Answer:
The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.
A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as the velocity vector.
a. True
b. False
Answer:
False
Explanation:
You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.
What is significant about the primary colors of pigments?
They can be mixed together to make almost any other color.
Any two primary colors of pigments combine to make white pigment.
Each primary color of pigment absorbs all other colors.
Any two primary colors of pigments combine to make black pigment.
Answer:
They can be mixed together to make almost any other color.
Explanation:
All the three primary colors can mix to form white color.
Blue and red mix to form a black color.
When a rigid body rotates about a fixed axis, all the points in the body have the same Group of answer choices linear displacement. angular acceleration. centripetal acceleration. tangential speed. tangential acceleration.
Answer:
angular acceleration.
Explanation:
Newton's law of universal gravitation states that the force of attraction (gravity) acting between the Earth and all physical objects is directly proportional to the Earth's mass, directly proportional to the physical object's mass and inversely proportional to the square of the distance separating the Earth's center and that physical object.
Generally, when a rigid body is made to rotate about a fixed axis, all the points in the body would typically have the same angular acceleration, angular displacement, and angular speed.
g As they reach higher temperatures, most semiconductors... Selected Answer: have an increased resistance. Answers: have a constant resistance. have an increased resistance. have a decreased resistance.
Answer:
have an increased resistance
Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 900-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
Required:
What is the magnitude of the dipole moment of the arrangement?
Answer:
The right solution is "[tex]4.5\times 10^{-10} \ Cm[/tex]".
Explanation:
Given that,
q = 0.50 nC
d = 900 mm
As we know,
⇒ [tex]P=qd[/tex]
By putting the values, we get
⇒ [tex]=0.50\times 900[/tex]
⇒ [tex]=(0.50\times 10^{-9})\times 0.9[/tex]
⇒ [tex]=4.5\times 10^{-10} \ Cm[/tex]
Answer:
The dipole moment is 4.5 x 10^-10 Cm.
Explanation:
Charge on each ball, q = 0.5 nC
Length, L = 900 mm = 0.9 m
The dipole moment is defined as the product of either charge and the distance between them.
It is a vector quantity and the direction is from negative charge to the positive charge.
The dipole moment is
[tex]p = q L\\\\p = 0.5 \times 10^{-9}\times 0.9\\\\p = 4.5\times 10^{-10} Cm[/tex]
An inductor of inductance 0.02H and capacitor of capatance 2uF are connected in series to an a.c. source of frequency 200 Hz- Calculate the Impedance in the circuit . TC
Explanation:
Given:
L = 0.02 H
C = [tex]2\:\mu \text{F}[/tex]
f = 200 Hz
The general form of the impedance Z is given by
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Since this is a purely inductive/capacitive circuit, R = 0 so Z reduces to
[tex]Z = \sqrt{(X_L - X_C)^2} = \sqrt{\left(\omega L - \dfrac{1}{\omega C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left(2 \pi L - \dfrac{1}{2 \pi f C} \right)^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{\left[2 \pi (200\:\text{Hz})(0.02\:\text{H}) - \dfrac{1}{2 \pi (200\:\text{Hz})(2×10^{-6}\:\text{F})} \right]^2}[/tex]
[tex]\:\:\:\:\:\:\:= \sqrt{(25.13\:\text{ohms} - 397.89\:\text{ohms})^2}[/tex]
[tex]\:\:\:\:\:\:\:=372.66\:\text{ohms}[/tex]
A submarine has a "crush depth" (that is, the depth at which
water pressure will crush the submarine) of 400 m. What is
the approximate pressure (water plus atmospheric) at this
depth? (Recall that the density of seawater is 1025 kg/m3, g=
9.81 m/s2, and 1 kg/(m-s2) = 1 Pa = 9.8692 x 10-6 atm.)
Answer:
P =40.69 atm
Explanation:
We need to find the approximate pressure at a depth of 400 m.
It can be calculated as follows :
P = Patm + ρgh
Put all the values,
[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]
So, the approximate pressure is equal to 40.69 atm.
In the following calculations, be sure to express the answer in standard scientific notation with the appropriate number of
significant figures.
3.88 x 1079 - 4.701 x 1059
x 10
g
Answer:
-45,597.07
Explanation:
if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh
There are two beakers of water on the table. We can compare the average kinetic energy of the water molecules in the two beakers by measuring their
A temperatures.
B volumes.
C densities.
D masses.
Answer: masses
Explanation:
Trust me
George Frederick Charles Searle
Answer:
George Frederick Charles Searle FRS was a British physicist and teacher. He also raced competitively as a cyclist while at the University of Cambridge. WikipediaExplanation:
GIVE BRAINLISTa certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigger than the object and the screen is 5 m away from the mirror as shown in fig 5.2, calculate the focal length of the mirror.
Answer:
f = 1 m
Explanation:
The magnification of the lens is given by the formula:
[tex]M = \frac{q}{p}[/tex]
where,
M = Magnification = 4
q = image distance = 5 m
p = object distance = ?
Therefore,
[tex]4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m[/tex]
Now using thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\[/tex]
f = 1 m
Hi can someon help me how to answer this?
Btw I'm from Philippines
Answer:
Test 1
1.True
2.True
3.True
4.False
5.True
6.True
7.False
8.True
9.True
10.True
yung iba nasa pic
If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact would it make? And could you also show me the equation to solve similar problems myself. Thank you.
Answer:
The impact force is 98000 N.
Explanation:
mass = 10 tons
The impact force is the weight of the object.
Weight =mass x gravity
W = 10 x 1000 x 9.8
W = 98000 N
The impact force is 98000 N.
A 64-ka base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
Required:
a. How much mechanical energy is tout due to friction acting on the runner?
b, How far does he slide?
Answer:
Explanation:
From the given information:
mass = 64 kg
speed = 3.2 m/s
coefficient of friction [tex]\mu =[/tex] 0.70
The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:
[tex]W = \Delta K.E[/tex]
[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]
[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]
Thus, the mechanical energy touted = 327.68 J
According to the formula used in calculating the frictional force
[tex]F_r = \mu mg[/tex]
= 0.70 × 64 kg× 9.8 m/s²
= 439.04 N
The distance covered now can be determined as follows:
d = W/F
d = 327.68 J/ 439.04 N
d = 0.746 m
A simple pendulum takes 2.00 s to make one compete swing. If we now triple the length, how long will it take for one complete swing
Answer:
3.464 seconds.
Explanation:
We know that we can write the period (the time for a complete swing) of a pendulum as:
[tex]T = 2*\pi*\sqrt{\frac{L}{g} }[/tex]
Where:
[tex]\pi = 3.14[/tex]
L is the length of the pendulum
g is the gravitational acceleration:
g = 9.8m/s^2
We know that the original period is of 2.00 s, then:
T = 2.00s
We can solve that for L, the original length:
[tex]2.00s = 2*3.14*\sqrt{\frac{L}{9.8m/s^2} }\\\\\frac{2s}{2*3.14} = \sqrt{\frac{L}{9.8m/s^2}}\\\\(\frac{2s}{2*3.14})^2*9.8m/s^2 = L = 0.994m[/tex]
So if we triple the length of the pendulum, we will have:
L' = 3*0.994m = 2.982m
The new period will be:
[tex]T = 2*3.14*\sqrt{\frac{2.982m}{9.8 m/s^2} } = 3.464s[/tex]
The new period will be 3.464 seconds.
A student claimed that thermometers are useless because a
thermometer always registers its own temperature. How would you
respond?
[
In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a 120-V receptacle on a wall. The picture tube of the television set uses 76 W, and there is 5.5 mA of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio Ns/Np of the transformer.
Ns/Np = ______.
Answer:
c) N_s / N_p = 115.15
Explanation:
Let's look for the voltage in the secondary, they do not indicate the power dissipated
P = V_s i
V_s = P / i
V_s = 76 / 5.5 10⁻³
V_s = 13.818 10³ V
the relationship between the primary and secondary of a transformer is
[tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]
[tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]
Ns / Np = 13,818 10³ /120
N_s / N_p = 115.15
A wheel has a diameter of 10m and weight 360N what minimum horizontal force is necessary to pull the wheel over a brick 0.1m when a force is applied at the wheel
An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance
Answer:
0.857 cm
Explanation:
We are given that:
The focal length for a convex lens to be (f) = 1.5cm
The object distance (u) = - 2.0 cm
We are to determine the image distance (v) = ??? cm
By applying the lens formula:
[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]
By rearrangement and making (v) the subject of the above formula:
[tex]v = \dfrac{uf}{u-f}[/tex]
replacing the given values:
[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]
[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]
v = 0.857 cm
a vechile having a mass of 500kg is moving with a speed of 10m/s.Sand is dropped into it at the rate of 10kg/min.What force is needed to keep the vechile moving with uniform speed
Answer:
1.67 N
Explanation:
Applying,
F = u(dm/dt)+m(du/dt)................ Equation 1
Where F = force, m = mass of the vehicle, u = speed.
Since u is constant,
Therefore, du/dt = 0
F = u(dm/dt)............... Equation 2
From the question,
Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s
Substitute these values into equation 2
F = 10(10/60)
F = 100/60
F = 1.67 N
Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle θ if the resultant force is directed vertically upward.
Answer:
how to solve this problem ???????
The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.
Resultant of the two forces
The resultant of the two forces is determined by resolving the force into x and y component as shown below;
[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]
where;
F1 = 500 NF2 = 600 NValue of Angle θThe value of Angle θ is determined from equation (1)
-500sinθ + 600sin(30) = 0
500sinθ = 600sin(30)
500sinθ = 300
sinθ = 3/5
θ = 36.87⁰
Resultant of the two forcesThe resultant of the forces is determined using the second equation;
500cosθ + 600cos(30) = R
500 x cos(36.87) + 600 x cos(30) = R
919.6 N = R
Learn more about resultant forces here: https://brainly.com/question/25239010
When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
v₂ = 53.23 m/s
Explanation:
Given that,
The mass of a golf club, m₁ = 158 g = 0.158 kg
The initial speed of a golf club, u₁ = 48.2 m/s
The mass of a golf ball, m₂ = 46 g = 0.046 kg
It was at rest, u₂ = 0
Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s
We use the conservation of energy to find the speed of the golf ball just after impact as follows :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]
So, the speed of the golf ball just after the impact is equal to 53.23 m/s.
Explain what a circuit breaker is and how it helps protect your house?
Explanation:
A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.
Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down
true or false A permanent magnet and a coil of wire carrying a current both produce magnetic fields
Answer:
True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
[tex]A=2.01×10^{16}\:\text{nuclei}[/tex]
Explanation:
Given:
[tex]\lambda = 4.96×10^3 s[/tex]
[tex]A_0 = 3.21x10^{17}[/tex] nuclei
t = 1.98×10^4 s
[tex]A=A_02^{-\frac{t}{\lambda}}[/tex]
[tex]A=(3.21×10^{17}\:\text{nuclei}) \left(2^{-\frac{1.98×10^4}{4.96×10^3}} \right)[/tex]
[tex]\:\:\:\:\:\:\:=2.01×10^{16}\:\text{nuclei}[/tex]
A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
[tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]
1 / p = 0.107375
p = 93.13 cm
A body initially at rest travels a distance 100 m in 5 s with a constant acceleration. calculate
(i) Acceleration
(ii) Final velocity at the end of 5 s.
Answer:
(i)8m/s²(ii)40m/s
Explanation:
according to the formula
½at²=s.
then substituting the data
½a•5²=100
a=8m/s²
v=at=8•5=40m/s
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2} a {t}^{2} }} \\ 100 = (0 \times 5) + \frac{1}{2} \times a \times {5}^{2} \\ 200 = 25a \\ { \tt{acceleration = 8 \: m {s}^{ -2} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ v = 0 + (8 \times 5) \\ { \tt{final \: velocity = 40 \: m {s}^{ - 1} }}[/tex]
A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?
Answer:
7.8% of the original volume.
Explanation:
From the given information:
Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C
Pressure [tex]P_1[/tex] = 240 kPa
Temperature [tex]T_2[/tex] = 45° C
At initial temperature and pressure:
Using the ideal gas equation:
[tex]P_1V_1 =nRT_1[/tex]
making V_1 (initial volume) the subject:
[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]
[tex]V_1 = \dfrac{nR*295}{240}[/tex]
Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:
the final volume [tex]V_2[/tex] can be computed as:
[tex]V_2 = \dfrac{nR*318}{240}[/tex]
Now, the change in the volume ΔV = V₂ - V₁
[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]
[tex]\Delta V = \dfrac{23nR}{240}[/tex]
∴
The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:
[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]
[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]
[tex]= 0.078[/tex]
= 7.8% of the original volume.
why is the water drawn from the bottom of the dam rather than the top?
Answer:
because minerals can be gotten from the bottom
Explanation:
it's self explanatory
