[tex]\sf\underline{Solution:}[/tex]
Here , the density of the material is 4g cm³ but it is not given in CGS system.
$\sf{As\:we\:know\:that:}$
$\sf\bold{Density=}$ $\sf\dfrac{Mass}{Volume}$
$\space$
[tex]\sf{Now,according \: to \:the\:question:}[/tex]
$\sf\small{Density\:of\:the\:material=4}$ $\sf\dfrac{g}{cm^2}$
$\space$
$\sf{It\:is\:given\:that:}$
In the system of units the mass is 100gram.
$\space$
Hence,
$\sf{The\:mass\:unit\:for\:4g=}$ $\sf\dfrac{4}{100}$ $\sf{units}$
$\space$
In the system of units,the length is 10cm.
Henceforth,
$\sf\small{The\:length \:for\:1cm\:units=}$ $\sf\dfrac{1}{10}$ $\sf{units}$
$\space$
☆ Substitute the required values in the given formula-
$\sf\purple{Density=}$ $\sf\dfrac\purple{Mass}\purple{volume}$
$\space$
$\sf\underline\bold{Density\:of\:the\:material:}$
= $\sf\dfrac{4/100}{1/10^3}$ $\sf\bold{units}$
$\space$
= $\sf\dfrac{4/100}{1/1000}$ $\sf\bold{units}$
$\space$
= $\sf\dfrac{4000}{100}$ $\sf\bold{units}$
$\space$
$\sf\underline\bold\blue{=40\:units}$
$\sf\small{Therefore,option\:2nd\:is\:correct!}$
_______________________________
What does electrical energy transform into when used by objects in a home?
Answer:
What can electrical energy be transformed into?
Electrical energy is transformed into many forms - mechanical/kinetic, sound, heat, light, and other forms of electromagnetic radiation - by everyday appliances.
Explanation:
Answer:
none of the other answers could be correct.
Explanation:
The sun nuclear energy transfers to radiant energy A fire Chemical energy transfers to thermal and radiant energy. wind turbine Kinetic motion energy transfers to electrical energy. Lightbulb Electrical energy transfers to radiant and thermal energy. Radio Electrical energy transfers to sound energy.
What is centripetal force and elastic force
Answer:
centripetal force is a net force that acts on an object to keep it moving along a circular path and elastic force acts to return a spring to its natural length
Two billiard balls having the same mass and speed roll directly toward each other. What is their combined momentum after they meet
Their combined momentum after they meet is 0.
What is law of conservation of momentum?According to the law of conservation of momentum, absent an external force, the combined momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.
Let the mass of each billiard balls = m.
If the initial velocity of first ball is u; according to the question, the initial velocity of the second ball be -u.
Hence, initial momentum of the first ball is: P₁ = mu
initial momentum of the second ball is: P₂ = -mu
Hence, total initial momentum of the two balls = P₁ + P₂ = mu + ( -mu) = 0.
So, according to law of conservation of momentum:
Their combined momentum after they meet is = 0.
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The combined momentum of the two billiard balls after they meet is zero.
When two pool balls of the same size and speed roll directly at each other and collide, their momentum will increase.
Let us denote the speed of the first ball as [tex]\rm p_1[/tex] and the speed of the second ball as [tex]\rm p_2[/tex]. The momentum of the balls are equal in magnitude but opposite in direction because they have the same mass and velocity.
[tex]\rm p_1= -p_2[/tex]
When the balls collide, their momentum increases:
Total momentum after the collision= [tex]\rm p_1+ p_2[/tex]
Substituting the relationship
[tex]\rm p_1= -p_2[/tex]
Total momentum after the collision
[tex]\rm p_1- p_1= 0[/tex]
Therefore, the combined momentum of the two billiard balls after they meet is zero.
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what is the velocity of 8.01
Answer: The velocity of what?
1 point
A hairdryer uses 10 A of current when plugged into a 120 V outlet. How
much power does it use?
Answer:
1200 watt
Explanation:
P= I*V=10*120=1200watt
A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure.
a) What is the range of values of the initial velocity so that the projectile falls between points M and N?
Answer: A
Explanation:
I want my points so yea
An eagle flying at a constant 120 km/h and has kinetic energy of 2,800 J. What is the mass of the eagle?
Answer:
The mass of the eagle is about 5 Kg
Explanation:
1/2 M= Ke/V^2
120 km/h = 33.3333m/s
1/2 M = 2,800/33.3333^2
1/2 M = 2,800/ 1111.10888889
1/2 M = 2.52000504001
(2) 1/2 M = (2) 2.52000504h001
= M = 5.04001008002
About 5 Kilograms
If a 75 kg box collides with a stationary 35 kg box with a force of 110 N, what must be true of the magnitude of the reaction force?
According to Newton third law of motion, what must be true of the magnitude of the reaction force is that the reaction force will be equal and opposite to the action force of 110 newtons
From Newton third law of motion which state that in every action of force, there will be equal and opposite reaction.
If a 75 kg box collides with a stationary 35 kg box with a force of 110 N, the mass 35 kg will produce a reaction force which is equal and opposite to the action force 110 newton received from mass 75 kg.
Therefore, what must be true of the magnitude of the reaction force from mass 35 kg is that the mass will produce an equal and opposite force equal to 110 Newtons.
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A plane flies from New York to LA at a constant speed of 800 km an hour how long will it take the plane to fly to 4,200 kilometers
Answer:
3.5 hours
Explanation:
Does the coefficient of kinetic friction depend on the speed?
Answer:
yes because if the speed/K.E. of body increases the friction will also apply more instantly
In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg. After the two balls collide, the final momentum of the cue ball is 0.01 kg. what is the magnitude of the final momentum of the eight ball
This question involves the concepts of the law of conservation of momentum.
The magnitude of the final momentum of the eight ball is "0.22 N.s".
According to the law of conservation of momentum:
[tex]P_{i1}+P_{i2}=P_{f1}+P_{f2}[/tex]
where,
[tex]P_{i1}[/tex] = initial momentum of the cue ball = 0.23 N.s
[tex]P_{i2}[/tex] = initial momentum of the eight ball = 0 N.s (since ball is initially at rest)
[tex]P_{f1}[/tex] = final momentum of the cue ball = 0.01 N.s
[tex]P_{f2}[/tex] = final momentum of the eight ball = ?
Therefore,
[tex]0.23\ N.s + 0\N.s = 0.01\ N.s+P_{f2}\\\\P_{f2} = 0.22\ N.s[/tex]
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Snow on the ground prevents polar climate systems from gaining heat by what mechanism?
Question 7 options:
Heating by greenhouse gases
Snow reflecting solar radiation
Heat spread from the equator
Release of heat from Earth's core
The characteristics of the thermal heating mechanisms allow to find the correct answer for how snow prevents soil heating is:
b) Snow reflecting solar radiation.
The heating of the system occurs by absorption of energy by three basic mechanisms:
Conduction. It occurs when the bodies are in constant and energy is transferred by the oscillations of the elements of the body, but there is no physical displacement of parts of it. Convection. This mechanism occurs mainly with liquid and gases, it is due to the displacement of masses of the element due to its greater energy. Radiation. In these mechanisms there is no physical contact and the energy is transferred by the emission of electromagnetic waves.
In this case, the snow receives electromagnetism waves from the sun, a large part of these waves are reflected by the snow, therefore they do not reach the ground and cannot be absorbed, thus decreasing its heating.
Let's analyze the claims:
a) False. Greenhouse gases do not interfere with snow on the ground.
b) True. This mechanism reduces radiation heating.
c) False. The heat that comes from the equator warms the air.
d) False. Heating by the Earth's core occurs everywhere and comes from the center of the earth, so it does not interact with snow.
In conclusion, with the characteristics of the thermal heating mechanisms, we can find the correct answer for how snow prevents soil heating is:
b) Snow reflecting solar radiation
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How far does an object travel if it is moving at a speed of 4 m/s for 3.2 seconds?
SOMEONE PLEASE HELP ME
Answer:
12.8 metres
Explanation:
distance = speed x time
distance = 4 x 3.2
distance = 12.8
You plug your microwave into an outlet and then you heat up a piece of pizza in it. This is an example of an...
Microwave Radiation????????????
The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what velocity must the ISS be moving in order to stay in its orbit?(1 point)
A) 7.91 × 10^3 m/s
B) 3.12 × 10^4 m/s
C) 7.66 × 10^3 m/s
D) 8.17 × 10^3 m/s
Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface. How does the velocity of Satellite A compare to the velocity of Satellite B?(1 point)
A) It depends on the masses of the satellites.
B) The velocity of Satellite A is greater than the velocity of Satellite B.
C) The velocity of Satellite B is equal to the velocity of Satellite A.
D) The velocity of Satellite A is less than the velocity of Satellite B.
What is the direction of the net force acting on a satellite as it orbits Earth at a constant speed?(1 point)
A) in the direction the satellite is moving
B) toward the center of Earth
C) away from the center of Earth
D) opposite the direction the satellite is moving
What happens to the gravitational force and orbital velocity of a satellite as the satellite transfers to an orbit that is closer to Earth?(1 point)
A) The gravitational force decreases and the velocity increases.
B) The gravitational force increases and the velocity increases.
C) The gravitational force decreases and the velocity decreases.
D) The gravitational force increases and the velocity decreases.
Newton's second law and universal gravitation law allow to find the results for questions about the motion of satellites in orbit are:
1) The correct answer is D: v = 8.17 10³ m/s
2) The correct answer is B
The velocity of Satellite A is greater than the velocity of Satellite B.
3) The correct answer is B
Toward the center of Earth
4) The correct answer is B
The gravitational force increases and the velocity increases.
Newton's second law establishes a relationship between force, mass, and the acceleration of bodies.
F = m a
In the case of the satellite the force is given by the law of universal gravitation.
[tex]F = - G \frac{Mm}{r^2 }[/tex]
Where G is the constant of universal gravitation. M and m the mass of each object and r the distance between them.
In this case the satellite is in a circular orbit, therefore the acceleration is centripetal.
[tex]a = \frac{v^2}{r}[/tex]
We substitute.
[tex]G \frac{Mm}{r^2} = m \frac{v^2}{r}[/tex]
[tex]\frac{GM}{r} = v^2[/tex]
Let's analyze the answers to find the correct one.
1) They indicate the height of the space station r = 4.08 10⁵ m and ask the speed.
[tex]v= \sqrt{ \frac{6.67 \ 10 ^{-11} 5.9 \ 10^{24}}{4.08 \ 10^6 } }[/tex]
v = 9.82 10³ m / s
The correct answer is D.
2) Satellite A has an orbit of hₐ = 500 km and satellite b an orbit of
h_b = 800 km
The distance from the center of the earth to each satellite is:
rₐ = R + hₐ
r_b = R + h_b
rₐ = 6.37 106 + 500 10³ = 6.87 10⁶ m
r_b = 6.37 10⁶ + 800 10³ = 7.17 10⁶ m
Let's find the ratio of the speeds
[tex]\frac{v_a}{v_b} = \sqrt{ \frac{r_b}{r_a} } \\\frac{v_a}{v_b} = \sqrt{ \frac{7.17}{6.87} }[/tex]
[tex]\frac{v_a}{v_b}[/tex] = 1,022
we see that the speed of satellite a is slightly greater than the speed of satellite b.
Let's analyze the claims.
A) False. The speed does not depend on the mass of the satellites.
B) True. The velocity of a is slightly greater than the velocity of b.
C) False. The speed of a is greater.
D) False. The speed of a is greater.
3) As the orbit is circular, the force must be radial, that is, it points towards the center of the earth.
Let's analyze the claims.
A) False. The speed modulus does not change, therefore there is no acceleration in the direction of the satellite.
B) True. Aim for the center of the Earth, change the direction of the velocity.
C) False. Aim for the scepter of the earth.
D) false. The modulus of velocity is constant and the direction changes towards the center of the earth, therefore the force must go towards the center of the earth.
4) The force in the law of universal gravitation increased as the distance decreased.
When a satellite approaches the earth its speed must increase since the speed is proportional in inverse of the square root of the distance.
Let's examine the claims.
A) False. The attraction force increases.
B) True. You agree with the explanation.
C) False. The gravitational force increases.
D) False. Speed increases.
In conclusion, using Newton's second law and the universal law of gravitation we can find the results for the questions about the movement of satellites in orbit are:
1) The correct answer is D: v = 8.17 10³ m/s
2) The correct answer is B
The velocity of Satellite A is greater than the velocity of Satellite B.
3) The correct answer is B
Toward the center of Earth
4) The correct answer is B
The gravitational force increases and the velocity increases.
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Answer:
1. 7.66 × 10^3 m/s
2. The velocity of Satellite A is greater than the velocity of Satellite B.
3. toward the center of Earth
4. The gravitational force increases and the velocity increases.
Explanation:
Please help if you can!
Answer:
a
Explanation:
bc i said so its right lol
Find the area of the given figure
Answer:
77 sq.cm
Explanation:
Solution,
Trapezium
Perpendicular (P1)=8cm
Perpendicular(P2)=14cm
height(h)=7cm
Now,
We know that,
Area(A)= 1/2 ×h (P1+P2)
= 1/2 × 7cm (8cm+14cm)
= 1/2 × 7cm(22cm)
= 1/2 × 154 squarecm
=77 sq.cm
4) What is the kinetic energy of o 5 kg object moving at a velocity of 25 m/s?
Ek = 1/2 mv^2
= 1/2 × 5 × (25)^2
= 1562.5 J
How did Thomson's model get its name?
Answer:
thats actually a really good question. sadly i don't know the answer to it. but im sure that there are others who can!
draw the following vector quantity Using the coordinate system.
a. 190 newton east
b. 120km/hr, 250 north of east
c. 60 meters southwest
The given vectors quantities can be described by their properties of both
magnitude and direction.
a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.b. The velocity vector extending from (0, 0) to (108.76, 50.714) on the coordinate plane is attached.c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.Reasons:
a. The magnitude of the vector = 190 N
The direction in which the vector acts = East
Therefore, in vector form, we have;
[tex]\vec{F}[/tex] = 190 × cos(0)·i + 190 × sin(0)·j = 190·i
The vector can be represented by an horizontal line, 190 units long
Coordinate points on the vector = (0, 0) and (190, 0)
The drawing of the vector with the above points using MS Excel is attached.
b. Magnitude of the velocity vector = 120 km/hr. 25° North of east
Solution;
The vector form of the velocity is; [tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j, which gives;
[tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j
[tex]\vec{v}[/tex] ≈ 108.76·i + 50.714·j
Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)
The drawing of the vector is attached
c. The magnitude of the vector = 60 m
The direction of the vector is southwest = West 45° south
The vector form of the displacement is [tex]\vec{d}[/tex] = 60 × cos(45°)·i + 60 × sin(45°)·j
Which gives;
[tex]\vec{d}[/tex] = 30·√2·i + 30·√2·j
Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)
The drawing of the vector is attached
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A 0.40 kg bead slides on a straight frictionless wire with a velocity of 3.50 cm/s to the right. The
bead collides elastically with a larger 0.60 kg bead initially at rest. After the collision the smaller
bead moves to the left with a velocity of 0.70 cm/s. Find the distance the larger bead moves along
the wire in the first 5.0 s following the collision?||
Answer:
Total momentum before collision
P1 =.4 * 3.5 = 1.4 ignoring units here
Total momentum after collision
P2 = .6 * V - .4 * .7 = .6 V - .28
.6 V = 1.4 + .28 momentum before = momentum after
V = 2.8 cm/sec
In 5 sec V moves 2.8 cm/sec * 5 sec = 14 cm
rube goldberg machine steps
Answer:
Rube Goldberg Machine is "a comically involved, complicated invention, laboriously contrived to preform a simple operation." 2. What are the 6 Simple Machines? A. The 6 Simple Machines are: wedge, screw, lever, wheel and axel, inclined plane and pulley.
If you move two objects with opposite charges apart, what happens to the potential energy between them? Explain your response.
I have read it in the ncert science book
That any material charged with the same energy they repels each other while if they have opposite energy they attracts each other.
Hope it helps.
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Two horizontal curves on a bobsled run are banked at the same angle, but one has twice the radius of the other. The safe speed (no friction needed to stay on the run) for the smaller radius curve is v. What is the safe speed on the larger radius curve?
A) v/2
B) v /√ 2
C) 2v
D) v√2
I shall answer it here so this can help other students out
First, FN = mg/cos
On a banked curve, FNx is the only acceleration force
FNx = mv^2/r
FNx = FNsin
mg/cos * sin = mv^2/r
cancel out the mass (m) and sin*1/cos is tan
gtan = v^2/r
rgtan = v^2
√(rgtan) = v (this is the velocity for the smaller radius.)
call larger radius velocity X
for larger radius, it becomes 2r, so √(2rgtan) = X
understand you can do √xy = √x * √y
√(rgtan) * √2 = X (√(rgtan) = v)
v * √2 = X (where X is velocity with larger radius)
Answer:
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, \sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}∑F=
r
1
mv
2
=
r
2
mu
2
................ii
form i and ii we can write
v^2= \frac{1}{2} u^2v
2
=
2
1
u
2
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Please help with this ski question to find Delta y
Jeff Gordon leads his race and must drive into a curve at top speed to win it all. The radius of the curve is 1000 m and the coefficient of static friction between his tires and the dry pavement is 0.50.
Answer:
Apparently the curve is not banked.
The maximum frictional force is -
Ff = m g * .5
Fc = m v^2 / R where Fc is the centripetal force that must be present in
order for the car not to slip.
.5 g = v^2 / R equating forces
v^2 = .5 R g = .5 ^ 1000 * 9.80
v = (500 * 9.80)^1/2 = 70 m/s
(about 230 ft/sec = 230/88 * 60 = 157 mph
A 10.0-kg crate slides along a raised horizontal frictionless surface at a constant speed of 4.0 m/s. The crate then slides down a frictionless incline and across a second, roughened horizontal surface as shown in the figure. What is the kinetic energy of the crate as it reaches the lower surface?
The kinetic energy of the crate as it reaches the lower surface is 80 J
Kinetic energy is the energy possed by an object in motion. Mathematically, the kinetic energy can be expressed as follow:
KE = ½mv²
With the above formula, the kinetic energy of the crate can be obtained as follow:
Mass (m) = 10 KgVelocity (v) = 4 m/sKinetic energy (KE) =?KE = ½mv²
KE = ½ × 10 × 4²
KE = 5 × 16
KE = 80 J
Therefore, the kinetic energy of the crate is 80 J
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What determines the amount of gravitational force between
objects?
Answer:
The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them. This means that the force of gravity increases with mass, but decreases with increasing distance between objects.
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Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 10 m/s.
Hi there!
Kinetic energy can be calculated using the following:
[tex]\large\boxed{KE = \frac{1}{2}mv^2}}[/tex]
Where:
KE = Kinetic energy (J)
m = mass (kg)
v = velocity (m/s)
Plug in the given values:
[tex]KE = \frac{1}{2}(625)(10^2) = \boxed{31250J}[/tex]
A projectile is fired with a velocity of 320 ms at an angle of 30 degree to a horizontal.1 Find the time to reach the maximum height. 2 it's horizontal range.
(a) The time taken for the projectile to reach the maximum height is 32.65 s.
(b) The horizontal range of the projectile is 9,049.1 m.
The given parameters:
Initial velocity of the projectile, u = 320 m/sAngle of projection, = 30 degreesThe time taken for the projectile to reach the maximum height is calculated as follows;
[tex]v_f = u- gt\\\\0 = 320 - 9.8t\\\\9.8t = 320\\\\t = \frac{320}{9.8} \\\\t = 32.65 \ s[/tex]
The horizontal range of the projectile is calculated as follows;
[tex]R = \frac{u^2 sin(2\theta)}{g} \\\\R = \frac{320^2 \times sin(2\times 30)}{9.8} \\\\R = 9,049.1 \ m[/tex]
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