This question involves the concepts of Gravitational Force and Weight force.
The value of acceleration is "b. 7.32 m/s²".
At the given height the weight of the rocket must be equal to the gravitational force between rocket and the Earth:
[tex]W=F_G\\mg=\frac{GmM}{R^2}\\\\g=\frac{GM}{R^2}[/tex]
where,
g = acceleration = ?
G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = mass of earth = 5.97 x 10²⁴ kg
R = Radius of Earth + Height = 6.378 x 10⁶ m + 1 x 10⁶ m = 7.378 x 10⁶ m
Therefore,
[tex]g=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{(7.378\ x\ 10^6\ m)^2}[/tex]
g = 7.32 m/s²
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As a truck rounds a curve, a box in the bed of the truck slides to the side farthest from the center of the curve. This movement of the box is a result of
Answer:
inertia .
because yes
the conduction of heat from hot body to cold body is an example of what thermodynamics process?
Answer:
Heat flow
Explanation:
Two steel guitar strings have the same length. String A has a diameter of 0.489 mm and is under 410 N of tension. String B has a diameter of 1.27 mm and is under a tension of 809 N. Calculate the ratio of the wave speeds, vA/vB, in these two strings.
Answer:
Explanation:
vA / vB = √(TA/(m/L)) / √(TB/(m/L))
The lengths are the same, so the L divides out to 1
The material is identical so the mass will be directly proportional to the cross sectional area of the string
vA / vB = √(TA/(πdA²/4)) / √(TB/(πdB²/4))
π/4 is common so divides out to 1
vA / vB = √(TA/dA²) / √(TB/dB²)
vA / vB = √(410/0.489²) / √(809/1.27²)
vA / vB = 41.407 / 23.396
vA / vB = 1.8488961...
vA / vB = 1.85
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 :)
A ball has the energy to move 30 m/s with the mass of 5. What is the energy
Need help ASAP, 1 MC
Answer:
The first one is the only one that is true all the time
Explanation:
The second one may be true if friction is high enough.
The other three are false all the time
Someone please help with this question. From my knowledge the answer I believe to be correct is 4Em but I’m still not so sure. Please explain!
Answer choices:
1/2 Em
Em
2Em
4Em
Answer:
Explanation:
For an ideal spring over a frictionless horizontal surface, stored energy is only a function of the spring constant k and the distance of compression. The mass of the block doing the compressing is irrelevant
Energy stored in the first example is
Em = ½kd²
Energy stored in the second example is
E₂m = ½k(2d)² = 4(½kd²) = 4Em
So the second situation has four times as much stored spring potential energy as the first situation
4 Em is correct
Good job!
A block slides on a rough 45 degree incline. The coefficient of friction is µk what is the ratio of acceleration when the block accelerates down the incline to the acceleration when the block is projected up the incline
Answer:
[tex]\frac{a_{d}}{a_{i}} = \frac{(1 -mu)}{mu}[/tex]
= (1 - μ)/μ
Explanation:
Always draw a diagram!
Up the incline:
[tex]Fr_{max}[/tex] = maximum friction
[tex]Fr_{max}[/tex] = μk
k = R = mg.cos(45) = mg.sin(45)
Resolution of forces parallel to the slope:
F (Fp in the diagram) = force of propulsion
g = gravity
[tex]F - Fr_{max} = ma_{i}[/tex]
[tex]F -[/tex] μ.mg.cos(45) [tex]= ma_{i}[/tex]
Down the decline:
Resolution of forces:
[tex]mg.sin(45) - Fr_{max} = ma_{d}[/tex]
[tex]mg.sin(45) -[/tex] μ.mg.cos(45) [tex]= ma_{d}[/tex]
Then, find the ratio:
[tex]\frac{ma_{d}}{ma_{i}} = \frac{mg.sin(45) - mu.mg.cos(45)}{-F + mu.mg.cos(45)} \\\\ \frac{a_{d}}{a_{i}} = \frac{k - k.mu}{-F + k.mu} \\\\ = \frac{k(1 -mu)}{-F + k.mu}[/tex]
Potentially, there is no need to consider F in this situation, in which case:
[tex]\frac{a_{d}}{a_{i}} = \frac{k(1 -mu)}{k.mu} \\\\ = \frac{(1 -mu)}{mu}[/tex]
= (1 - μ)/μ
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
How large is the acceleration of a 25 kg mass with a net force of 75 N applied horizontally to it?
Answer:
Explanation:
F = ma
a = F/m
a = 75/25
a = 3 m/s²
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
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
A student connects a 21.0 V battery to a capacitor of unknown capacitance. The result is that 52.8 µC of charge is stored on the capacitor. How much energy (in J) is stored in the capacitor?
Answer:
1.108 × [tex]10^{-3}[/tex]J
Explanation:
v=21.0v
Q=52.8× [tex]10^{-6}[/tex]
E=?
V=E/Q
E=v ×Q
=21 ×52.8 ×[tex]10^{-6}[/tex]
=1108.8 ×[tex]10^{-6}[/tex]
E= 1.108 × [tex]10^{-3}[/tex]J
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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(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.
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
A teacher took two latex balloons and blew them up with helium gas to the same size. She took one and labeled it Balloon A and placed it in a -15o C freezer. The second one she labeled BALLOON B, and she took it outside and tied it to the railing in the sun on a 30o C day. After a half hour, she had the students measure the circumference of each balloon. Which TWO outcomes do you predict the students will find and why?
Answer:
n
Explanation:
TWO outcomes can be predicted the students will find:
The size of balloon A becomes smaller.The size of balloon B becomes larger.What is the relation between temperature and volume of the gases?When a constant mass of gas is cooled, its volume falls, and when the temperature is raised, its volume grows. The volume of the gas rises by 1/273 of its initial volume at 0 °C for every degree of temperature rise.
In layman's words, the volume of a fixed mass of gas is exactly proportional to temperature at constant pressure.
The teacher took two latex balloons and blew them up with helium gas to the same size. As she placed Balloon A in a -15° C freezer, its temperature decreases and that's why, the size of balloon A becomes smaller. Again she placed Balloon B in the sun on 30° C day, its temperature increases and that's why, the size of balloon B becomes larger.
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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.
5. Layer of Earth consisting of crust & upper layer of mantle ________
Answer:
lithosphere
Explanation:
hope this helps you!!
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.
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
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.
explain the different conditions that can result in hot and cold lahars, and explain how lahars change the earth's surface?
As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.
a. Use this information to determine σ and ω0.
σ = _______s−1
ωo = ______rad/s
b. Determine the magnitude of the angular acceleration at t = 3.00 s.
______rad/s2
c. Determine the number of revolutions the wheel makes in the first 2.50 s
_______rev
d. Determine the number of revolutions it makes before coming to rest.
_______rev
Hi there!
a.
We can use the initial conditions to solve for w₀.
It is given that:
[tex]\frac{d\theta}{dt} = w_0e^{-\sigma t}[/tex]
We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:
[tex]\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}[/tex]
Now, we can solve for sigma using the other given condition:
[tex]2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}[/tex]
b.
The angular acceleration is the DERIVATIVE of the angular velocity function, so:
[tex]\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}[/tex]
c.
The angular displacement is the INTEGRAL of the angular velocity function.
[tex]\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.[/tex]
[tex]\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}[/tex]
[tex]\theta = 8.471 rad[/tex]
Convert this to rev:
[tex]8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}[/tex]
d.
We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.
[tex]0 = 3.7e^{-0.0714t}\\\\t = \infty[/tex]
Evaluate the improper integral:
[tex]\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.[/tex]
[tex]\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad[/tex]
Convert to rev:
[tex]51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}[/tex]
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.
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%
An astronaut uses a pendulum with a mass of 0.200 kg to measure the acceleration due to gravity on Planet X. He lifts the pendulum's mass a vertical height of 0.500 m and is able to determine that it gains 15.0 J of gravitational potential energy as it is lifted. Using this information, calculate the acceleration due to gravity (g) on Planet X
Answer:
Explanation:
PE = mgh
g = PE/mh
g = 15.0 / (0.200(0.500))
g = 150 m/s²
This is one strong astronaut if he can work in an environment where gravity is more than 15 times stronger than on earth.
Can someone help me solve this problems please? It's a physics problem.
Answer:
i cant see
Explanation:
but im smart
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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a car moves at a speed of 30m/s to the west of 3hr, what is its displacement of the car in km?
Answer:
Explanation:
30 m/s • 3 hr •3600 s/hr / 1000 m/km = 324 km west