball has a mass of 4.65kg and approximates a ping pong ball of mass 0.060kg that is at rest by striking it in an elastic collision. The initial velocity of the bowling ball is 5.00m/s, determine the final velocities of both masses after the collision.
Answer:
Look at work
Explanation:
Elastic Collision: Ki=Kf
M1=4.65kg
M2: 0.060kg
v1=5m/s
v2=0m/s
4.65*5+0.060*0=4.65*v1'+0.060*v2'
23.25+0=4.65v1'+0.060v2'
Also since it is an elastic collision we can use
v1+v1'=v2+v2'
4.65+v1'=v2'
4.65+v1'=v2'
Substitute into the earlier equation
23.25=4.65v1'+0.060(4.65+v1')
Expand
23.25=4.65v1'+0.279+0.06v1'
Solve for v1'
22.971=4.71v1'
v1'=4.88m/s
v2'=4.65+4.88=9.53m/s
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.
What is the magnitude of the velocity after it hits the ground?
Answer:
9.25 m/s
Explanation:
Which physical phenomenon is illustrated by the fact that the prism has different refractive indices for different colors
Answer:
The incoming white light is composed of light of different colors,
Since these different colors have different refractive indices they are refracted at different angles from one another.
The output light is then separated by color creating a color spectrum.
Since n is greater for shorter wavelengths (violet colors) these wavelengths are refracted thru the larger angles.
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g
Answer:
w = 1,066 rad / s
Explanation:
For this exercise we use Newton's second law
F = m a
the centripetal acceleration is
a = w² r
indicate that the force is the mass of the body times the acceleration
F = m 0.58g = m 0.58 9.8
F = 5.684 m
we substitute
5.684 m = m w² r
w = [tex]\sqrt{5.684/r}[/tex]
To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m
w = [tex]\sqrt{ 5.684/5}[/tex]
w = 1,066 rad / s
A 2.90 m segment of wire supplying current to the motor of a submerged submarine carries 1400 A and feels a 2.00 N repulsive force from a parallel wire 4.50 cm away. What is the direction and magnitude (in A) of the current in the other wire? magnitude A direction
Answer:
[tex]I_2=30.9A[/tex]
Explanation:
From the question we are told that:
Wire segment [tex]l_s=2.9m[/tex]
Initial Current [tex]I_1=1400A[/tex]
Force [tex]F=2.00N[/tex]
Distance of Wire [tex]d=4.50cm=>0.0450m[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{\mu_0 * I_1*I_2*l_s}{2 \pi *r}[/tex]
[tex]F=\frac{4 \pi*10^{-7} *1400 I*I_2*2.9}{2 \pi *0.0450}[/tex]
[tex]I_2=\frac{22.5*10^-2}{2*10^{-7}*1400*2.6}[/tex]
[tex]I_2=30.9A[/tex]
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigger than the object and the screen is 5 m away from the mirror as shown in fig 5.2, calculate the focal length of the mirror.
Answer:
f = 1 m
Explanation:
The magnification of the lens is given by the formula:
[tex]M = \frac{q}{p}[/tex]
where,
M = Magnification = 4
q = image distance = 5 m
p = object distance = ?
Therefore,
[tex]4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m[/tex]
Now using thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\[/tex]
f = 1 m
describe the cause of earth's magnetism ?
Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 2 m/s, exactly how fast (in m2/s) is the area of the spill increasing when the radius is 39 m?
Explanation:
The area of a circle of radius r is given by
[tex]A = \pi r^2[/tex]
Taking the derivative of A with respect to time t, we get
[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]
We also know that
[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]
[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]
Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass 5.00 kg and radius 0.120 m. For each the translational speed of the center of mass is 4.00 m/s. Sphere A is a uniform solid sphere and sphere B is a thin-walled, hollow sphere. Part B How much work, in joules, must be done on the solid sphere to bring it to rest? Express your answer in joules. VO AE4D ? J WA Request Answer Submit Part C How much work, in joules, must be done on the hollow sphere to bring it to rest? Express your answer in joules. Wa Request
Answer:
Explanation:
Moment of inertia of solid sphere = 2/5 m R²
m is mass and R is radius of sphere.
Putting the values
Moment of inertia of solid sphere I₁
Moment of inertia of hollow sphere I₂
Kinetic energy of solid sphere ( both linear and rotational )
= 1/2 ( m v² + I₁ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/5 m R² ω²)
= 1/2 ( m v² + 2/5 m v²)
=1/2 x 7 / 5 m v²
= 0.7 x 5 x 4² = 56 J .
This will be equal to work to be done to stop it.
Kinetic energy of hollow sphere ( both linear and rotational )
= 1/2 ( m v² + I₂ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/3 m R² ω²)
= 1/2 ( m v² + 2/3 m v²)
=1/2 x 5 / 3 m v²
= 0.833 x 5 x 4² = 66.64 J .
This will be equal to work to be done to stop it.
If the cornea is reshaped (this can be surgically done or with contact lenses) to correct myopia, should its curvature be made greater or smaller? Explain. Also, explain how hyperopia can be corrected?
Answer:
Myopia curvature of the cornea if it is negative the curvatures are positive,
hypermetry,
Explanation:
Myopia is the visual defect that does not allow to see distant objects, which is why it is corrected with a divergent lens so that the image is formed on the retina, therefore, by reforming the curvature of the cornea if it is negative
therefore the curvature must decrease
To correct hypermetry, the curvatures are positive, consequently the curvature of the lens must increase
Could you show detailed steps in how to solve this problem please
Answer: See attached pic. Hope this helps.
Explanation:
A frictionless piston-cylinder device contains 10 kg of superheated vapor at 550 kPa and 340oC. Steam is then cooled at constant pressure until 60 percent of it, by mass, condenses. Determine (a) the work (W) done during the process. (b) What-if Scenario: What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
Answer:
a) the work (W) done during the process is -2043.25 kJ
b) the work (W) done during the process is -2418.96 kJ
Explanation:
Given the data in the question;
mass of water vapor m = 10 kg
initial pressure P₁ = 550 kPa
Initial temperature T₁ = 340 °C
steam cooled at constant pressure until 60 percent of it, by mass, condenses; x = 100% - 60% = 40% = 0.4
from superheated steam table
specific volume v₁ = 0.5092 m³/kg
so the properties of steam at p₂ = 550 kPa, and dryness fraction
x = 0.4
specific volume v₂ = v[tex]_f[/tex] + xv[tex]_{fg[/tex]
v₂ = 0.001097 + 0.4( 0.34261 - 0.001097 )
v₂ = 0.1377 m³/kg
Now, work done during the process;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.1377 - 0.5092 )
W = 5500 × -0.3715
W = -2043.25 kJ
Therefore, the work (W) done during the process is -2043.25 kJ
( The negative, indicates work is done on the system )
b)
What would the work done be if steam were cooled at constant pressure until 80 percent of it, by mass, condenses
x₂ = 100% - 80% = 20% = 0.2
specific volume v₂ = v[tex]_f[/tex] + x₂v[tex]_{fg[/tex]
v₂ = 0.001097 + 0.2( 0.34261 - 0.001097 )
v₂ = 0.06939 m³/kg
Now, work done during the process will be;
W = mP₁( v₂ - v₁ )
W = 10 × 550( 0.06939 - 0.5092 )
W = 5500 × -0.43981
W = -2418.96 kJ
Therefore, the work (W) done during the process is -2418.96 kJ
Consider a tall building of height 200.0 m. A stone A is dropped from the top (from the cornice of the building). One second later another stone B is thrown vertically up from the point on the ground just below the point from where stone A is dropped.Birthstones meet at half the height of the tower. (a) Find the initial velocity of vertical throw of stone B.(b) Find the velocities of A and B, just before they meet.
Answer:
a) v₀ = 44.27 m / s, b) stone A v = 44.276 m / s, stone B v = 0.006 m / s
Explanation:
a) This is a kinematics exercise, let's start by finding the time it takes for stone A to reach half the height of the building y = 100 m
y = y₀ + v₀ t - ½ gt²
as the stone is released its initial velocity is zero
y- y₀ = 0 - ½ g t²
t = [tex]\sqrt{ -2(y-y_o)/g}[/tex]
t = [tex]\sqrt{ -2(100-200)/9.8}[/tex]
t = 4.518 s
now we can find the initial velocity of stone B to reach this height at the same time
y = y₀ + v₀ t - ½ g t²
stone B leaves the floor so its initial height is zero
100 = 0 + v₀ 4.518 - ½ 9.8 4.518²
100 = 4.518 v₀ - 100.02
v₀ = [tex]\frac{100-100.02}{4.518}[/tex]
v₀ = 44.27 m / s
b) the speed of the two stones at the meeting point
stone A
v = v₀ - gt
v = 0 - 9.8 4.518
v = 44.276 m / s
stone B
v = v₀ -g t
v = 44.27 - 9.8 4.518
v = 0.006 m / s
g a horizontal wheel of radius is rotating about a vertical axis. What is the magnitude of the resultant acceleration of a bug that is hanging tightl on the rim of the wheel
Answer:
a = w² r
Explanation:
In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.
a = v² / r
angular and linear variables are related
v = w r
we substitute
a = w² r
where r is the radius of the wheel
The primary coil in a transformer has 250 turns; the secondary coil has 500. Which is correct?
a. This is a step-down transformer.
b. The voltage in the secondary coil will be higher than in the primary.
c. The power in the secondary coil is greater.
d. The power in the primary coil is greater.
Explanation:
option b is the correct one
a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration
Answer:
0
Explanation:
a = dv/dt
if v is constant than the slope of the v graph will be 0, so dv/dt is 0
a= 0
How many neutrons are in an isotope of selenium-83
A. 34
B. 83
C. 49
D. 117
Answer:
C.49 is yr ans...
hope it helpsstay safe healthy and happy....If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact would it make? And could you also show me the equation to solve similar problems myself. Thank you.
Answer:
The impact force is 98000 N.
Explanation:
mass = 10 tons
The impact force is the weight of the object.
Weight =mass x gravity
W = 10 x 1000 x 9.8
W = 98000 N
The impact force is 98000 N.
If there are no other changes, explain what effect reducing the mass of the car will have on its acceleration when starting to move.
Answer:
when the mass of an object is decreased, the acceleration will increase
when mass is increased, acceleration decreases
Warm air rises because faster moving molecules tend to move to regions of less
A) density.
B) pressure.
C) both of these
D) none of the above
Answer:
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Explanation:
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How are Newton’s 1 and 2 law related?