Answer: its B I jus took it
Explanation:
3.5 kg object experience an acceleration of o.5 m/s^. What net force does the object experience?
A. 1.5 N
B. 1.75 N
C. 3.5 N
D. 7 N
Answer:A heavy box has a mass of 30 kg. The box experiences a net force of 90 N when pushed along the floor. Use
this information to answer questions 17 and 18
Explanation:
17. What is the acceleration of the box?
a. 3 m/s 2 b. 2 m/s 2 c. 0.3 m/s 2 d. 0.5 m/s 2
18. If the mass of the box was changed to one third as much, and the force was held constant, how would
the acceleration change?
a. It would be one third as much
b. It would be 3 times as much
c. It would be one half as much
d. It would be 2 times as much
Problem 5. A bullet leaves a rifle with a velocity of 452 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.93 m. Determine the acceleration of the bullet.
Answer:
1.1 x 10⁵m/s²
Explanation:
Given parameters:
Velocity = 452m/s
distance = 0.93m
Unknown:
Acceleration of the bullet = ?
Solution:
To solve this problem, we use one of the kinematics equation which is given below:
V² = U² + 2aS
V is the final velocity
U is the initial velocity = 0m/s
a is the unknown acceleration
S is the distance traveled
So;
452² = 0² + (2 x a x 0.93)
204304 = 1.86a
a = 1.1 x 10⁵m/s²
Nistu is cooking rice in a pot. When she leaves the pot on the stove for a while, the temperature of the rice increases. What happens to the molecules of the rice when the temperature of the rice increases?
Answer:
Nistu is cooking rice in a pot. When she leaves the pot on the stove for a while, the temperature of the rice increases. What happens to the molecules of the rice when the temperature of the rice increases? The energy of the molecules in the rice increases.
Explanation:
a sprinter starts from rest and accelerates at a rate of 0.16 ms ^-2 over a distance of 50.0 meters . How fast is the athlete traveling at the end of the 50.0 meters?
Answer:
Final velocity v = 4 m/s
Explanation:
Given:
Initial velocity u = 0 m/s
Acceleration a = 0.16 m/s²
Distance s = 50 m
Find:
Final velocity v
Computation:
Using; v² = u² + 2as
v² = (0)² + 2(0.16)(50)
v² = 16 m/s
v = 4 m/s
Final velocity v = 4 m/s
I’ll give brainliest and 10 points
Answer:
what's these its D okkkk
What is the downward pull on an object due to gravity?
mass
weight
acceleration
velocity
Answer:
Weight
Explanation:
WEIGHT - downward pull to the center of earth due to gravity
ACCELERATION - the rate of change of velocity
VELOCITY - rate of change if displacement in time
The major difference between the mass and weight is that mass of an object can never be zero but weight can be zero. The downward pull on an object due to gravity is called the weight. The correct option is B.
What is weight?The measure of the force of gravity acting on a body is defined as the weight. Actually weight is a force and its SI unit is Newton (N). The weight of an object depends on the mass and acceleration due to gravity.
The formula used to calculate the weight is:
w = mg
m = Mass of an object
g = Acceleration due to gravity
Weight is found to be a vector quantity. Because it has both magnitude and direction. It is directed towards the center of the earth or other gravity well.
When objects fall to the ground, it is the gravity which accelerates them. So the force of gravity which pulls the object downward is called the weight.
Thus the correct option is B.
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FILL IN THE BLANKS here on earth, the pull of gravity on a mass of 1 kg is ......... newtons
Answer:
9.8 Newton
Explanation:
At average gravity on earth (conventionally, g=9.80665m/s2),
a kilogram mass exerts a force of about 9.8 newton
I hope this answer helps you
A crate is dragged up a ramp at constant speed. The work done on the system can be accounted for by:
- Eel
-Eg
-Ek
-Eint
-both Eg and Eint
Answer:
The work done on the system can be accounted for by;
Both [tex]E_g[/tex] and [tex]E_{int}[/tex]
Explanation:
The speed of the crate = Constant
Therefore, the acceleration of the crate = 0 m/s²
The net force applied to the crate, [tex]F_{NET}[/tex] = 0
Therefore, the force of with which the crate is pulled = The force resisting the upward motion of the crate
However, we have;
The force resisting the upward motion of the crate = The weight of the crate + The frictional resistance of the ramp due to the surface contact between the ramp and the crate
The work done on the system = The energy to balance the resisting force = The weight of the crate × The height the crate is raised + The heat generated as internal energy to the system
The weight of the crate × The height the crate is raised = Gravitational Potential Energy = [tex]E_g[/tex]
The heat generated as internal energy to the system = [tex]E_{int}[/tex]
Therefore;
The work done on the system = [tex]E_g[/tex] + [tex]E_{int}[/tex].
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Explanation:
cookies
HOPE IT HELPS U....
Working to ensure that women get paid the same as a man for the same work is an example of? Social dysfunction, soical conflict, social justice, or social engineering?
Answer:
social justice is the correct answer
Explanation:
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A Student missed the bus and had to walk 2 miles West to school. They got detention and missed the bus back, so they had to walk 2 miles East to get home. What is the total Distance the student traveled?
If you add 2 miles from west then 2 miles east then it would 4 miles all together.
Question 5 of 10
When dim blue light shines on a piece of metal, electrons are ejected from its
surface, but not when bright red light shines on the metal. Does this
phenomenon provide evidence for the wave model or the particle model of
light? Why?
A. The particle model, because only high-energy frequencies of light
can remove electrons
O B. The wave model, because increasing the brightness causes red
light to remove electrons
C. The wave model, because only high-energy frequencies of light
can bend around a barrier
D. The particle model, because increasing the brightness causes red
light to remove electrons
Answer:
A. The particle model, because only high-energy frequencies of light can remove electrons .
Explanation:
Each photon of blue light has higher energy than each photon of red light has . So when each photon strikes each electron , it gets ejected . But the photon of red light has not sufficient energy to eject electron . Once the photon of red light strikes the electron , the energy is wasted off . Energy of photon can not be accumulated . Thus photon behaves like particle .
Food labels often indicate energy in joules and calories. A jar of jam contains 1030 kJ (245 kcal).
Check if that is correct.
Answer:
It is not correct.
The correct conversion is:
1030 KJ = 246.18 KCal.
Explanation:
To know if 1030 KJ is equivalent to 245 KCal, we shall convert 1030 kJ to kcal. This can be obtained as follow:
4.184 KJ = 1 KCal
Therefore,
1030 KJ = 1030 KJ × 1 KCal / 4.184 KJ
1030 KJ = 246.18 KCal.
Thus, 1030 KJ is equivalent to 246.18 KCal.
From the above calculation, we can see that:
1030 KJ ≠ 245 KCal
Rather,
1030 KJ = 246.18 KCal.
1. A heat engine has a cold reservoir that is 285 degrees Kelvin which accepts 370 J of heat, and a hot reservoir that is 371 degrees Kelvin which gives off 425 J of heat. Calculate the maximum and actual efficiencies of this engine.
2. A system that has a total change in entropy of 675 J/K and has received 243 J of heat energy will be at what temperature?
3. Assume that the above system maintains a constant pressure, and increases in volume by 2.5 m 3 while doing this work. What is the pressure of the system?
I can't help with this one.
The maximum efficiency of the heat engine is 23%.
The actual efficiency of the heat engine is 13%.
The temperature of the system is 273.36 ⁰C.
The pressure of the system is 607.5 Pa.
The given parameters:
Cold reservoir temperature, Tc = 285 KHeat, qc = 370 JHot reservoir temperature, Th = 371 KHeat of the hot reservoir, qh = 425 JThe maximum efficiency of the heat engine is calculated as follows;
[tex]\eta = 1- \frac{T_c}{T_h} \\\\ \eta = 1 - \frac{285}{371} \\\\ \eta = 0.23\\\\ \eta = 23 \ \%[/tex]
The actual efficiency of the heat engine is calculated as follows;
[tex]\eta = 1 - \frac{q_c}{q_h} \\\\ \eta = 1 - \frac{370}{425} \\\\ \eta = 0.13\\\\ \eta = 13\ \%[/tex]
The temperature of the system is calculated as follows;
[tex]\Delta S = \frac{\Delta H}{T} \\\\ T = \frac{\Delta H}{\Delta S}\\\\ T = \frac{243}{675} \\\\ T = 0.36 \ K\\\\ T = 273.36 \ ^0C[/tex]
The pressure of the system is calculated as follows;
[tex]W = P \Delta V\\\\ P = \frac{W}{\Delta V} \\\\ P = \frac{243}{2.5} \\\\ P = 607.5 \ Pa[/tex]
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If the density of a diamond is 3.5 /cm^ 3 what would be the mass of a diamond whose volume is 0.5 cm ^ 3 ?
Answer:
1.75g
Explanation:
Given parameters:
Density = 3.5g/cm³
Volume = 0.5cm³
Unknown:
Mass of the diamond = ?
Solution:
Density is the mass per unit volume of a substance
Density = [tex]\frac{mass}{volume}[/tex]
So;
Mass = 3.5 x 0.5 = 1.75g
A car accelerates northwest from 35 to 62 m/s in a time of 14 seconds. What is the acceleration?
a = Vf - Vi / t
A) 27 m/s2 northwest
B) 1.9 m/s2 northwest
C) -1.9 m/s2 northwest
D) 61 m/s
Answer:
B) 1.9 m/s2 northwest
Explanation:
Given parameters:
Initial velocity = vi = 35m/s
Final velocity = vf = 62m/s
Time = 14s
Unknown:
Acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time.
Acceleration = [tex]\frac{Vf - Vi}{t}[/tex]
Vf is the final velocity
Vi is the initial velocity
t is the time taken
Acceleration = [tex]\frac{62 - 35}{14}[/tex] = 1.93m/s² northwest
What is the magnetosphere, and how does it help the earth?
can someone plss help me???
What is the magnetosphere, and how does it help the earth?
The magnetosphere shields our home planet from solar and cosmic particle radiation, as well as erosion of the atmosphere by the solar wind - the constant flow of charged particles streaming off the sun. ... It is generated by the convective motion of charged, molten iron, far below the surface in Earth's outer core.
Explanation:
What is the definition of endothermic and exothermic?
pls help on a timer will give brainlist
Answer:
An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. While an endothermic process absorbs heat and cools the surroundings
Explanation:
Answer:
Basically, endothermic feels cooler to touch than exothermic.
Explanation:
Endothermic reactions are just reactions that emit energy (in form of heat) to its surroundings. Exothermic ones, on the other hand, draw energy from its surroundings to occur.
Always think - "Endothermic is cool. Exothermic is not cool."
A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air friction is disregarded, how far advance of the starving explorer’s drop zone should she release the package?
Answer:
Approximately [tex]2.1\; \rm km[/tex], assuming that [tex]g = -9.8\; \rm m \cdot s^{-2}[/tex].
Explanation:
Let [tex]t[/tex] denote the time required for the package to reach the ground. Let [tex]h(\text{initial})[/tex] and [tex]h(\text{final})[/tex] denote the initial and final height of this package.
[tex]\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})[/tex].
For this package:
Initial height: [tex]h(\text{initial}) = 2500\; \rm m[/tex].Final height: [tex]h(\text{final}) = 0\; \rm m[/tex] (the package would be on the ground.)Solve for [tex]t[/tex], the time required for the package to reach the ground after being released.
[tex]\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}[/tex].
[tex]\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}[/tex].
Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at [tex]95\; \rm m \cdot s^{-1}[/tex].) From calculations above, the package would travel forward at that speed for about [tex]22.588\; \rm s[/tex]. That corresponds to approximately:[tex]95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km[/tex].
Hence, the package would land approximately [tex]2.1\; \rm km[/tex] in front of where the plane released the package.
The variable plotted on the horizontal or x-axis is called the
Answer:
Independent Variable
Explanation:
The variables that are plotted on X-axis, the horizontal line of the graph are termed as independent variables. The variables plotted on Y-axis which is vertical side of the graph are dependent variables.
3.5 x 10^-3 as a regular number (convert from scientific notation to a real number)
Answer: 0.0035
============================================
Explanation:
The exponent is -3, so we move the decimal point 3 spots to the left to go from 3.5 to 0.0035
If the exponent was positive, then we would move the decimal point to the right that amount of spaces.
Which of the following employees derives their income from a fee for the items sold?
a commission
b. piece rate
c.salaried
d. tip
Please select the best answer from the choices provided
Answer:
i think its b because i learned this before if im wrong sorry it could be a or d
Explanation:
A badminton player hits a 0.010 kg shuttlecock that is initially at rest in his hand. The net force on the shuitlecock over time is shown below. What is the shuttlecock's speed at t = 30ms ?
Answer: 10 m/s
Explanation: I got the answer wrong, then it showed it to me
Calculate the momentum of a boat with a mass of 95kg moving at 2m/Ss
Answer:
190 kg m/s
Explanation:
Momentum = Velocity × Mass
→ Substitute in values
2 × 95 = 190 kg m/s
The momentum of the boat of mass 95 Kg moving at 2 m/s will be 190 Kg m/s.
What is Momentum?Momentum is defined as the product of mass of velocity. Momentum is a vector quantity. Mathematically, we can write momentum as -
p = mv
It is also the measure of the impact force imparted by any moving object, when it comes in contact with some other body.
Given is a boat with a mass of 95 Kg moving at 2 m/s.
Mass [m] = 95 Kg
Velocity [v] = 2m/s
We can write the momentum as -
p = mv
Substituting values, we get -
Momentum [p] = 95 x 2 = 190 Kg m/s
p = 190 Kg m/s
Therefore, the momentum of the boat of mass 95 Kg moving at 2 m/s will be 190 Kg m/s.
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0.0135 kg of steam at 100°C is
cooled, which turns it into water at
50.0°C. How much heat has been
removed? (Remember, it has to
change phase before cooling
down.)
Boil Pt (°C)
Material Melt Pt (°C) 4 0/kg)
Water
0
3.33.105
L(/kg) c (J/(kg*C))
2.26.106 4186
100
(Unit = J)
Answer:
33333.35 kg
Explanation:
I got it right on Acellus, rounded to 33300 sigfigs
When 0.0135 kg of steam at 100°C is cooled, turns it into water at 50.0°C. Then 33,342.45 J of heat has been removed.
What is latent heat?Latent heat is the heat absorbed or released by a substance during a change in its physical state without a corresponding change in temperature. This means that the heat energy is being used to break or form intermolecular bonds between the particles in the substance, rather than raising or lowering the temperature. Examples of processes that involve latent heat include melting, freezing, vaporization, and condensation. The amount of heat required or released during these phase transitions is dependent on the substance and the amount of material undergoing the change.
Here in the question,
To solve this problem, we need to use the formula for the heat required to change the phase of a substance, which is given by:
Q = m * L
where Q is the heat required, m is the mass of the substance, and L is the specific latent heat of the substance.
In this case, we need to find the heat required to change 0.0135 kg of steam at 100°C to water at 100°C, and then cool it down to 50°C.
First, we need to find the heat required to change the phase of the steam to water at 100°C. Using the table given, we can see that the specific latent heat of vaporization for water is 2.26 × 10^6 J/kg. So the heat required to change 0.0135 kg of steam to water at 100°C is:
Q1 = m * L = 0.0135 kg * 2.26 × 10^6 J/kg = 30,510 J
Next, we need to find the heat required to cool the water from 100°C to 50°C. Using the specific heat capacity of water, which is 4186 J/(kg°C), we can calculate the heat required as:
Q2 = m * c * ΔT = 0.0135 kg * 4186 J/(kg°C) * (100°C - 50°C) = 2832.45 J
The total heat removed is the sum of the heat required to change the phase of the steam and the heat required to cool the water:
Q = Q1 + Q2 = 30,510 J + 2832.45 J = 33,342.45 J =33.34245 kJ.
Therefore, 33,342.45 J of heat has been removed.
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The bottom of a ladder must be placed 4 feet from a wall. The ladder is 12 feet long. How far above the ground does the ladder touch the wall? Estimate your answer to the nearest tenth.
Answer:
Height of wall (Perpendicular) = 11.31 ft (Approx)
Explanation:
Given:
Base Distance (Base) = 4 feet
Height of ladder (Hypotenuse) = 12 feet
Find:
Height of wall (Perpendicular)
Computation:
Perpendicular = √Hypotenuse² - Base²
Height of wall (Perpendicular) = √12² - 4²
Height of wall (Perpendicular) = √144 - 16
Height of wall (Perpendicular) = √128
Height of wall (Perpendicular) = 11.31 ft (Approx)
List these things in order of size, starting with the smallest thing:
Universe, Earth, Galaxy, Moon, Solar System, Sun
What is the net force in this picture?
hellllllllllllllllllloppppppppppppppppppp mmmmmmmmmmmmeeeeeee
Answer:
I'M SORRY I CAN'T SEE THE PICTURE BUT IT WILL MOST LIKELY BE THE DIFFERENCE BETWEEN THE NUMBERS
Explanation:
for instance if there was someone pushing a desk with a net force of 9 towards the right, the net force would come to a total of 9 since there is no one on the other side pushing the desk. Meaning the desk would go right. I'm sorry if this isn't what you needed D:
The electromagnetic spectrum has waves of varying levels of energy. Which
wave has the least energy?
Answer:
Radio waves, I believe
Explanation:
Answer:
radio yes yup yup yup yup
A horizontal net force F is exerted on an object at rest. The object starts at x= 0 m and has a speed of 4.0 m/s after moving 6.0 m along a horizontal frictionless surface. The net force F as a function of the object's position x is shown below.
what is the mass of the object?
A horizontal net force f is exerted on a 4.0 kg object at rest. The object starts at x = 0 m and travels 6.0 m along a horizontal frictionless surface. The net force F as a function of the object's position x is shown below.
What is the velocity of the object?
Answer: 6.7 m/s