Explanation
U uniformly accelerated motion is the one in which the acceleration of the particle throughout the motion is uniform,the formula to find the distance is as follows:
[tex]\begin{gathered} x=v_ot+\frac{1}{2}at^2 \\ where \\ v_o\text{ is the initial velocity} \\ t\text{ is the time} \\ a\text{ is the acceleration} \end{gathered}[/tex]so
Step 1
a)let
[tex]\begin{gathered} v_o=21.3\text{ }\frac{m}{s} \\ a=3.6\frac{m}{s^2} \\ t=5\text{ s} \end{gathered}[/tex]b) now,replace in the formula and calculate
[tex]\begin{gathered} x=v_{o}t+\frac{1}{2}at^{2} \\ x=21.3\frac{m}{s}*5s+\frac{1}{2}3.6\frac{m}{s^2}*(5\text{ s\rparen}^2 \\ x=106.5\text{ m}+45\text{ m} \\ x=151.5\text{ m} \end{gathered}[/tex]therefore, the answer is 151.5 meters
I hope this helps you
How much heat transfer is required to completely boil 3500 g of water (already at its boiling point of 100°C) into a gas?answer in joules
Answer:
7910000 J
Explanation:
The heat transfer to phase change is calculated as:
[tex]Q=mH_v[/tex]Where m is the mass and Hv is the heat of vaporization.
For water, Hv = 2260 J/g.
So, replacing m = 3500 g and Hv = 2260 J/g, we get:
Q = 3500 g (2260 J/g)
Q = 7910000 J
Therefore, the answer is:
7910000 J
Select all of the following that are equal to an impulse of 30 units.
A. Force = 30, time = 1.2
B. Force = 0.1, time = 300
I C. Force = 6, time = 4
D. Force = 10, time = 3
ANSWER
B and D
EXPLANATION
The impulse is the product of the force applied to an object and the amount of time the force is applied
[tex]\Delta p=F\cdot\Delta t[/tex]Let's find the impulse with the information on each option.
A. F = 30, Δt = 1.2
[tex]\Delta p=30\cdot1.2=36[/tex]B. F = 0.1, Δt = 300
[tex]\Delta p=0.1\cdot300=30[/tex]C. F = 6, Δt = 4
[tex]\Delta p=6\cdot4=24[/tex]D. F = 10, Δt = 3
[tex]\Delta p=10\cdot3=30[/tex]The options that have an impulse equal to 30 are B and D
A 5.0g bullet is fired straight up from the ground, The bullet leaves the ground with a speed of 85 m/s taking to the ground as the reference levelA) what is the gravitational P.E of the bulletB) what is the initial KE of the bulletC) what is the total initial ME of the bullet when the bullet is 12m high above the groundD) what is the gravitational PE of the bulletE) what should the ME be at this height F) how fast is the bullet moving at this height
Given information:
Mass of the bullet;
[tex]\begin{gathered} m=5\text{ g} \\ =5\times10^{-3}\text{ kg} \end{gathered}[/tex]Initial velocity of the bullet;
[tex]u=85\text{ m/s}[/tex]Part (a),
Taking ground as the reference level. So, the initial height of the bullet is 0 m.
The gravitational potential energy is given as,
[tex]U=\text{mgh}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} U=(5\times10^{-3}\text{ kg})\times(9.8\text{ m/s}^2)\times0 \\ =0\text{ J} \end{gathered}[/tex]Therefore, the gravitational potential energy at the ground is 0 J.
Part (B)
The initial kinetic energy of the bullet is given as,
[tex]K=\frac{1}{2}mu^2[/tex]Substituting all known values,
[tex]\begin{gathered} K=\frac{1}{2}\times(5\times10^{-3}\text{ kg})\times(85\text{ m/s})^2 \\ =18.0625\text{ J} \end{gathered}[/tex]Therefore, the initial kinetic energy of the bullet is 18.0625 J.
Part (C).
According to the conservation of energy, the total mechanical energy (ME) of the bullet will remain conserved. Therefore, the total initial ME of the bullet when the bullet is 12 m high above the ground is 18.0625 J.
Part (D)
The gravitational potential energy when the bullet is 12 m high is,
[tex]U_h=mgh[/tex]Substituting all known values,
[tex]\begin{gathered} U_h=(5\times10^{-3})\times(9.8\text{ m/s}^2)\times(12\text{ m}) \\ =0.588\text{ J} \end{gathered}[/tex]Therefore, the gravitational potential energy when the bullet is 12 m high is 0.588 J.
Part (E).
The velocity of the bullet when it reaches the height of 12 m is given as,
[tex]v^2=u^2-2gh[/tex]Here, v is the velocity when the bullet is 12 m high.
Substituting all known values,
[tex]\begin{gathered} v^2=(85\text{ m/s})^2-2\times(9.8\text{ m/s})\times(12\text{ m}) \\ =6989.8\text{ m}^2\text{ /s}^2 \end{gathered}[/tex]The total mechanical energy when the bullet is 12 m high is,
[tex]ME=U_h+\frac{1}{2}mv^2[/tex]Substituting all known values,
[tex]\begin{gathered} ME=(0.588\text{ J})+\frac{1}{2}\times(5\times10^{-3}\text{ kg})\times(6989.8\text{ m}^2\text{ /s}^2) \\ =18.0625\text{ J} \end{gathered}[/tex]Therefore, the total mechanical energy ME when the bullet is 12 m high is 18.0625 J.
Part (F),
The velocity when the bullet is 12 m high is given as,
[tex]v=\sqrt[]{u^2-2gh}[/tex]Substituting all known values,
[tex]\begin{gathered} v=\sqrt[]{(85\text{ m/s})^2-2\times(9.8\text{ m/s}^2)\times(12\text{ m})} \\ \approx83.605\text{ m/s} \end{gathered}[/tex]Therefore, the velocity when the bullet is 12 m high is 83.605 m/s.
A 230-g mass hangs from a string that is wrapped around a pulley, as shown in the figure. The pulley is suspended in such a way that it can rotate freely. When the mass is released, it accelerates toward the floor as the string unwinds. Model the pulley as a uniform solid cylinder of mass 1.00 kg and radius 5.00 cm. Assume that the thread has negligible mass and does not slip or stretch as it unwinds.
Determine the magnitude of the pulley's angular acceleration.
Determine the magnitude of the acceleration of the descending weight.
Calculate the magnitude of the tension in the string.
The 230-g mass hanging from the 5.00 cm, 1.00 kg pulley and later accelerating as the string unwinds indicates that the parameters of the forces on the mass and pulley are as follows;
The angular acceleration of the pulley is approximately 61.8 m/s²The magnitude of the acceleration of the weight is approximately 3.09 m/s²The tension in the string is approximately 1.5456 NWhat is an angular acceleration of a rotating body?Angular acceleration is the rate at which angular velocity changes with time.
Mass attached to the pulley string, M = 230-g = 0.23 kg
Mass of the pulley, m = 1.00 kg
Radius of the pulley = 5.00 cm = 0.05 m
The equations that can be used are;
a = r·αT·r = I·αm·g - T = m·av = ω·rWhere;
a = The acceleration of the weight
α = The angular acceleration of the pulley
m = Mass of the attached weight
T = The tension in the string
v = The linear velocity
ω = The angular velocity of the pulley
Moment of inertia of the pulley, I = M·r²/2
Therefore, T = ((M·r²/2)/r) × (a/r) = M·a/2
m·g - M·a/2 = m·a
m·g = M·a/2 + m·a = a·(m + M/2)
a = m·g/(m + M/2)
The acceleration of the weight, a = 0.23 × 9.81/(0.23 + 1/2) ≈ 3.09
The acceleration of the weight, a ≈ 3.09 m²/sThe angular acceleration of the pulley, α = a/r
Therefore;
The angular acceleration of the pulley, α = 3.09/0.05 ≈ 61.8 m²/sThe tension in the string, T = m·g - m·a
Therefore;
T = 0.23 × 9.81 - 0.23 × 3.09 = 1.5456
The tension in the string, T ≈ 1.5456 NLearn more about the moment of inertia of a rotating body here:
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An ocean A. Open system B. Closed system C . Isolated
ANSWER:
A. Open system
STEP-BY-STEP EXPLANATION:
The ocean is a component of the hydrosphere, and the ocean surface represents the interface between the hydrosphere and the atmosphere above it. Which means that it generates an exchange of matter and energy, so the ocean is an example of an open system.
REMOTE GR 8 SCIENCE / SECTION1 / 9 OF 50Which transformation must occur when a dam uses water to generate energy?O A. Mechanical energy to potential energyоB. Chemical energy to potential energyC. Mechanical energy to electrical energyOD. Chemical energy to electrical energy
As water flows through the dam, its kinetic energy is used to turn a turbine, a generator convert's the turbine mechanical energy into electrical energy.
C. Mechanical energy to electrical energy
Part AA space vehicle accelerates uniformly from 80m/s att 0 to 167 m/s att 10.0 sHow far did it move between t = 2.0 s and t - 6.08 ?Express your answer to two significant figures and include the appropriate units.
Given:
Velocity at t = 0: 80 m/s
Velocity at t = 10.0 s: 167 m/s
Let's find the distance it covered between t = 2.0s and t = 6.0 s
Apply the kinematics formula:
[tex]v=u+at[/tex]Where:
v is the final velocity ==> 167 m/s
u is the initial velocity ==> 80 m/s
a is the acceleration
t is the time ==> 10.0 - 0 = 10.0 s
Now, let's find the acceleration.
Rewrite the formula for a:
[tex]\begin{gathered} a=\frac{v-u}{t} \\ \\ a=\frac{167-80}{10} \\ \\ a=\frac{87}{10} \\ \\ a=8.7m/s^2 \end{gathered}[/tex]The acceleration is 8.7 m/s²
To find the distance, apply the formula:
[tex]s=ut+\frac{1}{2}at^2[/tex]Thus, we have:
• At t = 2.0 s:
[tex]\begin{gathered} d_1=80(2)+\frac{1}{2}(8.7)(2)^2 \\ \\ d_1=160+17.4 \\ \\ d_1=177.4\text{ m} \end{gathered}[/tex]• At t = 6.0 s:
[tex]\begin{gathered} d_2=ut+\frac{1}{2}at^2 \\ \\ d_2=80(6)+\frac{1}{2}(8.7)(6)^2 \\ \\ d_2=480+156.6 \\ \\ d_2=636.6\text{ m} \end{gathered}[/tex]To find the distance traveled between t = 2.0 s and t = 6.0 s, we have:
d = d2 - d1 = 636.6 m - 177.4 m
d = 459.2 m
Therefore, the distance traveled betwen t = 2.0 s and t = 6.0 s is 459.2 m
ANSWER:
459.2 m
An eagle goes straight up with an initial velocity of 75m/s toward its food. Its food is located 250m above the ground. How fast will the eagle be moving when she reaches her food?
The vertical distance covered by the eagle can be given as,
[tex]h=ut+\frac{1}{2}gt^2[/tex]Plug in the known values,
[tex]\begin{gathered} 250m=(75\text{ m/s)t+}\frac{1}{2}(-9.8m/s^2)t^2 \\ -(4.90ms^{-2})t^2+(75\text{ m/s)t-250m=0} \\ (4.90ms^{-2})t^2-(75\text{ m/s)t+250 m=0} \end{gathered}[/tex]The above equation can be further solved as,
[tex]\begin{gathered} t=\frac{75\text{ m/s}\pm\sqrt[]{(75m/s)^2-4(4.90ms^{-2})(250\text{ m)}}}{2(4.90ms^{-2})^{}} \\ =\frac{75\text{ m/s}\pm26.9\text{ m/s}}{9.80m/s^2} \\ =10.4\text{ s, }4.91\text{ s} \end{gathered}[/tex]Therefore, the time taken by eagle to reach at food is 10.4 s or 4.91 s.
In the summer time fruit flies are the worst. One morning I woke up to 5 fruit flies in my kitchen. Three hours later there were 13 fruit flies. If the growth rate is continuous, at what rate are the flies increasing? Round answers to four decimal places.
We are asked to determine the continuous growth rate. To do that we will use the following function:
[tex]P(t)=P_0e^{kt}[/tex]Where "P0" is the initial population, "k" is the growth rate, and "t" is time. Replacing the initial population of 5 we get:
[tex]P(t)=5e^{kt}[/tex]Now we are told that the population is 13 when the time is 3 hours. Replacing we get:
[tex]13=5e^{3k}[/tex]Now we solve for "k". First, by dividing both sides by 5:
[tex]\frac{13}{5}=e^{3k}[/tex]Now we take natural logarithm to both sides:
[tex]\ln (\frac{13}{5})=\ln (e^{3k})[/tex]Now we use the following property of logarithms:
[tex]\ln x^y=y\ln x[/tex]Applying the property:
[tex]\ln (\frac{13}{5})=3k\ln e[/tex]We have that the value of ln(e) is 1, therefore:
[tex]\ln (\frac{13}{5})=3k[/tex]Now we divide both sides by 3:
[tex]\frac{1}{3}\ln (\frac{13}{5})=k[/tex]Solving the operation we get:
[tex]0.319=k[/tex]Therefore, the growth rate is 0.319.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!!
Answer:
the answer is B
Explanation:
because the hot air is transferred through heating being irrational from a thermal burning, because how the oven works
Explain the relation between the period and the other variable below in spring, show your point using formula.
Explanation:
The period in spring can be calculated as
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]Where m is the mass attached and k is the spring constant.
Answer:
Taking into account this formula, we can conclude the following:
- Period vs. Stretch distance.
The formula doesn't include the stretch distance, so the period is independent of the stretch distance. No matter what is the stretch distance the period will not change.
- Period vs. Mass attached
The period is proportional to the square root of the mass, so when the mass is greater the period is greater.
-Period vs. Spring constant
The period is inversely proportional to the square root of the spring constant, so when the spring constant increases, the period decreases.
A net constant force of 1500 N gives a toy rocket an acceleration of 2.5 m/s squared. what is the the mass of the rocket?
We have
F=1500N
a=2.5 m/s^2
We use the next formula
[tex]F=ma[/tex]Where F is the force, m is the mass and a is the acceleration
Then we substitute
[tex]1500=2.5\text{ m}[/tex]Then we isolate the m
[tex]m=\frac{1500}{2.5}=600kg[/tex]ANSWER
the mass of the rocket is 600kg
Describe what values you could solve for if you were given kinetic energy?
The kinetic energy is given by:
[tex]K=\frac{1}{2}mv^2[/tex]From the above equation,
we can solve from the mass m if velocity is given
or for the velocity if mass m is given
Hello there, to start off. I am BLIND literely and can't read the graphics on the website so the information about how to answer these problems needs to be written out. I am sorry but just how it is. Now the question. You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close together. Joe's tent is 21.5 m from yours, in the direction 21.5 ∘ north of east. Karl's tent is 44.5 m from yours, in the direction 40.5 ∘ south of east.Part AWhat is the distance between Karl's tent and Joe's tent?
The three tents form a triangle with two sides given by the distances between your tent and the tents of your friends; the remaining side is the distance between Joe's and Karl's tents.
To determine this side (and hence this distance) we need to notice that the angle between the two known sides is the addition of the angles between those sides and your position, hence the angle between the sides is 62°.
Now that we know this we can use the cosine law to determine the distance between the tents. The cosine law states that:
[tex]c^2=a^2+b^2-2ab\cos \gamma[/tex]this means that the squared of the distance we are looking for is equal to the sum of the squares of the other two distances minus twice the product of the other two sides and the cosine of the angle between them.
Then we have that:
[tex]\begin{gathered} c=\sqrt[]{21.5^2+44.5^2-2(21.5)(44.5)\cos (62)} \\ c=39.3 \end{gathered}[/tex]the distance we are looking for is the square root of twenty one point five squared plus forty four point five squared minus two multiplied by twenty one point five by forty four point five by the cosine of 62.
By making the operations we conclude that the distance between Joe's and Karl's tents is thirty nine point three meters
After traveling for 6.0 seconds, a runner reaches a speed of 12 m/s after startingfrom rest. What is the runner's acceleration?
Given data:
Initial speed of the runner as he starts from the rest;
[tex]u=0[/tex]Final speed of the runner;
[tex]v=12\text{ m/s}[/tex]Time taken;
[tex]t=6.0\text{ s}[/tex]The acceleration of the runner is given as,
[tex]a=\frac{v-u}{t}[/tex]Substituting all known values,
[tex]\begin{gathered} a=\frac{(12\text{ m/s})-0}{(6.0\text{ s})} \\ =2\text{ m/s}^2 \end{gathered}[/tex]Therefore, the acceleration of the runner is 2 m/s².
When 2 identical charged particles get closer to each other the strength of theelectrical force between them.A. increasesB. stays the sameC. decreasesD. you have to know the amount of the charge on the particles to answer
Given
2 identical charged particles get closer to each other
To find
The strength of the electrical force between them.
Explanation
The electrostatic force is indirectly proportional to the square of the distance between the charges.
So as the distance decreases, the electric force increases.
Conclusion
The correct opttion is
A. increases
An athlete swings a 7.9 kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.9 m at an angular speed of 0.27 rev/s.
The tangential velocity of the ball will be 1.53 m/s
What is tangential velocity?Tangential velocity is the linear component of an object's velocity moving along a circular path. If an object moves in a circular orbit with a distance r from the center, the object's velocity is always directed tangentially. This is called tangential velocity. Any instantaneous linear velocity is also said to be its tangential velocity.
The rate of change of the object's angular displacement is the angular velocity. It is represented by ω and its standard unit is radians/second. It differs from linear velocity as it only deals with objects moving in circular motion. So we measure the speed at which the angular displacement is swept.
Tangential speed = angular speed × radius of the circle
[tex]V_{t}[/tex] = r×ω
For the given case,
Radius (r) = 0.9 m
Angular speed (ω) = 0.27 rev/sec = 1.70 rad/sec
tangential speed ([tex]V_{t}[/tex]) = 0.9 × 1.70 = 1.53 m/s
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"An athlete swings a 7.9 kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.9 m at an angular speed of 0.27 rev/s. What is the tangential velocity of ball?"
I am unsure of how to get the answer for this question. Please help!
ANSWER:
c) 5.4
STEP-BY-STEP EXPLANATION:
Given:
u = 54 km/h
v = 32 km/h
d = 65 m = 0.065 km
We can calculate the time starting from the following equation:
[tex]d=\mleft(\frac{u+v}{2}\mright)\cdot\: t[/tex]We solve for t:
[tex]t=\frac{d}{\mleft(\frac{u+v}{2}\mright)}[/tex]We substitute and calculate the time in hours, then convert that time to seconds, just like this:
[tex]\begin{gathered} t=\frac{0.065}{\frac{54+32}{2}}=0.00151\text{ h} \\ t=0.00151\text{ h}\cdot\frac{3600\text{ sec}}{1\text{ h}} \\ t=5.4\text{ sec} \end{gathered}[/tex]Therefore, the time it would take is 5.4 seconds
a radio station broadcast a program at 128.7 Mhz. calculate the wavelength of the radiowave at this frequency
The wavelength of the radio wave is 2.33 meters.
Given that the frequency of the radio wave is 128.7MHz.
[tex]128 MHz = 128 * 10^{6} Hz[/tex]
Radio waves are electromagnetic waves. All electromagnetic waves (including radio waves ) travel at a speed [tex](c)[/tex] of [tex]3*10^{8} m/s[/tex].
The speed of the radio wave is given by speed = frequency*wavelength.
Hence, the wavelength of the radio wave can be calculated as,
[tex]wavelength = \frac{speed}{frequency}[/tex]
Thus, [tex]wavelength = \frac{3*10^{8} }{128.7*10^{6} } meters.[/tex]
[tex]=2.33 meters.[/tex]
Hence, the wavelength of the radio wave is [tex]2.33 meters.[/tex]
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The net torque of a bod about certain axis of rotation is 150 Nxm. You recalculate the net torque about a different axis of rotation. If your work is correctyou have to find the same number.is this tstemnt true or false?
Every individual torque acting on the body depends on the axis of rotation, however the net torque is independent of the axis we choose. Therefore, the statement is true.
A 120 kg car accelerates from initial speed 5.0 m/s to final speed 10.0 m/s in 5.0 seconds. How much power (watts) does that require?
In order to determine the required power, use the following formula:
[tex]P=\frac{W}{t}[/tex]where
t: time = 5.0 s
W: work
calculate the work W as follow:
[tex]W=F\cdot d=m\cdot a\cdot d[/tex]where
m: mass = 120 kg
a: acceleration = ?
d: distance
calculate the acceleration as follow:
[tex]a=\frac{v-v_0}{t}=\frac{10.0\text{ m/s- 5.0m/s}}{5.0s}=1\frac{m}{s^2}[/tex]the distance can be obtained by using the following formula:
[tex]d=v_0\cdot t+\frac{1}{2}at^2=(\frac{5.0m}{s})(5.0s)+\frac{1}{2}(\frac{1m}{s^2})(5.0s)^2=37.5m[/tex]then, replace the previous values of a and d to calculate W:
[tex]W=(120kg)(\frac{1m}{s^2})(37.5m)=4500J[/tex]Finally, replace W and d into the formual for the power P:
[tex]P=\frac{4500J}{5.0s}=900W[/tex]Hence, 900 watts are required to accelerate the car.
Unpolarized light with intensity 455.16 W/m2 passes first through a polarizing filter with its axis vertical, then through a polarizing filter with its axis 33.33o from vertical. What light intensity emerges from the second filter ?
Given:
The intensity of unpolarized incident light is
[tex]I_0=455.16\text{ W/m}^2[/tex]It passes first through a polarizing filter with its axis vertical
The second polarising filter is at an angle of,
[tex]\theta=33.33\degree[/tex]from the vertical
To find:
What light intensity emerges from the second filter ?
Explanation:
According to the Malus law, the emergent light intensity is,
[tex]\begin{gathered} I=I_0cos^2\theta \\ =455.16\times cos^233.33\degree \\ =317.74\text{ W/m}^2 \end{gathered}[/tex]Hence, the required intensity is
[tex]317.74\text{ W/m}^2[/tex]3. What are the charges and location of each of the parts of the atom?
There are two parts main parts of the atoms they are the nucleus and the orbits.
The atom mains consists of three particles, protons, neutrons, and electrons.
The protons are positively charged particles. The charge of a proton is 1.602×10⁻¹⁹ C. And the neutron does not possess any charge. The neutrons are neutral particles. Protons and neutrons are located at the nucleus of an atom.
The electrons are negatively charged particles. The charge of an electron is -1.602×10⁻¹⁹ C. The electrons will be orbiting the nucleus. Thus they are located around the nucleus of an atom.
E. O-27.7 m/s9. An object of mass 2 kg moving with a speedof 26 m/s to the right collides with an objectof mass 13 kg moving with a speed of 2 m/sto the left. If the collision is completely inelastic,calculate their speed after collision. (1 point)A. O-0.831 m/sB. O 1.733 m/sC. 06.089 m/sD. O5.39 m/sE. O 0.388 m/s10. An object of mass 16 kg moving with a speed
Given
Mass of an object, m=2 kg
Velocity of the object, u=26 m/s towards right
Mass of the other object, m'=13 kg
Speed of the object, u'=2 m/s towards left
To find
If the collision is completely inelastic,calculate their speed after collision.
Explanation
Since the collision is perfectly inelastic, so after collision both the body would combine together and move as one
Let the velocity after collision be v.
Considering the direction towards right as positive.
Thus,
By conservation of momentum
[tex]\begin{gathered} mu+m^{\prime}u^{\prime}=(m+m^{\prime})v \\ \Rightarrow2\times26-13\times2=(2+13)v \\ \Rightarrow v=1.733\text{ m/s} \end{gathered}[/tex]Conclusion
The velocity is B. 1.733 m/s
An object weighing 88 N is pushed by a 100 N force. What is the acceleration of the object?
Given data:
* The weight of the object is 88 N.
* The force acting on the object is 100 N.
Solution:
The weight of the object in terms of the mass of the object is,
[tex]W=mg[/tex]where m is the mass of the object, and g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} 88=m\times9.8 \\ m=\frac{88}{9.8} \\ m=8.98\text{ kg} \end{gathered}[/tex]According to Newton's second law, the acceleration of the object in terms of the force acting on the object is,
[tex]F=ma[/tex]where a is the acceleration of the object,
Substituting the known values,
[tex]\begin{gathered} 100=8.98\times a \\ a=\frac{100}{8.98} \\ a=11.14ms^{-2} \end{gathered}[/tex]Thus, the acceleration of the object is 11.14 meters per second squared.
Explain why force, acceleration, and velocity are vectors.
Answer:
simply because they all have direction
Explanation:
vector quantities are quantities which have both magnitude and direction.Force,acceleration and velocity have both magnitude and direction.
Uncle Harry weighs 75 N. What would his mass be in kilograms?
Uncle Harry's mass is 7.5 kg.
The weight of a body is defined as the gravitational force with which a body is attracted toward the center of the earth.
If [tex]g[/tex] is the acceleration due to gravity at a place, then a body of mass [tex]m[/tex] is attracted towards the center of the earth with a force equal to [tex]mg[/tex] at the place. Whereas, the mass of a body remains the same everywhere.
Hence the weight of a body is given by
[tex]W = mg[/tex] -------------- (1)
As the value of [tex]g[/tex] varies from place to place, the weight of a body also varies from place to place.
Given that, the weight of Uncle Harry, [tex]W=[/tex] [tex]75 N.[/tex]
From equation (1), we get mass as,
[tex]m=\frac{W}{g}[/tex] --------------(2)
Taking acceleration due to gravity, [tex]g= 10 m/s^{2}[/tex],.
Substitute for [tex]W[/tex] and [tex]g[/tex] in equation (2), we have,
[tex]m=\frac{75}{10}\\ \\m=7.5 kg[/tex]
Thus, the mass of Uncle Harry is [tex]7.5 kilograms.[/tex]
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Which two notes are not an octave apart?1) 256 Hz and 512 Hz2) 262 Hz and 524 Hz3) 331 Hz and 622 Hz4) 277 Hz and 554 Hz
ANSWER
Option 3
EXPLANATION
Two notes are said to be an octave apart when the frequency of one note is exactly 2 times the other. In other words, it has twice as many waves.
Let us calculate the ratio of each option:
[tex]\begin{gathered} \Rightarrow\frac{512}{256}=2 \\ \Rightarrow\frac{524}{262}=2 \\ \Rightarrow\frac{622}{331}=1.88 \\ \Rightarrow\frac{554}{277}=2 \end{gathered}[/tex]As we can see, only in option 3 is the higher frequency not exactly 2 times the lower frequency.
Therefore, the answer is option 3.
Two forces act on an object as shown. Calculate the magnitude of the resultant force.[ Hint: Use the Pythagorean Theorem]
Using law of vector addition , the magnitude (R) of resultant force is given by
[tex]\begin{gathered} R=\sqrt{30^2+40^2+2\times30\times40\sin90\degree;} \\ R=\text{ }\sqrt{4900}=\text{ 70N}\begin{cases}sin90\degree={1} \\ \end{cases} \end{gathered}[/tex]Final answer is 70 N.
Note:- we need to use law of vector addition to get resultant of two forces but we cant use Pythagorean theorem for vector .
Find the final temperature of the thermometer assuming no heat flows to the surroundings
ANSWER
[tex]\begin{equation*} 71.51\degree C \end{equation*}[/tex]EXPLANATION
Parameters given:
Mass of thermometer, m = 300 g
Initial temperature of thermometer, t1 = 35°C
Volume of water, V = 258 cm³
Initial temperature of water, T1 = 80°C
First, let us find the mass of the water using the formula for density:
[tex]\begin{gathered} \rho=\frac{M}{V} \\ \Rightarrow M=\rho *V \end{gathered}[/tex]where ρ = density of water = 1 g/cm³
Therefore, the mass of the water is:
[tex]M=1*258=258g[/tex]According to the conservation of energy, the total heat flow (the sum of the heat energy of the thermometer and water) must be equal to 0 since no heat flows to the surroundings:
[tex]\begin{gathered} Q_g+Q_w=0 \\ mc(T-t_1)+MC(T-T_1)=0 \end{gathered}[/tex]where c = specific heat capacity of glass thermometer = 0.2 cal/g°C
C = specific heat capacity of water = 1 cal/g°C
T = final temperature of thermometer and water
Hence, solving for T, we have that:
[tex]\begin{gathered} T=\frac{mct_1+MCT_1}{mc+MC} \\ T=\frac{(300*0.2*35)+(258*1*80)}{(300*0.2)+(258*1)} \\ T=\frac{2100+20640}{60+258}=\frac{22740}{318} \\ T=71.51\degree C \end{gathered}[/tex]That is the final temperature of the thermometer.