Answer:
The correct answer is "12 m/s²".
Explanation:
Given:
[tex]F_{app} = 100 \ N[/tex]
As we know,
⇒ [tex]F_{app} = mg-ma[/tex]
Or,
⇒ [tex]a = g-(\frac{F_{app}}{m} )[/tex]
By substituting the values, we get
⇒ [tex]=10-(-\frac{100}{50} )[/tex]
⇒ [tex]=10+2[/tex]
⇒ [tex]=12 \ m/s^2[/tex]
A car driving down a road runs of gas and will eventually stop because of:
A. Friction
B. Thrust
C. It will remain in motion forever
OD. Gravity
the momentum of a spring coil when the external compressing force is removed b the difference between the final momentum and the initial momentum of the object c the backward momentum felt by an object or person exerting force on another object d the difference between the total momentum of the system after impact and the total momentum of the system before impact
Answer:
the correct one is b
the difference between the final moment and the initial moment
Explanation:
The momentum is related to the moment
I = ΔP
∫ F dt = p_f - p₀
where p_f and p₀ are the final and initial moments, respectively
When checking the different answers, the correct one is b
the difference between the final moment and the initial moment
Sarah and Maisie are analysing data from their school sports day. Looking at the 1500 m results for Stephen, Maisie believes that Stephen’s displacement from the start line is 1500 m. Sarah says that she is incorrect and that his displacement from the start is actually 0 m. Which of the students is correct? Give reasoning for your answer.
Answer:
Sarah is right
Explanation:
This is an exercise that differentiates between scalars and vectors.
A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.
In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero
consequently Sarah is right
If the depth of water in a well is 10 m, what is the pressure exerted by it on the
bottom of the well? (Use g = 10 m/s)
[Ans: 10 N/m]
Answer:
Let d be the density of the water (1000 kg / m^3 eq to 1 gm / cm^3)
P = d g h for the pressure due to a column at the bottom of the column.
P = 1000 kg / m^3 * 10 m/s^2 * 10 m = 10^5 kg / m * s^2 = 10^5 N/m
If a pendulum's length is 2.00 m and ag = 9.80 m/s, how many complete oscillations does the pendulum make in 5.00 min?
Answer:
Number of oscillation = 106 oscillations
Explanation:
Given the following data;
Length = 2 mAcceleration due to gravity, g = 9.8 m/s²Time = 5 minutesTo find how many complete oscillations the pendulum makes in 5.00 min;
First of all, we would determine the period of oscillation of the pendulum using the following formula;
[tex] T = 2 \pi \sqrt{\frac{l}{g}} [/tex]
Where;
T is the period.l is the length of the pendulum.g is acceleration due to gravity.Substituting into the formula, we have;
[tex] T = 2 * 3.142 \sqrt{\frac{2}{9.8}} [/tex]
[tex] T = 6.284 \sqrt{0.2041} [/tex]
[tex] T = 6.284 * 0.4518 [/tex]
Period, T = 2.84 seconds
Next, we would determine the number of complete oscillation in 5 minutes;
We would have to convert the time in minutes to seconds.
Conversion:
1 minutes = 60 seconds
5 minutes = X seconds
Cross-multiplying, we have;
X = 5 * 60 = 300 seconds
Mathematically, the number of oscillation of a pendulum is given by the formula;
[tex] Number \; of \; oscillation = \frac {Time}{Period} [/tex]
Substituting into the formula, we have;
[tex] Number \; of \; oscillation = \frac {300}{2.84} [/tex]
Number of oscillation = 105.63 ≈ 106 oscillations
Number of oscillation = 106 oscillations
Me Ayudan con este ejercicio por favor !!!
A wheel accelerates from rest to 20 rad/s at a uniform rate of 3.5 rad/s2. Through what angle (in radians) did the wheel turn while accelerating
Answer:
[tex]\theta=57.14rad[/tex]
Explanation:
From the question we are told that:
Angular Velocity [tex]\omega=20rad/s[/tex]
Acceleration [tex]a=3.5rads/s^2[/tex]
Generally the equation for Angular velocity is mathematically given by
[tex]\omega_2^2=\omega_1^2+2 a \theta[/tex]
Therefore
[tex]20^2=0+2*2.5*\theta[/tex]
[tex]\theta=\frac{400}{2*3.5}[/tex]
[tex]\theta=57.14rad[/tex]
A 34-m length of wire is stretched horizontally between two vertical posts. The wire carries a current of 68 A and experiences a magnetic force of 0.16 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of 72.0° with respect to the wire.
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (at 76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F ?
Answer:
The correct answer would be - Low pitch.
Explanation:
As it is known that if frequency increases then pitch will be increase as well as pitch depends on frequency, Now for the question it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases. As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
One ball is aprojected in the uapward directon with a certain velocity ‘v’ and other
is thrown downwards with the same velocity
Complete question is;
One ball is projected in the upward direction with a certain velocity ‘v’ and other is thrown downwards with the same velocity at an angle θ.
The ratio of their potential energies at highest points of their journey, will be:
Answer:
u² : (u cos θ)²
Explanation:
Maximum potential energy for the first ball will be at a maximum height of;
H = u²/2g
Thus;
PE = mg(u²/2g)
For second ball at an angle θ, maximum PE will occur at a max height of (u cos θ)²/2g
PE = mg((u cos θ)²/2g)
The ratios of the potential energies are;
mg(u²/2g) : mg((u cos θ)²/2g)
mg will will cancel out since they are of same mass.
Thus;
(u²/2g) : (u cos θ)²/2g
Again 2g will cancel out to give;
u² : (u cos θ)²
An evacuated tube uses an accelerating voltage of 55 kV to accelerate electrons to hit a copper plate and produce x rays. Non-relativistically, what would be the maximum speed of these electrons?
Answer:
v = 4.4 x 10⁷ m/s
Explanation:
The kinetic energy of the electrons will be equal to the energy supplied by the electric voltage:
Kinetic Energy = Electric Energy
[tex]\frac{1}{2}mv^2 = eV[/tex]
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
v = speed of electron = ?
e = charge on electron = 1.6 x 10⁻¹⁹ C
V =Voltage = 55 kV = 55000 V
Therefore,
[tex]\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(v)^2 = (1.6\ x\ 10^{-19}\ C)(55000\ V)\\\\v^2 = \frac{(2)(8.8\ x\ 10^{-16}\ J)}{9.1\ x\ 10^{-31}\ kg}\\\\v = \sqrt{19.34\ x\ 10^{14}\ m^2/s^2}[/tex]
v = 4.4 x 10⁷ m/s
A jet plane is speeding down the runway during takeoff. Air resistance is not negligible. Identify the forces on the jet.
Answer:
Weight , Drag , Thrust , Normal force
Explanation:
It is given that a jet plane is speeding during takeoff. The air resistance is negligible.
Therefore, the forces that acts on the jet plane are :
Weight
As the plane has some mass, the weight of the plane is acted upon it. The weight is acted in the downward direction.
Normal force or the Lift
The lift force is the force that helps the plane to move up in the air and it opposes the weight of the plane. It is the normal force to the weight of the plane and this force holds the plane in the air.
Thrust
The thrust force is the force which helps to move an aircraft in the direction of the motion. This force is created by the jet engine. This force moves the plane in the forward direction through the air.
Drag
Drag force is caused due to the difference in the velocity of the solid objects and the air. The drag force opposes the forward motion of the jet plane.
derive expression for pressure exerted by gas
A 1500kg car start from rest and increases it velocity to 30mls in a time of 25sec. calculate the distance the car travel, how much force was use, how much work was done.
Answer:
workdone= 1/2mv^2
1/2×1500×30^2
675000J
distance of distinct vision.
is placed at a distance less than the distance of near point, its image o
will be blurred. Hence human eye can not see such object clearly.
ADDITIONAL INFORMATION
distance of distinct vision for a normal eye of different age groups
Babies = 7 cm
Adults = 25 cm
erson of age 55 years and above = 100 cm
ever, in our discussion we are concerned with a normal eye of an adult so least
The foulart position of an ahiect from a human eve so that the sh
The least distance up to which we can see the objects clearly without any strain is called least distance of distinct vision. Least distance of distinct vision for a normal human being is 25cm. For young people, the least distance of distant vision will be within 25cm which however it varies with age.
Answer:
25 you said ? thats incorecct
Explanation:
Electricity is distributed from electrical substations to neighborhoods at 13000 V. This is a 60 Hz oscillating (AC) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the 120 V that is delivered to your house.
Required:
a. How many turns does the primary coil on the transformer have if the secondary coil has 130 turns?
b. No energy is lost in an ideal transformer, so the output power Pout from the secondary coil equals the input power Pin to the primary coil. Suppose a neighborhood transformer delivers 280 A at 120 V. What is the current in the 1.3×10^4 V line from the substation?
Answer:
a) N₁ = 14083 turns, b) I₁ = 2.58 A
Explanation:
The relationship that describes the relationship between the primary and secondary of the transformer is
[tex]\frac{V_2}{N_2} = \frac{V_1}{N_1}[/tex]
a) They indicate that the secondary has N2 = 130 turns, the turns of the primary are
N₁ = [tex]N_2 \frac{V_1}{V_2}[/tex]
N₁ = [tex]130 \ \frac{13000}{120}[/tex]
N₁ = 14083 turns
b) since there are no losses, the power of the neighboring transformer is
P = V I
P = 120 280
P = 33600 W
this is the same power of the substation
P = V₁ I₁
I₁ = P / V₁
I₁ = 33600/13000
I₁ = 2.58 A
A car travelling at 14.0 m/s approaches a traffic light. The driver applies the brakes and is able to come to halt in 5.6 s. Determine the average acceleration of the car during this time interval.
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the car, u = 14 m/s
Finally, it comes to rest, v = 0
Time, t = 5.6 s
We need to find the average acceleration of the car during this time interval. We know that,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2[/tex]
So, the acceleration of the car is [tex]2.5\ m/s^2[/tex] in the opposite direction of motion.
heat from the sun comes on the earth by
Answer:
heat from the sun comes on the earth by radiation
In a similar rolling race (no slipping), the two objects are a solid cylinder and hollow cylinder of the same radius and mass. Which reaches the bottom first
Answer:
solid cylinder
Explanation:
the object that arrives first is the object that has more speed, let's use the concepts of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point
Em_f = K = ½ mv² + ½ I w²
since the body has rotational and translational movement
how energy is conserved
m g h = ½ mv² + ½ I w²
linear and angular velocity are related
v = w r
w = v / r
we substitute
m g h = ½ mv² + ½ I (v/r) ²
mg h = ½ v² (m + I /r²)
v = [tex]\sqrt{2gh \ \frac{m}{m + \frac{I}{r^2} } }[/tex]
the tabulated moments of inertia for the bodies are
solid cylinder I = ½ m r²
hollow cylinder I = m r²
we look for the speed for each body
solid cylinder
v₁ = [tex]\sqrt {2gh} \ \sqrt{\frac{m}{m + m/2} }[/tex]
v₁ = [tex]\sqrt{2gh} \ \sqrt{2/3}[/tex]
let's call v₀ = [tex]\sqrt{2gh}[/tex]
v₁ = 0.816 v₀
hollow cylinder
v₂ = [tex]\sqrt{2gh } \ \sqrt{\frac{m}{m+ m} }[/tex]
v₂ = v₀ √½
v₂ = 0.707 v₀
Therefore, the body that has the highest speed is the solid cylinder and since time is the inverse of speed, this is the body that spends less time to reach the bottom, that is, it is the first to arrive
Use the image of Potential vs. position in 1D to match each scenario with subsequent motion.
A (+) charge is placed at A and can only move in the x-direction. When it is released, what will happen?
Correct answer:
It will move to the left
A (-) charge is placed at A and can only move in the x-direction. When it is released, what will happen?
Incorrect answer:
It remains at where it was placed.
A (-) charge is placed at B and can only move in the x-direction. When it is released, what will happen?
Correct answer:
It remains at where it was placed.
A (+) charge is placed at B and pushed slightly to the right; it can only move in the x-direction. What will happen?
Correct answer:
It will move to the right.
A (-) charge is placed at B and pushed slightly to the right; it can only move in the x-direction. What will happen?
Correct answer:
It will oscillate around B
Continuing the previous exercise, determine the nature of work (for each force listed, not net force), KE and PE for:
1. A + charge moving away from a + charge, from rest, under field force only.
KE
[ Select ]
0 PE
[ Select ]
0 Work
[ Select ]
0
2. A + charge moving away from a + charge, from rest, with applied force slowing it.
Work is
[ Select ]
0
3. A - charge moving toward a + charge under field force only.
KE
[ Select ]
0 PE
[ Select ]
0 Work is
[ Select ]
0
4. A - charge moving toward a + charge with applied force slowing it.
Work is
[ Select ]
0
5. An applied force pulls a negative charge away from a positive charge.
Work is
[ Select ]
6. An applied force pushes 2 like charges together.
Work is
[ Select ]
Answer:
incorporators and it is the one you for the delay to get it for now that the new to me to the same as last week to week in my opinion of your
A wire carrying a 23.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.45 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet?
Answer:
3.55 T
Explanation:
Applying,
F = BILsin∅.............. Equation 1
Where F = Force, B = magnetic Field, I = current, L = Length of the wire, ∅ = Angle between the wire and the magnetic field
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 2.45 N, L = 3.00 cm = 0.03 m, I = 23.0 A, ∅ = 90° (Perpendicular)
Substitute these values into equation 2
B = 2.45/(0.03×23×sin90)
B = 2.45/0.69
B = 3.55 T
A 6.90 kg block is at rest on a horizontal floor. If you push horizontally on the 6.90 kg block with a force of 12.0 N. It just
starts to move.
What is the coefficient of static friction?
Numeric Response
[tex]\mu = 0.177[/tex]
Explanation:
Let's look at the forces on the two axes:
[tex]x:\:\:\:F - f_n = F - \mu N = 0\:\:\:\:\:\;(1)[/tex]
[tex]y:\:\:\:N - mg = 0\:\:\:\:\:\:\:\:\:(2)[/tex]
Substituting (2) into (1) and solving for [tex]\mu[/tex], we get
[tex]F = \mu mg[/tex]
[tex]\mu = \dfrac{F}{mg} = \dfrac{12.0\:\text{N}}{(6.9\:\text{kg})(9.8\:\text{m/s}^2)} = 0.177[/tex]
Calculate the buoyant force due to the surrounding air on a man weighing 600 N . Assume his average density is the same as that of water. Suppose that the density of air is 1.20 kg/m3.
Answer:
[tex]F_b= 0.720 N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=600N[/tex]
Average density [tex]\rho=1.20kg/m^3[/tex]
Mass
[tex]m=\frac{W}{g}[/tex]
[tex]m=\frac{600}{9.81}[/tex]
[tex]m=61.22kg[/tex]
Generally the equation for Volume is mathematically given by
[tex]V =\frac{ mass}{density}[/tex]
[tex]V= \frac{61.22}{1000}[/tex]
[tex]V=0.06122 m^3[/tex]
Therefore
Buoyant force [tex]F_b[/tex]
[tex]F_b=\rho*V*g[/tex]
[tex]F_b= rho_air*V*g[/tex]
[tex]F_b= 0.720 N[/tex]
what can you do to keep you BMI under weight
Here are some healthy ways to gain weight when you're underweight:
Eat more frequently. When you're underweight, you may feel full faster. Eat five to six smaller meals during the day rather than two or three large meals.
Choose nutrient-rich foods. As part of an overall healthy diet, choose whole-grain breads, pastas and cereals; fruits and vegetables; dairy products; lean protein sources; and nuts and seeds.
Try smoothies and shakes. Don't fill up on diet soda, coffee and other drinks with few calories and little nutritional value. Instead, drink smoothies or healthy shakes made with milk and fresh or frozen fruit, and sprinkle in some ground flaxseed. In some cases, a liquid meal replacement may be recommended.
Watch when you drink. Some people find that drinking fluids before meals blunts their appetite. In that case, it may be better to sip higher calorie beverages along with a meal or snack. For others, drinking 30 minutes after a meal, not with it, may work.
Make every bite count. Snack on nuts, peanut butter, cheese, dried fruits and avocados. Have a bedtime snack, such as a peanut butter and jelly sandwich, or a wrap sandwich with avocado, sliced vegetables, and lean meat or cheese.
Top it off. Add extras to your dishes for more calories — such as cheese in casseroles and scrambled eggs, and fat-free dried milk in soups and stews.
Have an occasional treat. Even when you're underweight, be mindful of excess sugar and fat. An occasional slice of pie with ice cream is OK. But most treats should be healthy and provide nutrients in addition to calories. Bran muffins, yogurt and granola bars are good choices.
Exercise. Exercise, especially strength training, can help you gain weight by building up your muscles. Exercise may also stimulate your appetite.
Urgent please help me
1433 km
Explanation:
Let g' = the gravitational field strength at an altitude h
[tex]g' = G\dfrac{M_E}{(R_E + h)^2}[/tex]
We also know that g at the earth's surface is
[tex]g = G\dfrac{M_E}{R_E^2}[/tex]
Since g' = (2/3)g, we can write
[tex]G\dfrac{M_E}{(R_E + h)^2} = \dfrac{2}{3}\left(G\dfrac{M_E}{R_E^2} \right)[/tex]
Simplifying the above expression by cancelling out common factors, we get
[tex](R_E + h)^2 = \dfrac{3}{2} R_E^2[/tex]
Taking the square root of both sides, this becomes
[tex]R_E + h = \left(\!\sqrt{\dfrac{3}{2}}\right) R_E[/tex]
Solving for h, we get
[tex]h = \left(\!\sqrt{\dfrac{3}{2}} - 1\right) R_E= 0.225(6.371×10^2\:\text{km})[/tex]
[tex]\:\:\:\:\:= 1433\:\text{km}[/tex]
An American traveler in China carries a transformer to convert China's standard 220 V to 120 V so that she can use some small appliances on her trip.
a. What is the ratio of turns in the primary and secondary coils of her transformer?
Np / Ns = ____________
b. What is the ratio of input to output current?
Iin /Iout = ___________
c. How could a Chinese person traveling in the United States use this same transformer to power her 220 V appliances from 120 V?
Answer:
(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833
(b) The ratio of input to output current is 0.55
(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.
Explanation:
Given;
input voltage, [tex]V_p[/tex] = 220 V
output voltage, [tex]V_s[/tex] = 120 V
General transformer equation is given as;
[tex]\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}[/tex]
where;
Np is number of turns in the primary coil
Ns is number of turns in the secondary coil
Is - is the secondary current or output current
Ip - is the primary current or input current
(a) The ratio of turns in the primary and secondary coils of her transformer;
[tex]\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833[/tex]
(b) The ratio of input to output current;
[tex]\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55[/tex]
(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.
How much work does the electric field do in moving a proton from a point with a potential of 170 V to a point where it is -50 V
What is the meant of by renewable energy and non-renewrable with example of each.
Answer:
Renewable energy is a type of energy that can be renewed easily, such as sunlight. By using Solar panels to collect the suns energy, we are not depleting it, so this source is renewable.
Non-renewable energy is something that cannot easily be replenished. An example would be oil because oil takes millions of years to form and cannot be renewed easily.
A pitching machine is programmed to pitch baseballs horizontally at a speed of 126 km/h. The machine is mounted on a truck and aimed forward. As the truck drives toward you at a speed of 85 km/h, the machine shoots a ball toward you. For each of the object pairings listed below, determine the correct relative speed.
a. The speed of the pitching machine relative to the truck
b. The speed of the pitched bell relative to the truck
c. The speed of the pitching machine relative to you
d. The speed of the pitched ball relative to you
Explanation:
a) zero, since the machine is mounted on the truck
b) 126 km/hr
c) 85 km/hr
d) 126 km/hr + 85 km/hr = 211 km/hr
An alternating voltage is connected in series with a resistance R and an inductance L If the potential drop across the
resistance is 200 V and across the inductance is 100V
then the applied voltage is
V 223.6
V 2006
V 300
V50
Please help me
Answer:
oh my God I got really confused right now
