If we kick both sodas with the same force, the soda that has a higher weight will have a lower acceleration. This is explained by Newton's second law of motion
The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.
F = ma
where
m = mass
a = accelaration
F = Force
A baseball player pitches a fastball toward home plate at a speed of 41.0 m/s. The batter swings, connects with the ball of mass 195 g, and hits it so that the ball leaves the bat with a speed of 37.0 m/s. Assume that the ball is moving horizontally just before and just after the collision with the bat.A. What is the impulse delivered to the ball by the bat? Enter a positive value if the impulse is in the direction the bat pushes the ball and enter a negative value if the impulse is in the opposite direction the bat pushes the ball. (kg m/s)B. If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat? (kN)
Given:
Initial velocity, vi = 41.0 m/s
Mass of ball, m = 195 g = 0.195 kg
Final velocity, vf = 37.0 m/s
Assuming the ball is moving horizontally just before and after collision with the bat, let's solve for the following:
• (A). What is the impulse delivered to the ball by the bat?
To find the impulse, apply the change in momentum formula:
[tex]\Delta p=p_f-p_i[/tex]Where:
pi is the initial momentum = -mvi
pf is the final momentum = mvf
Thus, we have:
[tex]\begin{gathered} \Delta p=mv_f-(-mv_i) \\ \\ \Delta p=mv_f+mv_i \\ \\ \Delta p=m(v_f+v_i) \\ \\ \Delta p=0.195(37.0+41.0) \\ \\ \Delta p=15.21\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Impulse can be said to equal change in momentum.
Therefore, the impulse delivered to the ball by the bat is 15.21 kg.m/s away from the bat.
• (B). If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat?
Apply the formula:
[tex]\text{ Impulse = Force }\ast\text{ time}[/tex]Rewrite the formula for force:
[tex]\text{ Force=}\frac{impulse}{time}[/tex]Where:
time = 3.00 m/s
impulse = 15.21 kg.m/s
Hence, we have:
[tex]\begin{gathered} \text{ F=}\frac{15.21}{3} \\ \\ F\text{ = 5.07 kN} \end{gathered}[/tex]Therefore, the magnitude of the average force exerted on the ball by the bat is 5.07 kN away from the bat.
ANSWER:
(A). 15.21 kg.m/s away from the bat
(B). 5.07 kN.
The critical angle for a certain liquid-air surface is 20°. What is the index of refraction of this liquid?
ANSWER
[tex]\begin{equation*} 2.92 \end{equation*}[/tex]EXPLANATION
To find the index of refraction of the liquid, we have to apply the formula for critical angle:
[tex]\theta=\sin^{-1}(\frac{n_r}{n_i})[/tex]where nr = refractive index of air = 1
ni = refractive index of liquid
Hence, by substituting the given values into the equation, we have that the index of refraction of the liquid is:
[tex]\begin{gathered} 20=\sin^{-1}(\frac{1}{n_i}) \\ \sin20=\frac{1}{n_i} \\ n_i=\frac{1}{\sin20} \\ n_i=2.92 \end{gathered}[/tex]That is the answer.
Two blocks of mass M₁ and M₂ are connected by a massless
string that passes over a massless pulley as shown in the
figure. M₁ has a mass of 3.75 kg and rests on an incline of
0₁ = 63.5°. M2 rests on an incline of 0₂ = 15.5°. Find the
mass of block M₂ so that the system is in equilibrium (i.e.,
not accelerating). All surfaces are frictionless
The correct answer is 58.58 Kg. (Mass of M_2)
What is mass string and friction system?
A spring-mass system in simple calculation can be described as a spring system where a block is hung or attached at the free end of the spring. If the surface is frictionless so µ = 0 (we can assume)
To just begin to slide up the friction will be kinetic friction
Applying free body diagram on blocks (as diagram is not given in question so assumption is the basis on given data only)
Given, M_1 = 3.75 Kg., M_2 =?
O_1 = 63.5◦ and O_2 = 15.5◦, g = 9.8 m/s2
So, if we require to keep the system in equilibrium position
Then we can write an equation as follows:
M_1x g x Sin63.5◦ = M_2 x g x Cos 15.5◦ (To be in Equilibrium)
63.5 x 9.8 x 0.89101 = M_2 x 9.8 x 0.9659
M_2 = 63.5 x 0.89101 / 0.9659
M_2 = 58.58 Kg. (Mass of M_2)
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The current in the circuit shown is 5.0 A, and the resistor is 2.0 Ω. What is the potential across the battery? A.8.0 VB.2.5 VC.10 VD.3.0 V
Given:
The current is,
[tex]i=5.0\text{ A}[/tex]The resistance is,
[tex]R=2.0\text{ ohm}[/tex]To find:
The potential across the battery
Explanation:
Using Ohm's law, we can write the potential across the battery is,
[tex]\begin{gathered} V=iR \\ =5.0\times2.0 \\ =10\text{ V} \end{gathered}[/tex]Hence, the potential across the battery is 10 V.
A ball is thrown directly downward with an initial speed of 8.45 m/s, from a height of 29.9 m. After what time interval does it strike the ground? s
Given:
The initial speed of the ball is: u = 8.45 m/s.
The ball is thrown from the height: h = 29.9 m
To find:
The time ball takes to strike the ground.
Explanation:
The time taken by the ball to strike the ground can be determined by using the following equation.
[tex]x=ut+\frac{1}{2}at^2[/tex]Here, x = -29.9 m, u = -8.45 m/s and a = -9.8 m/s^2. The negative sign indicates that the ball is falling in the downward direction.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} -29.9=-8.45t-\frac{1}{2}\times9.8t^2 \\ \\ 4.9t^2+8.45t-29.9=0 \end{gathered}[/tex]Solving the above quadratic equation, we get:
t = 1.754 s and t = -3.478 s
But the time is never negative, thus t = 1.754 s.
Final answer:
The ball takes 1.754 seconds to strike the ground.
The Student Council conducted a vote to determine whether the homcoing dance should have live muscis ora DJ. The number of students voting for live music was 215. The number of students voting of a DJ was 645.What percent of the students voted for the DJ?(Hint: Find the total first)
Divide the number of students who voted for the DJ over the total number of students to find the percent of the students that voted for the DJ.
Since 215 students voted for live music and 645 students voted for the DJ, the total number of students was:
[tex]215+645=860[/tex]Then, the percent of the students who voted for the DJ was:
[tex]\frac{645}{860}=0.75=75\text{ \%}[/tex]Therefore, the answer is: 75%
A car traveling at 11.6 meters per second crashes into a barrier and stops in 0.287 meters. What force must be exerted on a child of mass 21.2 kilograms to stop him or her in the same time as the car? Answer must be in 3 significant digits.
The equation to obtain the final speed of car is,
[tex]v^2=u^2+2as[/tex]Substitute the known values,
[tex]\begin{gathered} (0m/s)^2=(11.6m/s)^2+2a(0.287\text{ m)} \\ a=\frac{-134.56m^2s^{-2}}{2(0.287\text{ m)}} \\ \approx-234.4m/s^2 \end{gathered}[/tex]The negative sign of acceleration indicates that the car is deaccelerating.
The force required to stop the car is,
[tex]F=ma[/tex]Substitute the magnitude of known values,
[tex]\begin{gathered} F=(21.2kg)(234.4m/s^2)(\frac{1\text{ N}}{1kgm/s^2}) \\ =4969.28\text{ N} \\ \approx4970\text{ N} \end{gathered}[/tex]Thus, the force required to stop the car is 4970 N.
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.
Four wires running through the corners of a square with sides of length 16.166 cm carry equal currents, 3.684 A. Calculate the magnetic field at the center of the square.
For practical reasons, we can consider each side of the square as an infinite wire. This can be seen on the following drawing:
This way, the field on the center will be the sum of the contribution of each wire. We can calculate the contribution of a single wire as:
[tex]B=\frac{\mu_0i}{2\pi d}=\frac{4\pi *10^{-7}*3.684}{2\pi(\frac{16.166*10^{-2}}{2})}=9.115*10^{-6}T[/tex]Then, the total field will be this, multiplied by the number of wires:
[tex]B_t=4*9.115*10^{-6}=36.46\mu T[/tex]Then, the resulting field will be Bt=36.46uT
Describe the mathematical relationship between the distance (d) and the attractive force (F) between protons and electrons.
The attractive force and the distance are inversely proportional.
[tex]F\propto\frac{1}{r}[/tex]This relation means that the attractive force decreases as the distance increases, and the attractive force increases as the distance decrease.
carts, bricks, and bands
4. Which two trials demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass?
a. Trials 2 and 4
b. Trials 2 and 6
c. Trials 4 and 7
d. Trials 6 and 7
A. The two trials that demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass are Trials 2 and 4.
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceAt a constant mass;
F₁/a₁ = F₂/a₂
When the force is doubled, the acceleration of the object is given as;
a₂ = F₂a₁/F₁
a₂ = (2F₁ x a₁) / F₁
a₂ = 2a₁
From the trials,
acceleration of trial 2 = 0.51 m/s²
acceleration of trial 4 = 1 m/s²
Thus, the two trials that demonstrate the effect of a doubling of force upon the acceleration of a cart of constant mass are Trials 2 and 4.
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Tall pacific coast redwood trees can reach heights of about 100 m. If air drag is negligibly small, how fast is a sequoia come moving when it reaches the ground if it dropped from the top of a 100 m tree?
Given data:
Height of the tree;
[tex]h=100\text{ m}[/tex]Initial velocity;
[tex]u=0\text{ m/s}[/tex]The velocity of sequoia when it reaches the ground is given as,
[tex]v=\sqrt[]{u^2+2gh}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} v=\sqrt[]{(0\text{ m/s})^2+2\times(9.8\text{ m/s}^2)\times(100\text{ m})} \\ \approx44.27\text{ m/s} \end{gathered}[/tex]Therefore, sequoia will reach the ground with a velocity of 44.27 m/s.
A steel cable on a bridge has a linear mass density of 15 kg/m. If the cable has been pulled taunt with a tension of 5536 N, what is the speed of a wave on it?
a student drops a pebble from the edge of a vertical cliff. the pebble hits the ground 4 s after it was dropped. what is the height of the cliff? a. 20 m b. 40 m c. 60 m d. 80 m
The object's speed shortly before it lands on the earth is 40 m/s.
What is an example of velocity?The speed at which something moves in a specific direction is known as its velocity. as the speed of a car driving north on a highway or the pace at which a rocket takes off. Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.
The parameters are as follows: the pebble's time, t = 4 s; the object's velocity right before impact;
The kinematic equation is as follows;
v = in which
v = 0+10 (4)
The object's speed right before impact with the earth is v = 40 m/s2, where g is the acceleration caused by gravity and an is a constant of 10 m/s2. As a result,
the object's final velocity before impact is 40 m/s.
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Two cars in opposite directions were going at 32 mph before a collision. They had a head on inelastic collision, i.e. the two cars stuck together afterward. The common speed of the combined piece right after the collision is 20 mph. The mass of Car 1 was 2,000 lb. Car 2 was heavier. The mass of Car 2 was ____ lb.
The mass of Car 2 was 3000 lb.
We need to apply the concept of conservation of momentum.
The velocity of both cars= 32mph
Combined velocity = 20mph
Mass of Car 1= 2000 lb
According to the conservation of momentum
M1V1+ M2V2= (M1+M2)
2000x32- (-M2x 32)=20(2000+M2)
64000+32M2=40000 +20M2
24000= 8M2
M2= 3000lb
Therefore the mass of Car 2 is 3000lb.
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I got first part correct but dont know how to solve second part: Two new particles with identical positive charge 3 are placed the same 0.0809 m apart. The force between them is measured to be the same as that between the original particles. What is 3 ?
Answer:
5.92 *10^-6 C
Explanation:
For the two charges q3 the force between them is given by
[tex]F=k\frac{q_3\times q_3}{d^2}[/tex]Now we know that
F = 48.1 N, d = 0.0809 m, and k = 8.99 *10^9 kg⋅m^3⋅s^−2⋅C^-2; therefore, the above gives
[tex]48.1=(8.99\times10^9)\frac{q_3\times q_3}{(0.0809)^2}[/tex][tex]\Rightarrow48.1=(8.99\times10^9)\frac{(q_3)^2}{(0.0809)^2}[/tex]Now we solve for q_3.
Dividing both sides by 8.99 * 10^9 gives
[tex]\frac{48.1}{(8.99\times10^9)}=\frac{(q_3)^2}{(0.0809)^2}[/tex]multiplying both sides by (0.0809)^2 gives
[tex]\frac{48.1}{(8.99\times10^9)}\times\mleft(0.0809\mright)^2=(q_3)^2[/tex]finally, taking the square root of both sides gives
[tex]\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}=\sqrt{(q_3)^2}[/tex][tex]q_3=\sqrt[]{\frac{48.1}{(8.99\times10^9)}\times(0.0809)^2}[/tex]Evaluating the right-hand side gives
[tex]\boxed{q_3=_{}5.92\times10^{-6}C\text{.}}[/tex]Hence, the charge q_3 is 5.92 x 10^-6 C.
A __________ is described as a device with specific resistance and is used to control current.batteryresistorparallel connectionseries connection
ANSWER:
resistor
STEP-BY-STEP EXPLANATION:
A device that has a specific resistance and is also used to control current is the resistor.
Therefore:
A resistor is described as a device with specific resistance and is used to control current.
Altering __________ will be sure to result in an alteration of the speed of the sound wave. Select all that apply.a- the wavelength of the waveb- the period of the wavec- the frequency of the waved- the properties of the mediume- the intensity of the wavef- the amplitude of the wave
ANSWER:
d- the properties of the medium
STEP-BY-STEP EXPLANATION:
The speed of a wave depends on the properties of the medium through which the wave is moving. A realization in the properties of the medium will result in a change in the speed at which the wave moves through that medium.
Therefore, the only correct option is:
d- the properties of the medium
what is the smallest amount of time in which the person can accelerate the car from rest to 23 m/s and still keep the coffee cup on the roof. The coefficient of the static friction is 0.21. The maximum acceleration of the car that is allowed so that the cup does not fall is 2.1 m/s^2
Given:
The coefficient of the static friction, μ=0.21
The maximum acceleration of the car so that the cup does not fall, a=2.1 m/s²
The initial velocity of the car, u=0 m/s
The final velocity of the car, v=23 m/s
To find:
The smallest amount of the time in which the car can accelerate so that the coffee cup will still be on the roof.
Explanation:
From the equation of motion,
[tex]v=u+at[/tex]Where t is the smallest amount of time in which the person can accelerate and still keep the cup on the car.
On rearranging the above equation,
[tex]t=\frac{v-u}{a}[/tex]On substituting the known values,
[tex]\begin{gathered} t=\frac{23-0}{2.1} \\ =10.95\text{ s} \end{gathered}[/tex]Final answer:
Thus the smallest amount of the time in which the person can accelerate the car at the given rate and still keep the cup on the roof of the car is 10.95 s
PLEASE HELP
Which is not an accurate statement about Earth's gravitational pull?
A) Earth's gravitational pull helps keep it in orbit.
B) Earth's gravitational pull is the same as Jupiter's gravitational pull.
C) Earth's gravitational pull is 9.8 m/s2.
D) Earth's gravity helps keep people from floating outside of the planet.
Every 5 seconds, the crest of a wave in the ocean travels 25 meters. What is the speed of the wave?
Given
Every 5 seconds
One wave travel 25m
Speed of waves
Explanation
[tex]\begin{gathered} v=\frac{25m}{5s} \\ v=5\text{ m/s} \end{gathered}[/tex]
The answer would be 5 m/s
carts, bricks, and bands
3. Which statement describes the effect of a doubling of force upon the acceleration of a cart of constant mass?
a. Doubling the force will cause the acceleration to be twice the original value.
b. Doubling the force will cause the acceleration to be one-half the original value.
c. Doubling the force will cause the acceleration to be four times the original value.
d. Doubling the force will cause the acceleration to be one-fourth the original value.
The statement that describes the effect of a doubling a force at a constant mass is "doubling the force will cause the acceleration to be twice the original value.
The correct answer is option A.
What is the applied force on an object?
The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceAt a constant mass;
F₁/a₁ = F₂/a₂
When the force is doubled, the acceleration of the object is given as;
a₂ = F₂a₁/F₁
a₂ = (2F₁ x a₁) / F₁
a₂ = 2a₁
Thus, when the force on the cart is doubled and the mass is constant, the acceleration of the cart will double as well.
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help, Asap!!!!!!!!!!!!!!!!!!
Answer:
0
Explanation:
all of the movement and opposite movement are the same
An intrepid hiker reaches a large crevasse in his hiking route. He sees a nice landing ledge 60.0 cm below his position but it is across a 2.3 m gap. He spends 1.2 s accelerating horizontally at 5.92 m/s2 [right] in an attempt to launch himself to the safe landing on the far side of the gap. Does he make it?
The hiker made it to a safe landing on the other side of the gap after travelling horizontally at 2.49 m.
What is the time motion from the vertical height?
The time taken for the hiker to fall from the given height is calculated as follows;
h = vt + ¹/₂gt²
where;
v is the vertical velocity = 0t is the time of motiong is acceleration due to gravityh is the height of fallh = ¹/₂gt²
t = √(2h/g)
t = √[(2 x 0.6) / (9.8)]
t = 0.35 seconds
The horizontal velocity of the hiker during the period of acceleration is calculated as follows;
Vₓ = at
Vₓ = (5.92 m/s²) x (1.2 s)
Vₓ = 7.104 m/s
The horizontal distance travelled during the time period of 0.35 seconds;
X = Vₓt
X = 7.104 x 0.35
X = 2.49 m
Thus, the hiker made it to a safe landing on the other side of the gap which is 2.3 m wide and smaller to his horizontal displacement of 2.49 m.
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The average radius of Earth is 6,371 km. If the average thickness of oceanic crust is 7.5 km and the average thickness of continental crust is 35 km, what fraction of Earth's radius is each type of crust? Please show how to solve .
If the average thickness of the oceanic crust is 7.5 km and the average thickness of the continental crust is 35 km, then the fraction of the Earth's radius is each type of crust would be 0.00117720922 and 0.005493 respectively.
What is the percentage of a number?It is the relative value that represents the hundredth part of any number for example 2% of any number represents, 2 multiplied by the 1/100th of that number.
As given in the problem average thickness of the oceanic crust is 7.5 kilometers and the average thickness of the continental crust is 35 km,
The fraction of the earth's radius as oceanic crust = 7.5 / 6371
= 0.00117720922
The fraction of the earth's radius as continental crust = 35 / 6371
= 0.005493
Thus, the fraction of the Earth's radius is each type of crust would be 0.00117720922 and 0.005493 respectively.
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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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Physics
Hello how to solve this
2.15 mL of water will spill from the beaker when a 400mL glass beaker at room temperature is filled to the brim with cold water.
What is temperature and how come 2 mL of water spills off the beaker?Temperature as studied always is the measure for the degree of hotness or coldness.Here in the question is given a glass beaker of 400 mL which is at room temperature .Now the beaker is warmed up to 30 degree celsius, and then some of amount of water spilled.Change in volume = beta x v1 x change in temperature = 210 x 10^-4 x 400 x ( 30 - 4.4) = 2.15 mL.Hence the amount of water that will spill from the beaker would be 2.15 mL.To know more about temperature visit:
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How does the work needed to stretch a spring 2 cm compare to the work needed to stretch it 1 cm.A.Same amount of workB.twice the workC.4 times the work D.8 times the work
The work required to stretch a string is given by the following equation:
[tex]W=\frac{1}{2}kx^2[/tex]Where:
[tex]\begin{gathered} k=\text{ string constant} \\ x=\text{ distance the string is stretched} \end{gathered}[/tex]If the string is stretched 2 cm then we substitute the value of "x = 2" in the formula, we get:
[tex]W_2=\frac{1}{2}k(2)^2[/tex]Solving the square and simplifying:
[tex]W_2=2k[/tex]Now, if the string is stretched 1 cm we get:
[tex]W_1=\frac{1}{2}k(1)^2[/tex]Solving the operations:
[tex]W_1=\frac{1}{2}k[/tex]Now, we determine the quotient between W2 and W1:
[tex]\frac{W_2}{W_1}=\frac{2k}{\frac{1}{2}k}[/tex]Simplifying we get:
[tex]\frac{W_2}{W_1}=4[/tex]Now, we multiply both sides by W2:
[tex]W_2=4W_1[/tex]Therefore, the work required to stretch the string 2 cm is 4 times the work to stretch it 1 cm.
If you see an object following a curve path, you can safely assume:There is a net force being applied The net force is zero
If an object is following a curve, it means that it is changing direction. This change in direction will cause a change in velocity. Change in velocity causes acceleration. Since the velocity keeps changing, the acceleration is not zero. This also means that the net force is not zero. Therefore, you can safely assume:
There is a net force being applied
Calculate the potential energy of a 2 kg ball that is about to be dropped of a height of 25 m
Given:
The mass of the ball, m=2 kg
The height from which the ball is about to be dropped, h=25 m
To find:
The potential energy of the ball.
Explanation:
The type of potential energy that is stored in this ball is called gravitational potential energy. The gravitational potential energy is the energy that an object possesses due to its height. The gravitational potential energy is directly proportional to the height at which the object is situated.
The potential energy of the ball is given by,
[tex]E=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E=2\times9.8\times25 \\ =490\text{ J} \end{gathered}[/tex]Final answer:
The potential energy of the ball is 490 J
