Answer:
The answer is "512 J".
Explanation:
bullet mass [tex]m_1 = 10 g= 10^{-2} \ kg\\\\[/tex]
initial speed [tex]u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\[/tex]
block mass [tex]m_2 = 4\ Kg[/tex]
initial speed [tex]v_2 =-4.2 \frac{m}{s}[/tex]
final speed [tex]v_2= 0[/tex]
Let [tex]v_1[/tex] will be the bullet speed after collision:
throughout the consevation the linear moemuntum
[tex]\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\[/tex]
[tex]= 320 \frac{m}{s}[/tex]
The kinetic energy of the bullet in its emerges from the block
[tex]k=\frac{1}{2} m_1 v_1^2[/tex]
[tex]=\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J[/tex]
HELP WITH SCIENCE STAAR QUESTIONS
Answer:
EASY PEASY LEMON-SQUEEZY.
Explanation:
B
Mark Brainliest.
un cubo de aluminio tiene un volumen de 45cm3 cuál es su masa en gramos
The energy of motion is called...?
The gravitational force between two objects is proportional to the square of the distance between the two objects.
a. True
b. False
Answer: True!
Explanation: The force is proportional to the square of the distance between 2 point masses
Spring is a time when animals reproduce. What might you see a heron do? Build a nest A. Grow feathers A. Hibernate Migrate
Answer:
A. Build a nest.
Explanation:
It wont do any of the other things and they need a place to sleep and lay eggs plus it is a bird
The heron birds belong to Ardeidae family. Some of the species among them are referred to as egrets or bitterns rather than a heron. A heron bird build a nest during Spring. The correct option is A.
What is Heron bird?The heron is known as a long-legged, long-necked, fresh water and coastal birds. It a medium to large size bird with long necks and legs. They exhibit sexual dimorphism in size. The two male and female sex of the same species show distinct attributes beyond their difference in sexual organs.
Almost all the heron species are found in water and they are essentially non-swimming water birds which feed on the margins of lakes, ponds, sea, rivers and swamps. The small species of heron are generally referred to as the dwarf bittern.
The nesting season indicates the time of year during which birds build nests and lay eggs in them. In most cases bring up their young. It is in the spring season lot of food are available.
Thus the correct option is A.
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An ampere is measured in
Seconds
Hours
Answer:
Seconds
Explanation:
Hope it helps you in your learning process.
The rate of flow of charge (per second) through a conductor is called electric current.
S. I unit of electric current is Ampere (named after French Physicist Andre Marie Ampere)
A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of mass 1.51 kg and length 1.79 meters spinning clockwise with an angular velocity of 5.12 rad/s is dropped on the spinning disk and stuck to it (the centers of the disk and the rod coincide). The combined system continues to spin with a common final angular velocity. Calculate the magnitude of the loss in rotational kinetic energy due to the collision
Answer:
The loss in rotational kinetic energy due to the collision is 36.585 J.
Explanation:
Given;
mass of the disk, m₁ = 1.64 kg
radius of the disk, r = 0. 61 m
angular velocity of the disk, ω₁ = 17.6 rad/s
mass of the rod, m₂ = 1.51 kg
length of the rod, L = 1.79 m
angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)
let the counter-clockwise be the positive direction
let the clock-wise be the negative direction
The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;
m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)
where;
ωf is the common final angular velocity
1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)
21.1328 = ωf(3.15)
ωf = 21.1328 / 3.15
ωf = 6.709 rad/s
The moment of inertia of the disk is calculated as follows;
[tex]I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2[/tex]
The moment of inertia of the rod about its center is calculated as follows;
[tex]I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2[/tex]
The initial rotational kinetic energy of the disk and rod;
[tex]K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J[/tex]
The final rotational kinetic energy of the disk-rod system is calculated as follows;
[tex]K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J[/tex]
The loss in rotational kinetic energy due to the collision is calculated as follows;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J[/tex]
Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.
A honey bee's wings beat at 230 beats per second. If the speed of sound in air is 340 m/s, what is the wavelength of
the sound waves?
Answer:
[tex]from \: the \: wave \: equation \\ velocity = frequency \times wavelength \\ 340 = 230 \times \lambda \\ \lambda = \frac{340}{230} \\ \lambda = 1.5 \: m[/tex]
A person is dragging a packing crate of mass 74.9 kg across a rough horizontal floor where the coefficient of kinetic friction is 0.35. He exerts a force F at and angle 43.0 degrees above the horizontal. What is the Force F such that the crate moves at a constant speed
Answer:
351.28 N
Explanation:
Let F be the force on the object and f be the frictional force. The component of the force acting in the horizontal direction causing the object to move is FcosФ where Ф is the angle between F and the horizontal = 43.0°. The frictional force on the packing crate f = μN where μ = coefficient of kinetic friction = 0.35 and N = normal force = W = weight of the packing crate = mg where m = mass of crate = 74.9 kg and g = acceleration due to gravity = 9.8 m/s². So, f = μN = μW = μmg
So, the net force on the packing crate is
FcosФ - f = ma
FcosФ - μmg = ma
Since the crate moves at constant speed, its acceleration a = 0
So, FcosФ - μmg = ma
FcosФ - μmg = m(0)
FcosФ - μmg = 0
FcosФ = μmg
F = μmg/cosФ
Substituting the values of the variables into the equation, we have
F = μmg/cosФ
F = 0.35 × 74.9 kg × 9.8 m/s²/cos43.0°
F = 256.907 kg-m/s²/0.73135
F = 351.28 kg-m/s²
F = 351.28 N
What is the difference between 1 celcius and 1 kelvin
Answer:
One degree unit on the Celcius scale is equivalent to one degree unit on the kelvin scale. The only difference between these two scales is the zero point. The zero point on the Celcius scale was defined as the freezing point of water, which means that there are higher and lower temperatures around it.
What is the rate of flow of electric charge around a circuit
Answer:
Current
Explanation:
hope it helps and your day will be full of happiness
A water pipe having a 4.00 cm inside diameter carries water into the basement of a house at a speed of 1.00 m/s and a pressure of 167 kPa. The pipe tapers to 1.4 cm and rises to the second floor 7.8 m above the input point. What is the speed at the second floor
Answer:
[tex]8.16\ \text{m/s}[/tex]
Explanation:
[tex]d_1[/tex] = Initial diameter = 4 cm
[tex]v_1[/tex] = Initial velocity = 1 m/s
[tex]d_2[/tex] = Final diameter = 7.8 m
[tex]v_2[/tex] = Final velocity
[tex]A[/tex] = Area = [tex]\pi\dfrac{d^2}{4}[/tex]
From the continuity equation we get
[tex]A_1v_1=A_2v_2\\\Rightarrow \pi\dfrac{d_1^2}{4}v_1=\pi\dfrac{d_2^2}{4}v_2\\\Rightarrow v_2=\dfrac{d_1^2}{d_2^2}v_1\\\Rightarrow v_2=\dfrac{4^2}{1.4^2}\times 1\\\Rightarrow v_2=8.16\ \text{m/s}[/tex]
The speed of water at the second floor is [tex]8.16\ \text{m/s}[/tex].
An astrophysicist mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them, and they appear concentric when viewed from either end). The lenses are identical, each with a positive (converging) focal length of 15.2 cm. They are separated by a distance of 40.2 cm. Lens 1 is to the left of Lens 2. In order to evaluate the lens combination as a single optical instrument, the teacher places an object 30.0 cm in front of (to the left of) Lens 1.
Required:
a. What is the final image's distance (in cm) from Lens 2? (Omit any sign that may result from your calculation.)
b. Where is the final image located?
c. What is the overall magnification of the lens pair, considered as a single optical instrument?
Answer:
1 / i + 1 / o = 1 / f thin lens equations
i = o f / (o - f) rearranging
Lens 1: object = 30 cm f = 15.2 cm
i1 = 30 * 15.2 / (30 - 15.2) = 30.8 cm
o2 = 40.2 - 30/8 = 9.4 cm distance of image 1 from lens 2
i2 = 9.4 * 15.2 / (9.4 - 15.2) = - 24.6 cm
The final image is 24.6 cm to the left of lens 2
The first image is inverted
The second image is erect (as seen from the first image)
So the final image is inverted
M = m1 * m2 = (-30.8 / 30) * (24.6 / 9.4) = -2.69
It can be deduced that the final image's distance from Lens 2 will be 30.8 cm.
How to calculate the distanceBy using the Lens formula, the distance will be calculated thus:
1/v + 1/u = 1/15.2
1/v + 1/30.0 = 1/15.2
v = 30.8cm
In this case, the image formed will be to the right.
Lastly, the overall magnification of the lens pair will be:
M = (30.8/30.0)[(-28.4/40.7 - 30.8)]
M = -2.94.
In conclusion, the magnification is -2.94.
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A uniform metre rule of mass 100g balance the 40cm mark when a mass x is placed at the 20cm mark
what is the value of x
Answer:
X = 50 g
Explanation:
Please see attached photo for explanation.
From the attached photo,
Anticlock–wise moment = X × 20
Clockwise moment = 100 × 10
Anticlock–wise moment = clockwise moment
X × 20 = 100 × 10
X × 20 = 1000
Divide both side by 20
X = 1000 / 20
X = 50 g
Therefore, the value of X is 50 g
11) A tank of kerosene with density of 750 kg/m3 has a syphon used to remove the fluid that then exits into the local atmosphere, with pressure of 101 kPa. The pressure above the kerosene in the tank is 120 kPa absolute. The syphon tube has a diameter of 2 cm, exits the tank rising to 10 cm above the level of the kerosene and then drops down to 15 cm below the level of the kerosene where it exits into the atmospheric pressure. Calculate the exit velocity from the tube.
Answer:
[tex]7.32\ \text{m/s}[/tex]
Explanation:
[tex]v_1[/tex] = Velocity at initial point = 0
[tex]P_1[/tex] = Pressure in tank = 120 kPa
[tex]P_2[/tex] = Pressure at outlet = 101 kPa
[tex]\rho[/tex] = Density of kerosene = [tex]750\ \text{kg/m}^3[/tex]
[tex]Z_1[/tex] = Tank height = 15 cm
[tex]Z_2[/tex] = Height of pipe exit = 0
[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From Bernoulli's equation we have
[tex]\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+Z_2\\\Rightarrow \dfrac{P_1}{\rho g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}\\\Rightarrow v_2=\sqrt{2g(\dfrac{P_1}{\rho g}+Z_1-\dfrac{P_2}{\rho g})}\\\Rightarrow v_2=\sqrt{2\times 9.81(\dfrac{120\times 10^3}{750\times 9.81}+0.15-\dfrac{101\times 10^3}{750\times 9.81})}\\\Rightarrow v_2=7.32\ \text{m/s}[/tex]
The exit velocity from the tube is [tex]7.32\ \text{m/s}[/tex].
The Earth Science students are making a human scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters (the length of a football field). How far from the school will the student representing Mars stand? A) 50 meters B) 105 meters 150 meters D) 1500 meters NEED HELP ASAP
In the model 1 AU = 100 meters (the length of a football field) , then mars would be 150 meters far from the school, therefore the correct answer is option C.
What is a unit of measurement?A unit of measurement is a specified magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind. It is determined by convention or regulation.
As given in the problem the Earth Science students are making a human-scale model of the solar system out on the school playground. The school itself represents the Sun. In the model 1 AU = 100 meters,
As given in the table marks is 1.5 AU far from Mars,
1 AU = 100 meters
1.5 AU = 1.5 × 100
= 150 meters
Thus, mars would be 150 meters far from the school, therefore the correct answer is option C.
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A man kicks the 150-g ball such that it leaves the ground at an angle of 60° and strikes the ground at the same elevation a distance of 12 m away (14 points). Determine: (a). The velocity of the ball instantly after the man kicks the ball; (6 points) (b). Determine the impulse of his foot on the ball at A. (8 points) Neglect the impulse caused by the ball’s weight while it’s being kicked.
Answer:
[tex]313.92\ \text{m/s}[/tex]
[tex]47.088\ \text{kg m/s}[/tex]
Explanation:
m = Mass of ball = 150 g
[tex]\theta[/tex] = Angle of kick = [tex]60^{\circ}[/tex]
[tex]x[/tex] = Displacement of ball in x direction = 12 m
Range of projectile is given by
[tex]x=\dfrac{u\sin^2\theta}{2g}\\\Rightarrow u=\dfrac{2xg}{\sin^2\theta}\\\Rightarrow u=\dfrac{2\times 12\times 9.81}{\sin^260^{\circ}}\\\Rightarrow u=313.92\ \text{m/s}[/tex]
The velocity of the ball instantly after the man kicks the ball is [tex]313.92\ \text{m/s}[/tex]
Impulse is given by
[tex]J=mu\\\Rightarrow J=0.15\times 313.92\\\Rightarrow J=47.088\ \text{kg m/s}[/tex]
The impulse of his foot on the ball is [tex]47.088\ \text{kg m/s}[/tex].
Question
A car is moving on a flat roadway, What can be said about this system
A. The Kecan change there is no conservation of energy
B. It is not an isolated system because the car moves,
c. The law of conservation of energy applies,
O D. There is no PE, so no conservation of energy.
s the gravitational force greater between the objects in Pair 1 or Pair 2? Explain why. (Picture shown below)
Answer:
Pair 1
Explanation:
The gravitational force greater between the objects in Pair 1 because in Pair 2 the objects are far apart and in pair 1 the object are more closer to each other.
So, Gravitational force directly proportional to distance
Thus, Increase in distance = Increased in gravitational force
-TheUnknownScientist
A car has a mass of 1.00x10 to the 3rd power kilograms, it has an acceleration of 4.5 meters/seconds, what is the net force on the car?
Explanation:
Net force on the car= mass of the car × acceleration
F=1×10^3×4.5
=4.5×10^3 N
Tendons are, essentially, elastic cords stretched between two fixed ends; as such, they can support standing waves. These resonances can be undesirable. The Achilles tendon connects the heel with a muscle in the calf. A woman has a 20-cm long tendon with a cross-section area of 130 mm^2. The density of tendon tissue is 1100 kg/m^3.
Required:
For a reasonable tension of 600 , what will be the fundamental resonant frequency of her Achilles tendon?
Answer:
161.938 Hz
Explanation:
the computation of the fundamental resonant frequency is shown below
p = 1100 kg/m^3
A = 130 mm^2
= 130 ×10^-6 m^2
T = 600 N
L = 20 cm
= 0.2 m
Now the linear density of tendon is
= 1100 kg/m^3 × 130 ×10^-6
= 0.143 kg/m
Now the wave of the string is
= √600 ÷ √0.143
= 64.775 m/s
Now finally the fundamental resonant frequency is
= 64.775 ÷ (2 × 0.2)
=161.938 Hz
why do substance takes long to condense than freezing?
To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as:
Answer:
dorsiflexion
Explanation:
To decrease the angle between the anterior surface of the foot and anterior surface of the lower leg is described as: dorsiflexion
Answer:
dorsiflexion
Explanation:
Hope this helps
What is the speed of an object moving around a 0.75 m radius circle that completes a revolution in 0.50 seconds?
Answer:
the speed of an object is 9.42 m/s
Explanation:
The computation of the speed of an object is given below:
v = 2πr ÷ t
where
v denotes the speed
r denotes the radius
r denotes the time
So,
= 2 × 3.14 × 0.75 ÷ 0.50
= 9.42 m/s
Hence, the speed of an object is 9.42 m/s
the same would be considered and relevant too
The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student builds a pendulum by attaching one end of a string of known length to the ceiling and the other end to a small solid sphere. The sphere is pulled back so the string makes a 5 angle with the vertical and released from rest with a theoretical amplitude of Oscillation At The student measures the resulting experimental period Tg and after the sphere makes several oscillations, measures the experiment oscillation. In the absence of air resistance, T T and A. Which of the following best explains how air resistance affects the results of the experiment?
A) with resistance because the sphere will speed up
B) With air resistance As , because although the sphere slows down, it also will travel for a greater time
C) With resistance because the sphere will travel a greater distance
D) With air resistance, 7>7y, because the sphere will slow down
E) with resistance because although the sphere slows down, also will travel shorter distance
Answer:
The answer is "Choice E".
Explanation:
In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.
1. Compared to sadness, depression:
a. Is more common
b. Lests longer
c. Has fewer symptoms
d. Is easier to recognize
A person hears sounds with a frequency in the range of about 16 Hz to 20000 Hz. What is the wavelength of sound in air and water corresponding to these frequencies?
Please let me know the answer and how you got the answer (what formula you used)
Answer:
v=f×λ
Explanation:
v=f×λ
v is the speed of sound
f is the frequency of sound
λ is the wavelength
The speed of sound in air and water is different, but the frequency can be calculated from the given range above.
HAVE FUN LEARNING!!!( sorry im just to lazy to get the number so...)
how do you define teaching?
Answer:
Teaching is the process of attending to people's needs, experiences and feelings, and making specific interventions to help them learn particular things.
Explanation:
Hope it helps✌Answer:
In education, teaching is the concerted sharing of knowledge and experience, which is usually organized within a discipline and, more generally, the provision of stimulus to the psychological and intellectual growth of a person by another person or artifact.
PLEASE HELP
A ball, 1.8 kg, is attached to the end of a rope and spun in a horizontal circle above a student's head. As the student rotates the ball in a horizontal clockwise circle their lab partner counts 53 rotations in one minute. What is the ball's angular velocity in radians per second?
Answer:
The angular speed of the ball in radians per second is 5.55 rad/s.
Explanation:
Given;
mass of the ball, m = 1.8 kg
number of the ball's rotation per minute, n = 53 RPM
The angular speed of the ball in radians per second is calculated as follows;
[tex]\omega = 53\frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s } \\\\\omega = 5.55 \ rad/s[/tex]
Therefore, the angular speed of the ball in radians per second is 5.55 rad/s.
Question 4 of 10 A student measures the time it takes for two reactions to be completed. Reaction A is completed in 39 seconds, and reaction B is completed in 50 seconds. What can the student conclude about the rates of these reactions? A. The rate of reaction B is higher B. The rate of reaction A is higher. C. The rates of reactions A and B are equal.
Answer:
B. the rate of reaction is higher
Explanation: