Answer:
[tex]v_0[/tex]=4.761 m/s
t=0.786 sec
Explanation:
In a projectile motion (or 2D motion), the object is launched with an initial angle [tex]\theta[/tex] and an initial velocity [tex]v_0[/tex]
The components of the velocity are
[tex]v_{ox}=v_ocos\theta[/tex]
[tex]v_{oy}=v_osin\theta[/tex]
Similarly the displacement has the components
[tex]x=v_{ox}.t=v_ocos\theta.t[/tex]
[tex]y=v_osin\theta.t-\frac{gt^2}{2}[/tex]
The last formula is valid only if the object is launched at ground level, as our frog does.
There are two times where the value of y is zero, when t=0 (at launching time) and when it lands back from the air. We need to find that time t by making y=0
[tex]0=v_osin\theta.t-\frac{gt^2}{2}[/tex]
Dividing by t (assuming t different from zero)
[tex]0=v_osin\theta-\frac{gt}{2}[/tex]
Then we find the total flight as
[tex]t=\frac{2v_osin\theta}{g}[/tex]
Replacing this time in the formula of x
[tex]x=v_ocos\theta\frac{2v_osin\theta}{g}[/tex]
We can solve for [tex]v_o[/tex]
[tex]\displaystyle v_0=\sqrt{\frac{xg}{sin2\theta}}[/tex]
Knowing that x=2.20 m and [tex]\theta=54[/tex]°
[tex]\displaystyle v_0=\sqrt{\frac{xg}{sin2\theta}}=4.761m/s[/tex]
We now compute t
[tex]t=\frac{2v_osin\theta}{g}=0.786\ sec[/tex]
Final answer:
The launch speed of the frog is 4.81 m/s, and the time the frog spends in the air is 0.737 seconds.
Explanation:
A) Using the kinematic equation for projectile motion, we can find the launch speed. The launch speed, in this case, would be 4.81 m/s.
B) To determine the time the frog spends in the air, we can use the time of flight formula for projectile motion. The time in the air for the frog is 0.737 seconds.
On his way off to college, Russell drags his suitcase 15.0 m from the door of
his house to the car at a constant speed with a horizontal force of 95.0 N.
how much work does Russell do to overcome the force of friction
Answer:work done can be calculated by the following formula;
Explanation:
To solve this we must be knowing each and every concept related to work. Therefore, 1,425J of work Russell do to overcome the force of friction.
What is work?Work is defined as "Energy transfer that happens when an item is moved across a distance by such an external force, at least a portion that is exerted inside the direction of displacement" in terms of physics.
If there is a constant force operating on the block, but a difference between the force's direction as well as the displacement it causes. In this situation, the force F applies at an angle of to the displacement d.
The formula for work can be given as work
Work = force ×displacement
force=95.0 N
displacement = 15.0 m
substituting all the given values in the above equation, we get
Work =95.0 × 15.0 m
Work =1,425J
Therefore, 1,425J of work Russell do to overcome the force of friction.
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a cannon fired horizontally at 20 m/s from the top of a cliff lands 80m away. how tall is the cliff
Answer:
The height of the cliff is, h = 78.4 m
Explanation:
Given,
The horizontal velocity of the projectile, Vx = 20 m/s
The range of the projectile, s = 80 m
The projectile projected from a height is given by the formula
S = Vx [Vy + √(Vy² + 2gh)] / g
Therefore,
h = S²g/2Vx²
Substituting the values
h = 80² x 9.8/ (2 x 20²)
= 78.4 m
Hence, the height of the cliff is, h = 78.4 m
How fast does a 3 kg rocket accelerate if a force of 140 N is applied?
46.67 m/s²
Explanation:We are given;
Mass of a rocket = 3 kg Force applied = 140 Nwe are supposed to calculate the acceleration of the rocket;
We are going to use the second Newton's law of motion;According to Newton's second Law of motion force is equivalent to the product of mass and acceleration.This is from the fact that the resultant force and the rate of change in linear momentum are directly proportional.Therefore;
F = Ma
In this case;
Rearranging the formula
a = F ÷ m
= 140 N ÷ 3 kg
= 46.67 m/s²
Thus, the acceleration of the rocket is 46.67 m/s²
A snowmobile is used to pull one sled across the ice. The mass of snowmobile and the rider is 315 kg . The mass of the sled is 150 kg. The coefficient of kinetic friction for the sled on ice is 0.15 and the coefficient of kinetic friction on the snowmobile is 0.25. The acceleration of the snowmobile and the sled is 1.9m/^2.
a. What is the net force acting on the system?
b. What is the tension in the rope between the sled and the snowmobile?
c. What is the applied force on the snowmobile?
a) The net force on the system is 883.5 N
b) The tension in the rope is 505.5 N
c) The applied force on the snowmobile is 1875.7 N
Explanation:
a)
To solve this first part, we just analyze all the forces acting in the horizontal direction on the snowmobile+sled system, and we apply Newton's second law, which states that:
[tex]\sum F = (m+M)a[/tex]
where
[tex]\sum F[/tex] is the net force on the system
m = 150 kg is the mass of the sled
M = 315 kg is the mass of the snowmobile+rider
[tex]a=1.9 m/s^2[/tex] is the acceleration
Therefore, solving for [tex]\sum F[/tex], we find the net force on the system:
[tex]\sum F = (150+315)(1.9)=883.5 N[/tex]
b)
We can write now the equation of the forces acting on the sled only. We have:
[tex]T-F_f = ma[/tex]
where:
T is the tension in the rope between the sled and the snowmobile, which is pulling the sled forward
[tex]F_f[/tex] is the force of friction acting on the sled
m = 150 kg is the mass of the sled
[tex]a=1.9 m/s^2[/tex] is the acceleration
The force of friction can be written as
[tex]F_f = \mu_k mg[/tex]
where
[tex]\mu_k = 0.15[/tex] is the coefficient of kinetic friction of the sled on ice
Substituting into the previous equation and solving for T, we find the tension:
[tex]T-\mu_k mg = ma\\T=ma+\mu_k mg=(150)(1.9)+(0.15)(150)(9.8)=505.5 N[/tex]
c)
We can now write the equation of the forces acting on the snowmobile, and we have:
[tex]F_a - T - F_F = Ma[/tex]
where:
[tex]F_a[/tex] is the applied force
T = 505.5 N is the tension in the rope, which pulls the snowmobile backward
[tex]F_F[/tex] is the force of friction on the snowmobile
M = 315 kg is the mass of the snowmobile+rider
[tex]a=1.9 m/s^2[/tex] is the acceleration
The force of friction can be written as
[tex]F_F = \mu_k Mg[/tex]
where
[tex]\mu_k = 0.25[/tex] is the coefficient of kinetic friction of the snowmobile on ice
Substituting into the previous equation and solving for [tex]F_a[/tex], we find:
[tex]F_a = Ma+T+\mu_k Mg=(315)(1.9)+505.5+(0.25)(315)(9.8)=1875.7 N[/tex]
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As the swing moves from point B to point A, which form of energy is increasing, kinetic or potential?
Answer:
The kinetic energy
Explanation:
The potential energy (P) is the energy that the swing has due to its position, and it can be calculated with the expression
P = m . g . h
where,
m: mass
g: gravity
h: height
The higher is the swing, the more potential energy it has.
On the other hand, the kinetic energy (K) is the energy which a body possesses by virtue of being in motion, and it can be calculated using the expression
K = 1/2 . m . v²
where
v: speed
In B, the swing is virtually still (v = 0), while in A, v is maximum and so is its kinetic energy.
Then, as the swing moves from point B to point A, the kinetic energy increases and the potential energy decreases.
What is the momentum of a 40kg object going at 100m/s?
Answer:
The momentum of the object is, p = 400 kg m/s
Explanation:
Given data,
The mass of the object, m = 40 kg
The velocity of the object, v = 100 m/s
The momentum of the object is defined as the product of the object's mass and its velocity. it is given by the formula,
p = m v
Substituting the values,
p = 40 kg x 100 m/s
= 400 kg m/s
Hence, the momentum of the object is, p = 400 kg m/s
The teacher told Fatima that all types of energy are kinetic energy, potential energy, or both. Fatima drew this diagram to help her organize this
information. She then sorted each type of energy into its correct category: chemical, electrical, electromagnetic, mechanical, nuclear, and
thermal
categories:
(Both)
(Potential
energy)
(Kinetic
energy)
Which type of energy belongs only in the Kinetic Energy category?
A. chemical
B. electromagnetic
C. mechanical
D. thermal
The type of energy belongs only in the Kinetic Energy category is thermal energy. The correct option is D.
The only type of energy that exclusively falls within the kinetic energy category is thermal energy. It stands for the energy generated from the movement of particles inside a substance, which produces temperature and heat.
Thermal energy is stored energy that directly corresponds to the movement of particles, causing them to vibrate and collide, as opposed to potential energy, which is stored energy based on an object's position or condition.
Depending on the situation, kinetic and potential components can be present in chemical, electromagnetic, mechanical, and nuclear energy.
In conclusion, because it has to do with particle motion and heat transfer, thermal energy is the only type of energy that specifically belongs to the kinetic energy category.
Thus, the correct option is D.
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A snowmobile is used to pull one sled across the ice. The mass of the snowmobile and the rider is 315 kg. The mass of the sled is 150 kg. The coefficient of kinetic friction for the sled on ice is 0.15 and the coefficient of kinetic friction on the snowmobile is 0.25. The acceleration of the snowmobile and the sled is 1.9/s^2.
a. What is the net force acting on the system?
b. What is the tension in the rope between the sled and the snowmobile?
c. What is the applied force on the snowmobile?
a) The net force on the system is 883.5 N
b) The tension in the rope is 505.5 N
c) The applied force is 1875.8 N
Explanation:
a)
We start by considering the whole system snowmobile+sled. We can apply Newton's second law:
[tex]\sum F = (m+M)a[/tex]
where
[tex]\sum F[/tex] is the net force on the system
m = 150 kg is the mass of the sled
M = 315 kg is the mass of the snowmobile+rider
[tex]a=1.9 m/s^2[/tex] is the acceleration
Substiting the values into the equation, we find the net force:
[tex]\sum F=(150+315)(1.9)=883.5 N[/tex]
b)
Now we consider only the forces acting on the sled. Again, we apply Newton's second law:
[tex]T-F_f = ma[/tex]
where:
T is the tension in the rope, which pulls the sled forward
[tex]F_f[/tex] is the frictional force acting on the sled, which acts backward
m = 150 kg is the mass of the sled
[tex]a=1.9 m/s^2[/tex] is the acceleration
The force of friction on the sled is given by
[tex]F_f = \mu mg[/tex]
where
[tex]\mu=0.15[/tex] is the coefficient of friction of the sled on ice
m = 150 kg
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting and solving for T, we find:
[tex]T-\mu mg = ma\\T=m(a+\mu g)=(150)(1.9+0.15 \cdot 9.8)=505.5 N[/tex]
c)
Now we consider instead the forces acting on the snowmobile+rider only. Applying again Newton's second law,
[tex]F-F_F-T = Ma[/tex]
where
F is the applied force, which pushes the snowmobile forward
[tex]F_F[/tex] is the force of friction acting backward on the snowmobile
T = 505.5 N is the tension in the rope, which pulls the snowmobile backward
M = 315 kg is the mass of the snowmobile+rider
[tex]a=1.9 m/s^2[/tex] is the acceleration
The force of friction on the snowmobile is given by
[tex]F_F = \mu Mg[/tex]
where
[tex]\mu=0.25[/tex] is the coefficient of friction of the snowmobile on ice
m = 315 kg
[tex]g=9.8 m/s^2[/tex]
Substituting and solving for F, we find the applied force:
[tex]F=Ma+\mu Mg+T=(315)(1.9)+(0.25)(315)(9.8)+505.5=1875.8 N[/tex]
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Flow of electrical current in a wire is analogous to the flow of water in a pipe. True False
Answer:
false
Explanation:
Answer:
TRUE
In a wire, negatively charged electrons move, and positively charged atoms don't. Electrical engineers say that, in an electrical circuit, electricity flows one direction: out of the positive terminal of a battery and back into the negative terminal.
A lot of people think of electron flow as electrons moving along a wire freely like cars go down a highway.
Explanation:
Which of these is a qualitative research method?
A.case studies
B.experiments
C.surveys
D.laboratory observation
Answer:
A. case studies, because it's one of the 5 groups that are part of the qualitative methods
This passage describes a chemical reaction. Read the passage. Then, answer the question below.
As the cells in your body break down food, a poisonous substance called ammonia (NH3) begins to build up in your blood. To reduce the amount of ammonia in your blood, your liver combines ammonia with carbon dioxide (CO2) and other substances to make urea (CH4N2O). The urea later leaves your body as waste.
In the chemical reaction described in the passage, which of the following are reactants? Select all that apply.
Answers: carbon dioxide (CO2), urea (CH4N2O), ammonia (NH3)
In the described chemical reaction, the reactants are ammonia (NH3) and carbon dioxide (CO2). These reactants are combined in the body to produce urea (CH4N2O), which is the product.
Explanation:In the chemical reaction described in the passage, the reactants are the substances that combine to produce another substance. From the passage, we can identify that ammonia (NH3) and carbon dioxide (CO2) are combined by the liver to produce urea (CH4N2O). Hence, in this chemical reaction, ammonia and carbon dioxide are the reactants, while urea is the end product or the product of the reaction.
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In physical changes, substances might change in ________ but not in ________.
Answer:
In physical changes, substances might change in the form of the matter(the shape) but not in it's chemical composition.
Explanation:
The shape can change as well as the size and look but it's chemical composition does not change........which means a paper folded into a plane remains a paper and doesn't become anything else.
In chemical changes.......a wood burned changes to charcoal in this it's a new substance not the wood we burned.
The making of paper plane is a physical change and burning of wood is a chemical change. Chemical change gives new substance and is irreversible which means u can't get the wood back again. While physical changes don't give new substances and the plane can again be unfolded into a sheet of paper.
In physical changes, substances might change in form but not in chemical composition.
A physical change is a type of change in which the physical form of the substance is changed but not the chemical composition.
These physical changes affect the form of the chemical substance, but not the chemical composition.
Examples of physical changes;
freezing of waterboiling of watermelting of iceThus, we can conclude that in physical changes, substances might change in form but not in chemical composition.
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2. An object experiences an acceleration, g, when it is on the surface of a planet of radius R. What will be
the acceleration on the object after it has been moved to a distance of 4R from the center of the planet?
A) 16g
B) 4g
C) 1/4 g
D) 1/16g
The new acceleration of gravity is D) 1/16 g
Explanation:
The magnitude of the acceleration of gravity in the gravitational field of a planet is given by
[tex]g=\frac{GM}{r^2}[/tex]
where
G is the gravitational constant
M is the mass of the planet
r is the distance of the object from the centre of the planet
In this problem, the acceleration of gravity g on the surface of the planet (when r=R) is
[tex]g=\frac{GM}{R^2}[/tex]
Then the object is moved to a distance of
r' = 4R
Substituting into the original equation, we can find what is the new acceleration of gravity:
[tex]g'=\frac{GM}{(4R)^2}=\frac{1}{16}(\frac{GM}{R^2})=\frac{1}{16}g[/tex]
So, the acceleration of gravity has decreased by a factor 16.
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Final answer:
When an object is moved from a distance of R to 4R from the center of a planet, the gravitational acceleration it experiences decreases to 1/16th of the original value, making the correct answer D) 1/16g.
Explanation:
The question deals with the change in gravitational acceleration experienced by an object when its distance from the center of a planet is increased. According to Newton's law of universal gravitation, the gravitational force, and hence the acceleration, is inversely proportional to the square of the distance between the objects. Therefore, if the distance is increased from R to 4R, the acceleration will decrease by a factor of the square of the increase in distance. This is calculated as (1/4)^2 or 1/16th of the original acceleration.
So, the correct answer is D) 1/16g. As a comparison, g on the surface of Earth is approximately 9.8 m/s², and when discussing acceleration in multiples of g, it helps to envision the effect of this force under various conditions, such as creating an acceleration of 1 g using centripetal forces.
1kg slab of concrete loses 12,000j of heat when it cools from 30 Celsius to 26 Celsius. Determine the specific heat capacity of concrete.
The specific heat capacity of concrete is [tex]3.0 J/(g^{\circ}C)[/tex]
Explanation:
When a certain amount of energy Q is supplied/given off to/from a sample of substance with mass m, the temperature of the substance increases/decreases by an amount [tex]\Delta T[/tex], according to the equation
[tex]Q=mC_s \Delta T[/tex]
where
m is the mass of the substance
[tex]C_s[/tex] is the specific heat capacity of the substance
[tex]\Delta T[/tex] is the change in temperature of the substance
In this problem, we have:
m = 1 kg = 1000 g is the mass of the concrete slab
[tex]Q = -12,000 J[/tex] is the amount of energy lost by the slab
[tex]\Delta T = 30-26= -4^{\circ}C[/tex] is the change in temperature of the slab
Solving the equation for [tex]C_s[/tex], we find the specific heat capacity of concrete:
[tex]C_s = \frac{Q}{m \Delta T}=\frac{-12,000}{(1000)(-4)}=3.0 J/(g^{\circ}C)[/tex]
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Final answer:
The specific heat capacity of concrete is calculated using the heat transfer formula Q = mcΔT, and the given values of 1 kg mass, a temperature change of 4°C, and a heat loss of 12,000 J. The specific heat capacity is found to be 3,000 J/kg/°C.
Explanation:
To determine the specific heat capacity of concrete, we start by using the formula for heat transfer:
Q = mcΔT
where Q is the heat energy transferred (in joules, J), m is the mass (in kilograms, kg), c is the specific heat capacity (in J/kg/°C), and ΔT is the change in temperature (in °C). In this scenario, we are given that a 1kg slab of concrete loses 12,000J of heat when it cools from 30°C to 26°C.
The change in temperature (ΔT) is the final temperature minus the initial temperature, that's 26°C - 30°C = -4°C. We are solving for c, the specific heat capacity of concrete. It is possible to solve for c by rearranging the heat transfer equation:
c = Q / (mΔT)
We substitute the known values into this equation:
c = 12,000 J / (1 kg × -4°C)
c = 12,000 J / -4 kg°C
c = -3,000 J/kg/°C
The negative sign indicates that the concrete is losing heat, however, the specific heat capacity is a positive value. Therefore, the specific heat capacity of concrete is 3,000 J/kg/°C.
A lightbulb is connected to a 60 Hz power
source having a maximum voltage of 156 V.,
What is the resistance of the light bulb that
uses an average power of 38.8 W?
The resistance of the lightbulb connected to a 60 Hz power source and using an average power of 38.8 W, with a maximum voltage of 156 V, is approximately 312.4 Ω.
Explanation:To find the resistance of the lightbulb, we must first establish the effective value of the voltage, since the given voltage is the maximal one. For sinusoidal alternating currents (like the standard electrical supply), the effective or 'root mean square' voltage is the maximum voltage divided by the square root of 2 (approximately 1.414). So, the effective voltage (Vrms) is 156 V / √2 = 110.3 V.
Next, we use the formula for power P = V^2 / R, where V is the effective (rms) voltage and R is the resistance. Looking for R, we rearrange the equation to get R = V^2 / P. Substituting the given values we get R = (110.3 V)^2 / 38.8 W = 312.4 Ω. Therefore, the resistance of the light bulb is approximately 312.4 Ω.
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The resistance of a light bulb is approximately 313.57 Ω. This is calculated by converting maximum voltage to RMS voltage and applying the power formula P = V²/R.
To solve this, we need to follow these steps:
First, convert the maximum voltage (Vmax) to the root mean square (RMS) voltage (Vrms) using the formula [tex]V_{rms} = V_{max} / \sqrt2[/tex]. So, [tex]V_{rms} = 156 V / \sqrt2 = 110.3 V[/tex]
Use the power formula that relates power (P), voltage (V), and resistance (R):
[tex]P = V_{rms}^2 / R[/tex]
Rearrange it to solve for resistance (R):
[tex]R = V_{rms}^2 / P[/tex]
Substitute the values:
R = (110.3 V)² / 38.8 W
R ≈ 313.57 Ω
Thus, the resistance of the light bulb is approximately 313.57 Ω.
pls help me yall (-｡-;
[Choose all the answers that apply!]
Earth's water:
is reused and recycled
is mostly freshwater
is strongly attracted to other materials
has a neutral pH value
can dissolve more substances than any other liquid
Answer:
1.yes
2.no
3.yes
4.no
5.yes
What is the potential energy for 65kg climber on top of Mount Everest (8,800 m high)
Answer:
The potential energy is related to the mass and height of the object from the ground and to the acceleration due to gravity g, by the expression, PE = m g h. Thus, PE = 65 kg × 8800 m × 9.8 m/s² = 5.6 × 10⁶ Joules. Therefore, the potential energy of the climber will be 5.6 × 10⁶ Joules.
Explanation:
A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at
an average velocity of 625 m/s. What is the velocity of the rocket after the fuel has
burned? •
Thiss is the answer I just need to show work>>>>>>>(7.8 m/s)
The velocity of the rocket is 7.8 m/s
Explanation:
We can solve the problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the rocket+fuel system must be conserved.
Before the launch, the total momentum of the system is zero, since the rocket and the fuel are at rest:
[tex]p=0[/tex]
After the launch, the total momentum is
[tex]p=MV+mv[/tex]
where
M = 4.0 kg is the mass of the rocket
V is the velocity of the rocket
m = 50.0 g = 0.050 kg is the mass of the fuel ejected
v = -625 m/s is the velocity of the fuel (taking "backward" as negative direction)
Since the total momentum is conserved, we have
[tex]0=MV+mv[/tex]
So we can solve the equation to find V, the velocity of the rocket:
[tex]V=-\frac{mv}{M}=-\frac{(0.050)(625)}{4.0}=+7.8 m/s[/tex]
And the positive sign means the rocket moves forward.
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The velocity of the rocket after the fuel has burned is 7.81 m/s.
From law of conservation of Momentum,
[tex]MV=mv[/tex]
Where M is mass of rocket and V is velocity of rocket
And m is mass of burned fuel and v is speed of fuel.
Given that, [tex]M=4Kg,m=50g=0.05Kg,v=625m/s[/tex]
Substitute values in above equation,
[tex]4*V=0.05*625\\\\V=\frac{0.05*625}{4} =7.81m/s[/tex]
Hence, the velocity of the rocket after the fuel has burned is 7.81 m/s.
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Callie did a lab during which she investigated the difference in cellular respiration rates between two different types of corn: germinating and non-germinating. A germinating seed is one from which plant has started to grow. A non-germinating seed is usually dry, and a new plant has not yet emerged.The data that she gathered is displayed in the graph below.
Which of the following statements are true concerning the data that Callie gathered during the lab? Choose the two that apply.
A.The germinating corn seed produced more energy than the non-germinating corn seed.
B.The non-germinating corn seed produced more carbon dioxide than the germinating corn seed.
C. The germinating corn seed consumed more oxygen than the non-germinating corn seed.
D. The non-germinating corn seed performed more cellular respiration and the germinating corn seed.
E. The non-germinating corn seed performed cellular respiration and the germinating corn seed performed fermentation.
Please I NEED the TWO ANSWERS ASAP!
Answer:
I just did this question and it's A and D
The correct options are :
(A) The germinating corn seed produced more energy than the non-germinating corn seed.
(C) The germinating corn seed consumed more oxygen than the non-germinating corn seed.
information from the graph:
We can clearly see in the graph that the slope of the germinating seed is steeper than that of the non-germinating seed.Since the germinating seed has a steeper graph, it shows that its rate of consumption of oxygen is higher than that of the non-germinating seed.As the germinating seed consumes more oxygen, it implies that it produces more energy than the non-germinating seed.Learn more about the slope:
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Select the correct answer.
Which of Newton's laws explains why your hands get red when you press them hard against a wall?
A
Newton's law of gravity
B. Newton's first law of motion
OC Newton's second law of motion
OD. Newton's third law of motion
Reset
Reset
Next
Next
Answer:
D. Newton's Third Law of Motion
Explanation:
Newton's law of gravity is definitely not applicable to your hands. So we can cross this bad boy out
Newton's First Law is F=MA (force equals mass times acceleration). This is basically the root of most physics but it isn't the reason for your hand being red after hitting a wall.
Newton's Second law deals with velocities and forces, so even though you are apply a force your are not changing the velocity of the wall much.
Newton's Third Law basically says that for whatever force you apply to an object, that object will apply an equal and opposite force back to you. This is why your hand gets red. When you slap the wall with all your strength, the wall hits your hand back with the same amount of force. The 2nd law can also be seen when you're trying to push a desk and it won't budge. You are pushing on it, but the desk is pushing back. (there are multiple other factors applicable like friction but we physicists like to ignore them :) )
I hope this helps!
According to newtons third law of motion action nd reaction forces are equal in magnitudes nd opposite in direction but they do not cancel each other why ??
Because the action and the reaction forces act on different objects
Explanation:
Newton's third law of motion states that:
"When an object A exerts a force on an object B (action), then object B exerts an equal and opposite force (reaction) on object A"
From the statement above, we clearly see that the two forces mentioned in the law act on different objects. In fact, the action is exerted on object B, while the reaction is exerted on object A.
When we are considering the free-body diagram of an object, we just represent all the forces acting on that object, but not the forces exerted on other objects: this means that the action and the reaction never appear in the same free-body diagram of the same object, so they do not cancel out, simply because they are applied to different objects.
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In what ways is force described?
by the distance and speed
by the height and the width
by the magnitude and direction
A force is described by its magnitude and direction.
Force is described in physics by its magnitude, which is the size or amount of the force, and its direction, which is where the force is heading.
Explanation:In physics, force is described in terms of its magnitude and direction. Magnitude refers to the size or quantity of the force. It's measured in units like newtons. Direction, on the other hand, tells us where the force is heading or pointing to. For example, if you are pushing a box, the magnitude of the force would be how hard you are pushing and the direction would be the direction in which the box moves.
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A hill has three paths up and sides to a flat summit area that three path Lanks are different but the vertical height from the bottom of the mountain to the top is the same
A) which path would be easiest for a car to claim
B) not including the energy used to overcome the internal friction of the car which path requires the most energy for a car to climb?
I’m just confused on the concept of work. How come work is not affected when there is the same height
Answer:
1- The lank with the least slope will be the easiest to climb because it will require the least force.
2- All three paths will require the same energy as the work done will be the same in all three cases.
Explanation:
1- Work done W = Force x Displacement i.e. W = F × cosФ where 'Ф' is the angle between force and displacement. Smaller the angle least is the force required to climb. Hence the path with least slope will be easier to climb.
2- Since W = F × cosФ as the angle increase cos Ф decrease which means the slope is increase and consequently the force required also increases but the total work done remains the same. Hence the energy required to climb all three Lanks will be the same.
Examine the unbalanced equation.
KCIO3 → KCl + O2
What is the best classification for the unbalanced equation's reaction?
decomposition
displacement
synthesis
combustion
Answer:
Not combustion
Explanation:
I just took a test and got it wrong
The answer is Decomposition
Thank god the answer was removed
I got it correct
Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy store? Assume Dane is on Earth, where g = 10 N/kg.
Answer:
The right answer is 160j
Dane is holding an 8 kg box 2 meters above the ground. The box's gravitational potential energy will be 160 joules.
What is Gravitational Potential Energy?The term gravitational potential energy refers to the energy that an item stores as a result of its elevation above the Earth's surface. This energy is a result of an object being subjected to gravity.
Potential energy is transformed into kinetic energy when a stone falling from a height impacts the earth's surface.
According to the question, the given values are :
Mass, m = 8 kg
Height, h = 2 meters above the ground and,
g = 10 N/kg
Gravitational potential energy = mgh
P = 8×10×2
P = 160 J.
Hence, gravitational potential energy is 160 J.
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A brick is dropped from a big scaffold. What is its velocity after 4.0s?
The velocity of the brick is 39.2 m/s downward
Explanation:
The motion of the brick is a free fall motion, since the object is affected only by the force of gravity. Therefore, it has a uniformly accelerated motion towards the ground, with constant acceleration of [tex]g=9.8 m/s^2[/tex].
So, we can find its velocity using the suvat equation:
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time
For the brick in this problem (taking downward as positive direction)
u = 0 (it is dropped from rest)
[tex]a=g=9.8 m/s^2[/tex]
Therefore, its velocity after t = 4.0 s is:
[tex]v=0+(9.8)(4.0)=39.2 m/s[/tex]
Downward, because the sign is positive.
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A 5.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. What is the tension in the string?
The tension in the string that accelerates a 5 kg cart to 2 m/s² is equal to 10 Newton.
Explanation:
Tension:
Tension is a force exerted by the rope, Tension refers to the pulling force transmitted axially by the means of a string, cable or a chain.
In the given case, Tension in the string is simply the pulling force it applies on the cart. There is no special formula for tension in the string.
Newton's second law is often stated as F = ma, which means the force (F) acting on an object is equal to the mass (m) of an object times the acceleration (a) produced in it due to that force.
Mass of the cart = 5 kg
Acceleration produced = 2 m/s²
Tension in the string = F = ma
= 5*2
= 10 kg m/s²
= 10 N
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Lf you exert a net force of 8 N on a 2-kg object, what will its acceleration be? a. 2 m/s2 b. 10 m/s2 c. 16 m/s2 d. 4 m/s2
Answer:
D) a = 4m/s2
Explanation:
[tex]f = ma \\ a = \frac{f}{m} \\ a = \frac{8}{2} \\ a = 4[/tex]
where f = 8N and mass = 2kg
If you exert a net force of 8 N on a 2-kg object, the acceleration of the object would be [tex]4 m / s^{2}[/tex].
Answer: Option D
Explanation:
According to Newton second law, it states the relation between mass and the force required to accelerate the object. The mathematical expression can be given as follows,
[tex]\text {Force}=\text {mass} \times \text {accelaration}[/tex]
The above expression shows that exerted force is equal to the product of mass and acceleration.
Where, in the given problem given the data as,
Force, F= 8 N
Mass, m = 2 kg
So, we can write the above equation as,
[tex]\text {Acceleration, } a=\frac{F}{m}[/tex]
By substituting the given values in the above equation, we get,
[tex]\text {acceleration, } a=\frac{8}{2}=4 \mathrm{m} / \mathrm{s}^{2}[/tex]
Therefore, the acceleration of the object would be [tex]4 m / s^{2}[/tex].
A plank rests on top of the axles of two identical wheels. Each wheel's outer radius is 0.25 meters and each wheel's axle has radius 0.05 meters. If the wheels roll 1.00 meter from their original position, how much does the plank move from its original position?
Assume that there is no slipping anywhere, and that the plank does not tip.
45 POINTS TO WHOEVER CAN ANSWER, THANK YOU VERY MUCH!
The plank moves the same distance as the point on the circumference of the wheels, which is 1.00 meter, because there is no slipping and the plank doesn't tip.
Explanation:When the wheels roll 1.00 meter from their initial position without slipping, the distance moved by the plank will be equivalent to the circumference of a circle traced by the outer edge of the wheels. The outer radius of each wheel is 0.25 meters, thus the circumference C can be calculated using the formula C = 2πr, which gives us C = 2π(0.25) = 1.57 meters. Since the wheels have rolled 1.00 meter, which is less than a full revolution, we need to find what fraction of a revolution 1.00 meter corresponds to and then calculate the corresponding linear distance that the plank would move.
To find the fraction of a revolution, we divide 1.00 meter by the circumference to get approximately 0.637 revolutions. Since the plank's movement corresponds to the rotation of the wheels and because there's no slippage, the plank will move the same linear distance as the edge of the wheels over this fraction of a revolution. Therefore, the plank moves approximately 0.637 times the wheel's circumference, which is 0.637 × 1.57 meters ≈ 1.00 meter.
The plank moves the same distance as the point on the circumference of the wheels, which in this case is 1.00 meter.
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Three balls which have equal masses are fired with equal speeds. One ball is fired up at an angle of 45 degrees, the other is fired up at an angle of 60 degrees, and the third ball is fired straight up at an angle of 90 degrees from the horizontal surface. Rank in order, from largest to smallest, their speeds Va, Vb, and Vc as they cross a dashed horizontal line approximately a couple meters away from the balls. Explain. (All three are fired with sufficient speed to reach the line.)
Answer:
[tex]V_{c} = V_{b} = V_{a}[/tex]
Explanation:
Let [tex]V_{a}[/tex], [tex]V_{b}[/tex], [tex]V_{c}[/tex] be the speed of the balls which was fired at an angle of 45°, 60°, 90° respectively from the horizontal surface, when the balls crosses the dashed horizontal line a few meters away from the place where it is fired initially
After firing the only force that is acting on the ball will be the force of gravity assuming that there is no air resistance, so the acceleration will be g in the downward direction
Using the below formula we can calculate the final velocity of the ball after travelling some distance
v² - u² = 2×a×s
where v is the final velocity of the ball
u is the initial velocity of the ball
s is the distance between the points where initial velocity is u and final velocity is v
a is the acceleration of the ball
From this we get
v² = u² + 2×a×s
Here in this case we are considering in vertical direction and in this direction for all balls a = -g as g is acted on the balls opposite to the direction of motion
s is same for all balls
Let u be the initial speed of all balls
speed is same for all balls but not initial velocity in the vertical direction
For the ball that is fired from an angle of 90°, initial velocity is u
For the ball that is fired from an angle of 60°, initial velocity is u×sin60° = u×(√3 ÷ 2)
For the ball that is fired from an angle of 45°, initial velocity is u ×sin45° = u×(1 ÷ √2)
As a and s in the formula are constant for all particles, the final velocity will depend on their initial velocities
We can observe that as angle with horizontal is increasing the initial velocity is also increasing
∴ Final velocity in the vertical direction of the ball fired from 90° > Final velocity in the vertical direction of the ball fired from 60° > Final velocity in the vertical direction of the ball fired from 45°
As there is no acceleration in horizontal direction initial velocity in horizontal direction = final velocity in horizontal direction
In horizontal direction
velocity of the ball fired from 90° is 0
velocity of the ball fired from 60° is u×cos60° = u÷2
velocity of the ball fired from 45° is u ×sin45° = u×(1 ÷ √2)
Speed = √((velocity in horizontal direction)² +(velocity in vertical direction)²)
For all balls we get speed² = u² - 2×g×s
∴ Final speed of all balls is same
Final answer:
The ball fired straight up at an angle of 90 degrees will have the highest speed when it crosses the horizontal line, followed by the ball fired at an angle of 45 degrees, and then the ball fired at an angle of 60 degrees.
Explanation:
The speeds of the three balls can be determined by analyzing their motion in the vertical and horizontal directions. Since the vertical motion is independent of the horizontal motion, the ball fired straight up at an angle of 90 degrees will have the highest speed when it crosses the horizontal line. This is because it is only influenced by gravity in the vertical direction and does not have any horizontal motion to slow it down. The ball fired at an angle of 45 degrees will have a slightly lower speed, while the ball fired at an angle of 60 degrees will have the lowest speed. This is because their horizontal velocities are not as high as the ball fired straight up. Therefore, the ranking from largest to smallest speed is: Vc (fired straight up), Va (fired at 45 degrees), and Vb (fired at 60 degrees).