Total heat removed = 473.04 kCal
Explanation:Heat removed to convert the 116°C to 100°C steam
[tex]\begin{gathered} H=mc(\theta_2-\theta_1) \\ \\ H=650(1.996)(116-100) \\ \\ H=20758.4J \end{gathered}[/tex]Heat removed from 100°C of steam to 100°C of water (Latent heat of condensation)
[tex]\begin{gathered} H_c=650\times2257 \\ \\ H_c=1467050J \end{gathered}[/tex]Heat removed from 100°C water to 0°C water
[tex]\begin{gathered} H_w=650\times4.2\times100 \\ \\ H_w=273000J \end{gathered}[/tex]Heat removed from 0°C water to 0°C ice
[tex]\begin{gathered} H_i=mL_f \\ \\ H_i=650(336) \\ \\ H_i=218400J \end{gathered}[/tex]Total heat removed = 20758.4J + 1467050 + 273000 + 218400
Total heat removed = 1979208.4 J
Convert to kilocalorie
Total heat removed = 1979208.4/4184
Total heat removed = 473.04 kCal
Writing Simple ExpressionsChoose all of the TRUE statement(s).Add 6 and 5, then multiply by 4 is the same as 4(6 + 5).4 times greater than 80 + 25 is the same as 4 x (80 + 25).Subtract 15 from 42 is the same as 15 − 42.9 times greater than 11 + 12 is the same as 9 + 11 + 12.8 times greater than 21 + 15 is the same as 8(21 + 15).
Add 6 and 5, then multiply by 4 is the same as 4(6 + 5). TRUE
4 times greater than 80 + 25 is the same as 4 x (80 + 25). TRUE
Subtract 15 from 42 is the same as 15 − 42. TRUE
9 times greater than 11 + 12 is the same as 9 + 11 + 12. FALSE
9 (11+12)
8 times greater than 21 + 15 is the same as 8(21 + 15). TRUE
A 50 kW tractor moves at a speed of 2.5 m/s. What is the traction force of the tractor?
In order to calculate the force, we can use the formula below:
[tex]P=F\cdot v[/tex]Where P is the power (in W), F is the force (in N) and v is the velocity (in m/s).
So we have:
[tex]\begin{gathered} 50000=F\cdot2.5\\ \\ F=\frac{50000}{2.5}\\ \\ F=20000\text{ N} \end{gathered}[/tex]Therefore the traction force is 20 kN.
In a system of 2 large round objects, R1 and R2 (R1 is larger), what properties will affect the force of gravity between them? (select all that apply)
According to Newton's Law of Universal Gravitation, if two particles with masses M and m are located a distance r from each other, an attractive force between them appears, and its magnitude is given by:
[tex]F=G\frac{Mm}{r^2}[/tex]Where G is the gravitational constant:
[tex]G=6.67\times10^{-11}N\frac{m^2}{\operatorname{kg}^2}[/tex]From the equation, we can see that the properties that affect the force between them are their masses and the distance between them.
Then, the correct choices are:
- Mass of R1
- Mass of R2
- Distance from the center of R1 to the center of R2.
A flywheel with a moment of inertia of 3.45 kg·m2is initially rotating. In order to stopits rotation, a braking torque of -9.40 N·m is applied to the flywheel. Calculate the initialangular speed of the flywheel if it makes 1 complete revolution from the time the brake isapplied until it comes to rest
Given data
*The given moment of inertia is I = 3.45 kg.m^2
*The given braking torque is T = -9.40 N.m
*The angular distance traveled is
[tex]\theta=(1\times2\pi)rad_{}[/tex]*The final angular speed is
[tex]\omega=0\text{ rad/s}[/tex]The angular acceleration of the flywheel is calculated by using the torque and moment of inertia relation as
[tex]\begin{gathered} T=I\alpha \\ \alpha=\frac{T}{I} \\ =\frac{-9.4}{3.45} \\ =-2.72rad/s^2 \end{gathered}[/tex]The formula for the initial angular speed of the flywheel is given by the rotational equation of motion as
[tex]\omega^2-\omega^2_0=2a\theta[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} (0)^2-\omega^2_0=2\times(-2.72)(2\pi) \\ \omega_0=\sqrt[]{2\times2.72\times2\pi} \\ =5.88\text{ rad/s} \end{gathered}[/tex]Hence, the initial angular speed of the flywheel is 5.88 rad/s
If a space ship traveling at a 1000 miles per hour enters and area free of Gravitational forces,it’s engine must run at some minimum level in order to maintain the ships velocity. is this statement true or false
The given statement is false.
If a spaceship traveling at 1,000 miles per hour enters an area free of gravitational forces, its engine must run at some maximum level in order to maintain the ship’s velocity
A wooden sphere of mass 4.0 kg is completely immersed in water. A pushing force of 20. N is
applied.
21°
20 N
4.2 ms²
19⁰
At the moment shown in the diagram, the sphere is stationary and it experiences an
acceleration upwards and to the right as shown.
Calculate the size of the upwards force due to the water (upthrust) acting on the sphere.
The size of the upwards force due to the water (upthrust) acting on the sphere is 12.64 N.
What is upthrust?
Buoyancy or upthrust, is an upward force exerted by a fluid on an object immersed in the fluid due to the weight of the object
Thus, upthrust is the upward force acting on an object immersed in a liquid.
Fu - Df = F(net_u)
where;
Fu is the upward forceDf is the downward force applied on the objectF(net_u) is the net upward forceFu - F x sin(21) = ma x sin(19)
where;
m is the mass of the wooden spherea is the upward acceleration of the wooden sphereFu - 20 x sin(21) = (4 x 4.2) x sin(19)
Fu - 7.17 = 5.47
Fu = 12.64 N
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Two Styrofoam blocks are brought near each other and are observed to repel each other. Each block has the same amount of charge on it. Which statement is true about this situation? A)the charge on each block must be positive B)the charge on each block must be negative C)the blocks are creating magnetic fields from non moving electric charges D)each block has the same kind of charge but we don't' know what kind it is
ANSWER:
D)each block has the same kind of charge but we don't' know what kind it is
STEP-BY-STEP EXPLANATION:
Two charges of the same type always repel each other.
In this case, it is not possible to know what type of load they are, only that they are the same.
Therefore, the correct answer is:
D)each block has the same kind of charge but we don't' know what kind it is
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You and some friends are duck pin bowling. You’re up and you roll your bowling ball (m = 1.6kg) down the lane. It collided with one pin (0.68kg) on the end in a perfectly elastic one dimensional collision. If the ball was moving at a velocity of 6 m/s just before it hits the pin, what is the velocity of the bowling pin after the collision?
Given:
The mass of the bowling ball is m1 = 1.6 kg
The mass of the pin is m2 = 0.68 kg
The initial velocity of the ball is
[tex]v_i=\text{ 6 m/s}[/tex]Required: Velocity of the bowling pin after the collision.
Explanation:
According to the conservation of momentum, the velocity after the collision will be
[tex]\begin{gathered} m1v_i+m2\times0\text{ =\lparen m1+m2\rparen v}_f \\ v_f=\frac{m1v_i}{m1+m2} \\ =\frac{1.6\times6}{1.6+0.68} \\ =4.21\text{ m/s} \end{gathered}[/tex]Final Answer: The velocity of the bowling pin after the collision is 4.21 m/s
Two plastic blocks of the same plastic material have the following characteristics:
Block A: Mass 10 g, Volume 25 mL, and
Block B: Mass 20 g, Volume 50 mL.
Which of the statements below is true? (Density of water is 1g/mL at room temperature.)
Group of answer choices
Both blocks will sink equally under water.
Block A will sink deeper under water.
Both blocks will float equally over water.
Block B will float more than Block A over water.
The true statement, given the data from the question is: Both blocks will float equally over water.
How to determine the true statementTo know which statement is true, we shall obtain the density of each blocks. Details below
For block A:
Mass of block A = 10 gramsVolume of block A = 25 mL Density of block A = ?Density = mass / volume
Density of block A = 10 / 25
Density of block A = 0.4 g/mL
For block B:
Mass of block B = 20 gramsVolume of block B = 50 mL Density of block B = ?Density = mass / volume
Density of block B = 20 / 50
Density of block B = 0.4 g/mL
From the above calculation, we can see that the two blocks have the same density.
Thus, we can conclude that the true statement is both blocks will float equally over water.
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A rock sample has a mass of 6 kg and a volume of 0.002 m3. Calculate the density of this rock sample.
The density of the rock sample is 3000 kg/m³.
Density is an important property of matter because it can be used to determine the weight of an object. For example, if you know the density of a rock sample and its volume, you can calculate its weight by multiplying the density by the volume. Density can also be used to determine the composition of a material. For example, if you know the densities of different materials, you can identify the material of a rock sample by measuring its density.
Density is a measure of how much mass is contained in a given volume.. The calculation of the density of the rock sample:
Density = Mass / Volume
Density = 6 kg / 0.002 m³
Density = 3000 kg/m³
As a result, the rock sample has a density of 3000 kg/m³.
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Question 10 of 10If one of two interacting charges is doubled, the force between the chargeswillO A. decrease by 4 timesO B. increaseC. decreaseD. stay the sameSUBMIT
The force between the two charges is
[tex]Force\text{ }\propto\text{ charge}[/tex]Thus, if the charge is doubled, then its force will also be doubled.
Hence the correct option is increase.
Which label goes through the horizontal axis?
A. Distance
B. Acceleration
C. Force
D. Mass
Answer:
A: Distance
Explanation:
You can not change gravity with acceleration or force in a weird way. Only mass and distance really affect gravity in normal physics.
Gravity is stronger when two objects are closer. Therefore gravity would decrease as distance increased.
Mass increases gravity.
The image below shows that gravity (depth) increases as the distance from the black hole decreases.
Will give brainliest!
A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stop-light to 28 m/s in 6.0 s, what angle theta does the string make with the vertical?
For the pair of fuzzy dice hanging by a string from the rearview mirror, while accelerating from a stop-light to 28 m/s in 6.0 s, the string makes an angle of 25.5 degrees with the vertical.
What is an angle?An angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.
Parameters given include:
velocity: 28,/s time= 6.0 seconds
a = v/t = 28m/s / 6.0s = 4.667 m/s²
Taking into account the forces acting on the dice...there is a force of gravity acting straight down with a magnitude of mg.
so we have that
Angle = tan-1(a/g)
Angle = tan-1(4.667 / 9.8) = 25.5 degrees.
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A Tour de France cyclist at a rate of 6.5 m/s and is 333 m
ahead of a crew van with powerdrink refills. The van is
traveling at 15 m/s and is accelerating at a constant rate of
0.4 m/s2
How much time will it take for the crew van to catch up
with the cyclist?
Given data:
* The speed of the cyclist is 6.5 m/s.
* The initial distance of the cyclist from the van is 333 m.
* The initial velocity of the van is 15 m/s.
* The acceleration of the van is,
[tex]a=0.4ms^{-2}[/tex]Solution:
Let x be the distance from the van initial position at which the cyclist and van meet.
Let the cyclist meet the van at time t.
By the kinematics equation, the position of the cyclist at time t is,
[tex]x-333=u_ct+\frac{1}{2}a_ct^2[/tex]where u_c is the speed of the cyclist, a_c is the acceleration of the cyclist, t is the time taken and 333-x is the distance traveled by the cyclist at time t,
The acceleration of the cyclist is zero.
Substituting the known values,
[tex]\begin{gathered} x-333=6.5t+0 \\ x-333=6.5t \end{gathered}[/tex]By the kinematics equation, the position of the van after time t is,
[tex]x=u_vt+\frac{1}{2}a_vt^2[/tex]where u_v is the velocity of the van, a_v is the acceleration of the van, and t is the time taken,
Substituting the known values,
[tex]\begin{gathered} x=15t+\frac{1}{2}\times0.4\times t^2 \\ x=15t+0.2t^2 \end{gathered}[/tex]Substituting this value of x in the kinematics equation of the cyclist,
[tex]\begin{gathered} (15t+0.2t^2)-333=6.5t \\ 15t+0.2t^2-6.5t-333=0 \\ 0.2t^2+8.5t-333=0 \end{gathered}[/tex]By solving the quadratic equation,
[tex]\begin{gathered} t=\frac{-8.5\pm\sqrt[]{8.5^2-(4\times0.2\times(-333))}}{2\times0.2} \\ t=\frac{-8.5\pm\sqrt[]{^{}8.5^2+(4\times0.2\times333)}}{2\times0.2} \\ t=\frac{-8.5\pm18.4}{0.4} \\ t=24.8\text{ s or-67.25 s} \end{gathered}[/tex]As the value of time cannot be negative.
Thus, the time at which the cyclist and van meet is 24.8 seconds.
What is the gravitational force between two trucks, each with a mass of 2.0 x 10^4 kg, that are 2.0m apart? (G=6.673 x 10^-11 N•m^2/kg^2)
Firs we will use the next formula
[tex]F=G\text{ }\frac{m_1\cdot m^{}_2}{r^2}[/tex]Where G is the gravitational force, m nd m2 are the masses and r is the distance between the masses
In our case
m1=m2= 2.0 x 10^4 kg
r= 2m
G=6.673 x 10^-11 N•m^2/kg^2
We substitute
[tex]F=6.673\times10^{-11}\cdot\frac{2.0x10^4\cdot2.0x10^4}{2^2}=0.0066743N=6.7\times10^{-3}[/tex]ANSWER
The gravitational force is 0.00667=6.7x10^-3N
cassy can get more force on the bricks she breaks with a blow of her bare hand when _______.
Answer:Her hand is made to bounce from the bricks
Explanation:
Cassy can get more force on the bricks that she breaks with a blow of her bare hand when her hand is made to bounce from the bricks,
What is Force?A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's speed can vary, or accelerate, as a result of a force. Intuitively, a pull or a push can also be used to describe force. Being a vector quantity, a force also has magnitude and direction. The SI unit of newton is used to measure it (N). The letter F stands for force.
According to Newton's second law's original formulation, an object's net force is equal to the speed at which momentum is changing over time.
Objects' velocities can be altered by the concepts of push, drag, and torque. Thrust causes an object's velocities to increase, while torque causes an object's velocities to decrease. Each part of an extended body typically exerts pressures on its neighboring sections, and the internal mechanical force is determined by how these stresses are distributed.
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Newton’s Second Law states “The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.” Explain how your observations in both a and b support this Law.
Newton’s Second Law states “The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.”
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
Force = mass × acceleration
Assuming the force constant the acceleration is inversely proportional to the mass of the object.
Thus, acceleration is directly proportional to force and inversely proportional to the mass of the body.
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Two identical point charges exert a repulsive force of 0.500 N on one another when separated by 1.5 m. What is the magnitude of the net charge of either point charge?
Given,
The repulsive force exerted by the charges, F=0.500 N
The distance between the charges, d=1.5 m
From Coulomb's law,
[tex]F=\frac{\text{kqq}}{r^2}[/tex]Where q is the magnitude of the charge of each point charge and k is the coulomb's constant.
On rearranging the above equation,
[tex]\begin{gathered} F=\frac{kq^2}{r^2} \\ \Rightarrow q=\sqrt[]{\frac{F}{k}}r \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} q=\sqrt[]{\frac{0.5}{9\times10^9}}\times1.5 \\ =1.1\times10^{-5}\text{ C} \end{gathered}[/tex]Thus the magnitude of the charge of each point charge is 1.1×10⁻⁵ C
Therefore the correct answer is option B.
On a standard day the speed of sound is 345 meters per second. A whistle whose frequency is 725 Hz is movingtoward an observer at a speed of 25.2 meters per second. What is the wavelength of the sound at the observer?(a) 0.367 m(b) 0.441 m(c) 0.511 m(d) 0.623 m
Take into account that this is a situation where the source moves presenting the Doppler effect.
In order to determine the wavelength of the sound generatd by the whistle at the observer, first calculate the frequency at the observer by using the following formula:
[tex]f=\frac{v}{v-v_s}f_s[/tex]where:
f: frequency at the observer = ?
fo: source frequency = 725 Hz
vs: source speed = 25.2 m/s
v: speed of sound = 345 m/s
replace the previous values of the parameters into the fomrula for f:
[tex]\begin{gathered} f=\frac{345m/s}{345m/s-25.2m/s}725Hz \\ f=782.1Hz \end{gathered}[/tex]Next, use the following formula to determine the wavelength of the sound at observer, by using the previous result:
[tex]\lambda=\frac{v}{f_s}=\frac{345m/s}{782.1Hz}=0.441m[/tex]Hence, the wavelength of the sound at the observer is 0.441 m
A bus is travelling along straight road at 100km/hr and the bus conductor walks a 6km/hr on the floor of the bus and in the same direction as the bus. Find the speed of the conductor relative to the road and relative to the bus. If the bus conductor now walks at the same rate but in opposite direction as the bus, find his new speed relative to the road.
Given data:
* The speed of the bus is 100 km/hr.
* The speed of the conductor is 6 km/hr.
Solution:
(a). If the bus and conductor are traveling in the same direction.
The net speed of the conductor relative to the road is,
[tex]v_1=v_b+v_c_{}[/tex]where v_1 is the velocity of the conductor relative to the road, v_b is the velocity of the bus, and v_c is the velocity of the conductor inside the bus,
Substituting the known values,
[tex]\begin{gathered} v_1=100+6 \\ v_1=106\text{ km/hr} \end{gathered}[/tex]Thus, the speed of the conductor relative to the road is 106 km/hr.
(b). The speed of the conductor relative to the bus is the speed of the conductor on the bus,
Thus, the speed of the conductor relative to the bus is 6 km/hr.
I dont understand this formula I need helpF = 6.67408 * 10^-11 * (1.5 * 10^5) (8.5 * 10^2) ------------------------------ 2500^2
Answer:
F = 1.36*10^-9
Explanation:
The given equation is
[tex]F=6.67\ast10^{-11}\ast\frac{(1.5\ast10^5)(8.5\ast10^2)}{2500^2}[/tex]To calculate the value of F, you need to multiply 1.5*10^5 by 8.5*10^2 as follows:
[tex]F=6.67\ast10^{-11}\ast\frac{1.275\times10^8}{2500^2}[/tex]Then, 2500² = 2500 x 2500 = 6.25 * 10^6, so replacing the value, we get:
[tex]F=6.67\ast10^{-11}\ast\frac{1.275\ast10^8}{6.25\ast10^6}[/tex]Now, we need to divide 1.275*10^7 by 6.25*10^6, so
[tex]F=6.67\ast10^{-11}\ast(20.4)[/tex]Finally, multiply 6.67*10^-11 by 20.4, so
[tex]F=1.36\ast10^{-9}[/tex]The source of the Sun’s heat and light energy is:A. combustion of helium gas.B. fusion of hydrogen nuclei.C. gravitational pressure.D. burning of fossil fuels.
To find
The source of the Sun’s heat and light energy is:
Explanation
The sun's core is very hot. So under pressure nuclear fusion takes place. Here hydrogen is changed to helium.
Conclusion
The correct option is
B. fusion of hydrogen nuclei.
When a rubber band with a force constant of 58.7 N/m is stretched a certain distance,
there is 1.94 J of elastic potential NRG stored in it. How far has the band been stretched
According to the given statement the band has been stretched far is 0.25 m.
What is the meaning of elastic potential?The energy that is stored when a force is used to bend an elastic object is known as elastic potential energy. Until the force is released as well as the object springs returns to its original form, doing work in the process, the energy is retained. The object may be squeezed, stretched, or bent during the deformation.
What's an example of elastic potential energy?For example, if you pull a spring, it will return to its initial form when you release it (energy input equals energy output.) Elastic potential energy is rendered possible by this.
Briefing:U= 1.94 J
k = 58.7 N/m
U = 1/2 kΔ x²
1.94 = 1/2 kΔ x²
3.88/58.7 = Δ x²
x = √(3.88 /58.7)
x = 0.25 m
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What is the momentum of a 934 kg car moving 10 m/s?
ANSWER:
9340 kg*m/s
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 934 kg
Speed (v) = 10 m/s
The formula to calculate the momentum is as follows:
[tex]\begin{gathered} p=m\cdot v \\ \text{ we replacing} \\ p=934\cdot10 \\ p=9340\text{ kg*m/s} \end{gathered}[/tex]The momentum of the car is 9340 kg*m/s.
An electric oven has a resistance of 50.0 Ω and a voltage of 220 V. How much current does it draw?
Given
Resistance of the oven, R=50.0 ohm,
Voltage is V=220 V
To find
The current drawn
Explanation
Let the current be I
By Ohm's law,
[tex]V=RI[/tex]Putting the values,
[tex]\begin{gathered} 220=50I \\ \Rightarrow I=4.4A \end{gathered}[/tex]Conclusion
The current drawn is 4.4 A
A student pushes a baseball of m = 0.13 kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of d = 0.18 meters from its original equilibrium point. The spring constant of the spring is k = 970 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring.
A. What is the maximum height, h, in meters, that the ball reaches above the equilibrium point?
B. What is the ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point?
The maximum height reached by the baseball above the equilibrium point is 12.14 m.
The ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point is 458.8 m/s.
What is the maximum height attained by the ball?The maximum height reached by the ball is determined as follows:
Data given:
compression of the spring is d = 0.18 m.mass of the baseball is m = 0.13 kgthe spring constant, K = 970 N/mThe gravitational potential energy at the compressed position is zero.
Based on the law of conservation of energy, the total energy of the system is conserved.
Let the final height from the bottom be h
m * g* h = ¹/₂ * K * d²
h = (k * d²) / (2 * m * g)
h = 970 * 0.18² / (2 * 0.13 * 9.81)
h = 12.32 m
Height above the equilibrium point = 12.32 - 0.18
Height above the equilibrium point = 12.14 m
Velocity is calculated1 as follows:
Half of the maximum height relative to the equilibrium point = 12.14/2 + (0.18) = 6.25 m
¹/₂ * m * v² = ¹/₂ * K * d²
m * v² = K * d²
v = √(K * d² / m)
v = √(970 * 6.25² / 0.13)
v = 458.8 m/s
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How much work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed? The coefficient of kinetic friction between the sofa and the floor is 0.23.
Work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed. The coefficient of kinetic friction between the sofa and the floor is 0.23 is 349 Nm.
given that :
mass = 74 kg
distance d = 2.1 m
coefficient of kinetic friction , μk = 0.23
work done is given as :
w = fd
f = μk m g
f = 0.23 × 74 × 9.8
f = 166 N
therefore ,
work = fd
w = 168 × 2.1
w = 349 Nm
Work does Scott do to push a 74 kg sofa 2.1 m across the floor at a constant speed. The coefficient of kinetic friction between the sofa and the floor is 0.23 is 349 Nm.
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When you apply a force F on an object of mass of m, it would produce acceleration a. If you apply the same force on another object of mass 2m, how would be the acceleration of the second object?
ANSWER
[tex]\frac{a}{2}[/tex]EXPLANATION
When a force is applied on an object of mass m, it produces an acceleration of a.
We can represent this relationship using Newton's second law of motion:
[tex]\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m} \end{gathered}[/tex]Now, the same force is applied on an object with a mass of 2m.
Let the acceleration experienced by the object be a1. This implies that:
[tex]\begin{gathered} F=\left(2m\right)\left(a_1\right) \\ \Rightarrow a_1=\frac{F}{2m} \end{gathered}[/tex]We can write this new acceleration in terms of a as follows:
[tex]\begin{gathered} a_1=\frac{1}{2}\left(a\right) \\ a_1=\frac{a}{2} \end{gathered}[/tex]That would be the acceleration of the second object.
The answer is the third option.
A sled of mass 26 kg has an 18 kg child on it. If big brother is pulling with a 30 N force to the right and 10 N up, and big sister is pushing with a 40 N force to the right and 16 N down, what is the normal force?
Given data:
* The mass of the sled is m_1 = 26 kg.
* The mass of the child is m_2 = 18 kg.
* The force in the upwards direction by the big brother is F_1 = 10 N.
* The force in the downwards direction by the big sister is F_2 = 16 N.
Solution:
The net mass on the sled along with the child is,
[tex]\begin{gathered} m=m_1+m_2 \\ m=26+18 \\ m=44\text{ kg} \end{gathered}[/tex]The net weight of the sled along with the child is,
[tex]\begin{gathered} w=mg \\ w=44\times9.8 \\ w=431.2\text{ N} \end{gathered}[/tex]The weight of the sled along the child is acting on the sled in the downwards direction.
Thus, the normal force acting on the sled (taking upward force as negative and downward force as positive) is,
[tex]\begin{gathered} N=w+F_2-F_1 \\ N=431.2+16-10 \\ N=437.2\text{ newton} \end{gathered}[/tex]Thus, the normal force acting on the sled is 437.2 N.