The average momentum of Cart 1 during the entire time shown before the collision is 0.24 kgm/s.
The given parameters:
Mass of the cart 1 = 500 g = 0.5 kgThe average velocity of the cart 1 before collision is calculated as follows;
[tex]v = \frac{\Delta x}{\Delta t} \\\\v = \frac{0.55 - 0.31}{0.5 - 0} \\\\v = 0.48 \ m/s\\\\[/tex]
The average momentum of Cart 1 during the entire time shown before the collision is calculated as follows;
[tex]P = mv\\\\P = 0.5 \times 0.48\\\\P = 0.24 \ kgm/s[/tex]
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A body with a uniform acceleration travels distances of 24m and 64m during the first two equal consecutive intervals of time, each of duration 4s. Determine the initial velocity and acceleration of the moving body.
Answer:
Explanation:
Average velocity in the 24 m interval is 24 / 4 = 6 m/s
Average velocity in the 64 m interval is 64 / 4 = 16 m/s
There is a 4 second interval between the two points where average velocity equals actual velocity
a = Δv/t = (vf - vi) / t = (16 - 6) / 4 = 2.5 m/s²
s = v₀t + ½at²
24 = v₀(4) + ½(2.5)4²
4v₀ = 24 - 20
v₀ = 1 m/s
Not asked for but the velocity at the end of the first segment and beginning of the second segment is 11 m/s and final velocity is 21 m/s
A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total energy.
Answer:
a = ω^2 A formula for max acceleration (ignoring sign)
V = ω A formula for max velocity
V^2 = ω^2 A^2 = a A from first equation
E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J
(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule
A block of wood
wood, with mass 1.34 kg rests stationary
on horizontal ground.
The coefficient of Kinetic
friction between the block and the ground is 0.966.
A bullet, with mass 0.250kg, moving horizontally
hits and sticks into the block of wood. We find
that the speed of the block of wood, with the
ballet embedded in it, just after collision is 11.9 m/s.
A) calculate the speed of the bullet before hitting the block of wood.
it, just after the collision
is 11-9mis.
as calculate the speed of the bullet before
s
hitting the block of wood.
Answer:
Explanation:
conservation of momentum
m(u) + M(0) = (m + M)v
u = (m + M)v/m
u = (0.250 + 1.35)(11.9) / 0.250
u = 76.16
u = 76.2 m/s
That's a fairly massive, and slow, bullet.
Which combination of three concurrent forces acting on a body could not produce equilibrium?
1
1 N, 3N, EN
2
2 N, 2N, 2N
.
3.
3 N, 4N, EN
4.
4N, 4N, 5N
All the three concurrent forces acting on a body will not produce equilibrium.
The given parameters:
1. 1 N, 3 N and 5 N
2. 2N, 2N and 2 N
3. 3N, 4N and 5 N
4. 4N, 4N and 5 N
Concurrent forces lie on the same plane and their line of action pass through a common point.
A body under concurrent forces is in equilibrium if the resultant of the forces on the body is zero.
[tex]\Sigma F = 0\\\\F_1 + F_2 + F_3 = 0\\\\F_1 + F_ 2 = - F_3[/tex]
where;
[tex]F_3[/tex] is the equilibrant force
First set of concurrent forces;
[tex]1 \ N \ + \ 3\ N = 4 \ N\\\\F_ 3 = 5 \ N\\\\5 \ N > 4 \ N[/tex]
Second set of concurrent forces;
[tex]2 \ N \ + \ 2 \ N = 4 \ N\\\\F_ 3 = 2 \ N\\\\4 \ N > 2 \ N[/tex]
Third set of concurrent forces;
[tex]3 \ N \ + \ 4 \ N = 7 \ N\\\\F_ 3 = 5 \ N\\\\7 \ N > 5 \ N[/tex]
Fourth set of concurrent forces;
[tex]4 \ N \ + \ 4 \ N = 8 \ N\\\\F_ 3 = 5 \ N\\\\8 \ N > 5 \ N[/tex]
Thus, we can conclude that all the three concurrent forces acting on a body will not produce equilibrium.
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5. Layer of Earth consisting of crust & upper layer of mantle ________
Answer:
lithosphere
Explanation:
hope this helps you!!
13) What property of matter would be measured with this piece of equipment?
A) The mass of an apple
hing
)
By the temperature of a room
The volume of water in a glass.
D) The length of a piece of string.
Answer:
a
Explanation:
i took the test 100%
Physical Science A 2021-2022
Why does increasing the number of trials increase confidence in the results of the experiment?
Answer:
Increasing the number of trials reduces the impact of any one imprecise measurement. … To increase the number of attempts, you can find an average result for the experiment, as well as find and discrepancies as human error if you perform an experiment several times.
Explanation:
hope it helps :)
Answer:
It is because the increase in the number of trials reduces the impact of any one imprecise measurement. Using an average value for data points provides a better representation of the true value.
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angular velocity ω of the wheel at the bottom of the incline?
Express your answer in radians per second.
The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.
For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.
As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.
i.e.
P.E = K.E + R.K.E
[tex]\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}[/tex]
[tex]\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}[/tex]
[tex]\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 + \omega^2}[/tex]
[tex]\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2} }[/tex]
[tex]\mathbf{\omega^2=\dfrac{39.24 }{2}}[/tex]
[tex]\mathbf{\omega=\sqrt{19.62 } \ rad/sec}[/tex]
[tex]\mathbf{\omega=4.429 \ rad/sec}[/tex]
Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec
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The angular velocity of the wheel depends on the mass, radius and the
mode of rotation of the wheel (with or without slipping).
The angle velocity at the bottom of the incline, ω ≈ 4.43 rad/secReasons:
The given parameters are;
Radius of the wheel, r = 2.0 m
Height of the incline, h = 8.0 m
Required:
Angular velocity of the wheel at the bottom of the incline.
Solution:
The potential energy of the wheel at the top of the hill, P.E. = m·g·h
[tex]Sum \ of \ the \ kinetic \ energy \ of \ the \ wheel, \ K.E. = \mathbf{\displaystyle \frac{1}{2} \cdot m \cdot v^2 + \frac{1}{2} \cdot I \cdot \omega ^2}[/tex]
Where;
v = The translational velocity of the wheel = ω·r
I = The moment of inertia of the wheel = m·r²
Therefore'
[tex]Sum \ of \ K.E. = \displaystyle \frac{1}{2} \cdot m \cdot (\omega \cdot r)^2 + \frac{1}{2} \cdot m \cdot r^2 \cdot \omega ^2 = \mathbf{m \cdot r^2 \cdot \omega^2}[/tex]
At the bottom of the hill, the potential energy is converted to kinetic energy
Therefore;
P.E. = Sum of K.E.
m·g·h = m·r²·ω²
g·h = r²·ω²
[tex]\displaystyle \omega = \sqrt{ \frac{g \cdot h}{r^2} } = \mathbf{ \frac{\sqrt{g \cdot h} }{r}}[/tex]
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]\displaystyle \omega = \frac{\sqrt{9.81 \times 8} }{2} \approx \mathbf{ 4.43}[/tex]
The angular velocity of the of the wheel at the bottom of the incline, ω ≈ 4.43 rad/secLearn more about the law of conservation of energy here:
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How large is the tension in a rope that is being used to accelerate a 100 kg box upward at 2m/s2?
[tex]\\ \sf\Rrightarrow T=F[/tex]
[tex]\\ \sf\Rrightarrow T=ma[/tex]
[tex]\\ \sf\Rrightarrow T=100(2)[/tex]
[tex]\\ \sf\Rrightarrow T=200N[/tex]
g A 24-gg bullet strikes and becomes embedded in a 1.50-kgkg block of wood placed on a horizontal surface just in front of the gun. Part A If the coefficient of kinetic friction between the block and the surface is 0.23, and the impact drives the block a distance of 9.5 mm before it comes to rest, what was the muzzle speed of the bullet
The muzzle speed of the bullet before the collision is 415.3 m/s.
The given parameters:
Mass of the bullet, m₁ = 24 gMass of the wood, m₂ = 1.5 kgCoefficient of kinetic friction, μk = 0.23Distance traveled by the block before stopping, d = 9.5 mApply the principle of work-energy theorem to determine the final velocity of the block-bullet system;
[tex]F_f \times d = \frac{1}{2} mv^2\\\\\mu_k F_n \times d = \frac{1}{2} mv^2\\\\\mu_ k (m_1 + m_2)g \times d = \frac{1}{2} (m_1 + m_2)v^2\\\\\mu_k g \times d= \frac{1}{2} v^2\\\\2\mu _k gd = v^2\\\\v= \sqrt{2\mu _k gd } \\\\v = \sqrt{2 \times 0.23 \times 9.8 \times 9.5} \\\\v = 6.54 \ m/s[/tex]
Apply the principle of conservation of linear momentum to determine the muzzle speed of the bullet;
[tex]m_1 u_1 \ + \ m_2u_2 = v(m_1 + m_2)\\\\0.024(u_1) \ + \ 1.5(0) = 6.54(0.024 + 1.5)\\\\0.024u_1 = 9.967\\\\u_1 = \frac{9.967}{0.024} \\\\u_1 = 415.3 \ m/s[/tex]
Thus, the muzzle speed of the bullet before the collision is 415.3 m/s.
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A 70 kg man is running up the stairs which is 3m high in 2s.(a)How much work is done by the man?(b)What is the power exerted by the man? (Use g = 10ms 2)
Explanation:
m = 70 kg
s = 3m
t =2s
g = 10 m/s²
(a)How much work is done by the man?
W = Fs
= mg X s
= 70 x 3 x 10
= 210 x 10
= 2100 Joule
(b)What is the power exerted by the man?
P = W/t
= 2100/2
P = 1050 Watt
(c) It is suggested that one side of the copper sheet cools to a lower temperature than the
other side.
Explain why this does not happen.
[2]
Answer:
Explanation:
The word "sheet" implies that the copper is quite thin.
Copper is also a very good conductor of heat.
Therefore, with a very short heat flow distance to cover and a high rate of heat transmission, temperature differences on either side of the sheet are almost instantaneously eliminated by heat flow.
esse is swinging Miguel in a circle at a tangential speed of 3.50 m/s. If the radius of the circle is
0.600 m and Miguel has a mass of 11.0 kg, what is the centripetal force on Miguel? Round to the nearest whole number.
Answer:
Explanation:
F = mv²/R
F = 11.0(3.50²)/0.600 = 225 N
Avery is experimenting with a simple circuit. She measures the current in the circuit three different times with a different battery each time. First, she uses a 1.5-volt battery. Next, she uses a 3-volt battery. Last, she uses a 9-volt battery. The resistance stays the same during each test. How did the current change for each test? Explain.
Answer: the current increases with each 3 volt and 9 volt. The relationship between resistance and current in a circuit is that the greater the resistance the less the current and the greater the current the less the resistance is. yayayay I could answer this I big brain :)
the conduction of heat from hot body to cold body is an example of what thermodynamics process?
Answer:
Heat flow
Explanation:
How many joules of energy does a 100-watt light bulb use per hour? How fast would a 70-kg person have to run to have that amount of kinetic energy?
Answer:
*1) 100 Joule energy
*2) 101.2 m/s
Explanation:
*1) 1J = 1w
100J = 100w
*2) A 70-kg person will have to run at a speed of 101.2 m/s to have that amount of kinetic energy.
I need help with this question!
Answer:
Explanation:
Conservation of momentum
In the x direction
2(15) + 2(-10) = 2(-5) + 2(vBfx)
vBfx = 10 m/s
In the y direction
2(30) + 2(5) = 2(20) + 2(vBfy)
vBfy = 15 m/s
vBf = 10i + 15j
KEi = ½(2)(15² + 30²) + ½(2)(-10² + 5²) = 1250 J
KEf = ½(2)(-5² + 20²) + ½(2)(10² + 15²) = 750 J
KEf - KEi = 750 - 1250 = -500 J
VERY EASY QUESTION FOR HIGH SCHOOL STUDENTS:
Which of the following frequencies would you expect a young person to be able to hear? 500 Hz, 6000 Hz, 25000 Hz, 15 Hz, 15000 Hz
Answer:
Explanation: 6000z
hey if you talk to me i will mark you as a brainliest and if you answer all my question
huh huh huh
Answer:
what will happen if i will answer ur questions?
Explanation:
is there gonna be a bad thing or a good thing
I'm reasking this because I keep getting links not a real answer and I need a proper answer soon please
Answer:
Adding salt to the water increases the density of the solution because the salt increases the mass without changing the volume very much.
Explanation: the explanation is in a file
12- Calculate the power when a force of 60N moves an object over a distance of 0.6 km in 20
minutes
A. 100watts
B. 6,000 watts
C. 0.25watts
D. 30 watts
Hi there!
To solve, we must begin by calculating the total WORK done on the object.
W = F · d (Force · displacement)
Plug in the given values. Remember to convert km to m:
1 km = 1000 m
0.6 km = 600 m
W = 60 · 600 = 36000 J
Now, we can solve for power:
P = W/t
Convert minutes to seconds:
1 min = 60 sec
20 min = 1200 sec
P = 36000/1200 = 30 W ⇒ Choice D.
That’s easy please tell me!
Would it be possible to predict the speeds that a coaster will reach before it’s ever placed on the track?
Yes, it's possible to predict the speeds that a coaster will reach before it’s
ever placed on the track.
This is usually calculated with the potential energy which is
Potential energy = m g h
where m is mass, g is acceleration due to gravity and h is height.
The given formula above is used in predicting the speeds that a coaster will
reach before it’s ever placed on the track.
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Can someone help me solve this problems please? It's a physics problem.
Answer:
i cant see
Explanation:
but im smart
Please answer the question in the picture
A car is moving north on a freeway. If a bug is flying south on the freeway, is the momentum of the bug positive or negative?
Neither
Positive
Negative
Can be both depending on the weather
Negative
Because the car is moving up and the bug is moving down. but it also depends on the weather so choice between one of those two I think is Negative but I may be wrong.
The 0.01 kg marble is dropped from rest at A through the smooth glass tube and accumulate in the basket at C as shown in Figure Q2(b). Determine: i) the velocity of the marble at B ii) the horizontal distance R of the basket from the end of the tube, and iii) the speed at which the marble falls into the basket.
Crazy Wally Ok Ok ok hhahahaha
i need help with the problem below
Answer:
Explanation:
a) F = ma
a = F/m
a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s
b) t = v/a
t = 200 / 7.2 x 10⁻⁶
t = 2.8 x 10⁷ s about 10½ months
c) v² = u² + 2as
s = (v² - u²) / 2a
s = (200² - 0²) / (2( 7.2 x 10⁻⁶))
s = 2.8 x 10⁹ m nearly 7 times around the earth
And all this assumes NO FRICTION.
while spinning in a centrifuge a 70.0 kg astronaut experiences an acceleration of 5.00 g, or five times the acceleration due to gravity on the earth. what is the centripetal force acting on her
Answer:
Explanation:
70.0(5.00)(9.81) = 3,433.5 = 3430 N
To solve this we must be knowing each and every concept related to centripetal force and its calculations. Therefore, the centripetal force acting on her is 3430 N.
What is centripetal force?The term centripetal relates to a propensity to gravitate toward the center. Centripetal refers to moving in the direction of the center. The force that maintains an item moving in a circular direction and helps it stay on the path is referred to as centripetal force.
Furthermore, centrifugal force is indeed the tendency of things to deviate from a circular route and fly in a straight line. People frequently confuse centripetal force with centrifugal force.
Mathematically,
F = m a
= 70 acceleration
= 70 × 5 × 9.81
= 3430 N
Therefore, the centripetal force acting on her is 3430 N.
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Explain how the removal of heat energy affects the speed of the particles in a substance
Answer:
The removal of heat energy slows the speed of particles
Explanation:
When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.