Answer:
in the direction of the applied force
Explanation:
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.
Answer:
Explanation:
F ×1 = 0.5×0.145×47×47
F = 160.15 N
describe the relation among density, temperature, and volume when the pressure is constant, and explain the blackbody radiation curve
Answer:
in all cases with increasing temperature the density should decrease.
Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body,
Explanation:
Let's start for ya dream gas
PV = nRT
Since it indicates that the pressure is constant, we see that the volume is directly proportional to the temperature.
The density of is defined by
ρ = m / V
As we saw that volume increases with temperature, this is also true for solid materials, using linear expansion. Therefore in all cases with increasing temperature the density should decrease.
Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body, since all the radiation that falls on it is absorbed or emitted.
This type of construction has a characteristic curve where the maximum of the curve is dependent on the tempera, but independent of the material with which it is built, to explain the behavior of this curve Planck proposed that the diaconate in the cavity was not continuous but discrete whose energy is given by the relationship
E = h f
If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?
Answer:
The sun will be located near the Gemini constellation at sunset
You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum
Answer:
The width is [tex]Z = 0.0424 \ m[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]
The wavelength of the light is [tex]\lambda = 721 \ nm[/tex]
The position of the screen is [tex]D = 2.83 \ m[/tex]
Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as
[tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]
Where m which is the order of the interference is 1
substituting values
[tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]
[tex]\theta = 0.5317 ^o[/tex]
Now the width of first minimum of the interference pattern is mathematically evaluated as
[tex]Y = D sin \theta[/tex]
substituting values
[tex]Y = 2.283 * sin (0.5317)[/tex]
[tex]Y = 0.02 12 \ m[/tex]
Now the width of the pattern's central maximum is mathematically evaluated as
[tex]Z = 2 * Y[/tex]
substituting values
[tex]Z = 2 * 0.0212[/tex]
[tex]Z = 0.0424 \ m[/tex]
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.
When you are told that the wind has a "Small Coriolis force" associated with it, what is that "small force" exactly
Answer:
Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force, air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.
Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.
Answer:
7.2N/C
Explanation:
Pls see attached file
As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94 rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Answer:
I₂/I₁ = 0.53
Explanation:
During the motion the angular momentum of the skater remains conserved. Therefore:
Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms
L₁ = L₂
but, the formula for angular momentum is:
L = Iω
Therefore,
I₁ω₁ = I₂ω₂
I₂/I₁ = ω₁/ω₂
where,
I₁ = Initial Moment of Inertia
I₂ = Final Moment of Inertia
ω₁ = Initial Angular Velocity = 3.14 rad/s
ω₂ = Final Angular velocity = 5.94 rad/s
Therefore,
I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)
I₂/I₁ = 0.53
A particle with charge q is to be brought from far away to a point near an electric dipole. Net nonzero work is done if the final position of the particle is on:__________
A) any point on the line through the charges of the dipole, excluding the midpoint between the two charges.
B) any point on a line that is a perpendicular bisector to the line that separates the two charges.
C) a line that makes an angle of 30 ∘ with the dipole moment.
D) a line that makes an angle of 45 ∘with the dipole moment.
Answer:
Net nonzero work is done if the final position of the particle is on options A, C and D
Explanation:
non zero work is done if following will be the final position of the charges :
A) Any point on the line through the charges of the dipole , excluding the midpoint between the two charges.
C) A line that makes an angle 30° with the dipole moment.
D) A line that makes an angle 45° with the dipole moment.
Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.0° above the horizontal, and the second arrow is fired straight upward. Assume an isolated system and choose the reference configuration at the initial position of the arrows.
(a) what is the maximum height of each of the arrows?
(b) What is the total mechanical energy of the arrow-Earth system for each of the arrows at their maximum height?
Answer:
a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
Explanation:
a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:
First arrow
[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
Now, the system is expanded and simplified:
[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]
[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]
Where:
[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.
[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
The initial vertical speed of the arrow is:
[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]
Where:
[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.
[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:
[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]
[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{2} - y_{1} = 56.712\,m[/tex]
Second arrow
[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]
[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{3} - y_{1} = 342.816\,m[/tex]
The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.
b) The total energy of each system is determined hereafter:
First arrow
The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:
[tex]E = U + K_{x}[/tex]
The expression is now expanded:
[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]
Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.
[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:
[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]
[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]
If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 201.720\,J[/tex]
Second arrow:
The total mechanical energy is equal to the potential gravitational energy. That is:
[tex]E = m\cdot g \cdot y_{max}[/tex]
[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]
[tex]E = 201.720\,J[/tex]
Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
Which jovian planet should have the most extreme seasonal changes? a. Saturn b. Neptune c. Jupiter d. Uranus
Answer:
D). Uranus.
Explanation:
Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.
As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, option D is the correct answer.
If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Complete question:
A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Answer:
The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.
Explanation:
Given;
magnitude of applied force, F = 1.5 N
Apply Newton's second law of motion;
F = ma
[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;
[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]
Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.
Two people play tug of war. The 100-kg person on the left pulls with 1,000 N, and the 70-kg person on the right pulls with 830 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons
Answer:
The tension on the rope is T = 900 N
Explanation:
From the question we are told that
The mass of the person on the left is [tex]m_l = 100 \ kg[/tex]
The force of the person on the left is [tex]F_l = 1000 \ N[/tex]
The mass of the person on the right is [tex]m_r = 70 \ kg[/tex]
The force of the person on the right is [tex]F_r = 830 \ N[/tex]
Generally the net force is mathematically represented as
[tex]F_{Net} = F_l - F_r[/tex]
substituting values
[tex]F_{Net} = 1000-830[/tex]
[tex]F_{Net} = 170 \ N[/tex]
Now the acceleration net acceleration of the rope is mathematically evaluated as
[tex]a = \frac{F_{net}}{m_I + m_r }[/tex]
substituting values
[tex]a = \frac{170}{100 + 70 }[/tex]
[tex]a = 1 \ m/s ^2[/tex]
The force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by a is mathematically represented as
[tex]m_l * a = F_r -T[/tex]
Where T is the tension on the rope
substituting values
[tex]100 * 1 = 1000 - T[/tex]
=> T = 900 N
A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
Answer:
nvbnncbmkghbbbvvvvvvbvbhgggghhhhb
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element
Answer:
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.Explanation:
The formula for calculating the power expended in a circuit is P = I²R... 1
i is the current (in amperes)
R is the resistance (in ohms)
If a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I
New power dissipated P₂ = (I₂)²R
P₂ = (2I)²R
P₂ = 4I²R ... 2
Dividing equation 2 by 1 will give;
P₂/P = 4I²R/I²R
P₂/P = 4
P₂ = 4P
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.
HELP ILL MARK BRAINLIEST PLS!!!!
A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.
Answer:
Yes, work has been done on the mud.
Explanation:
Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.
The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:
[tex]Q[/tex], the charge stored in the capacitor, and[tex]C[/tex], the capacitance of this capacitor.While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],
where
[tex]\epsilon[/tex] is the permittivity of the material between the two plates.[tex]A[/tex] is the area of each of the two plates.[tex]d[/tex] is the distance between the two plates.Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.
However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:
[tex]V = \displaystyle \frac{Q}{C}[/tex].
The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?
Explanation:
The collision forces are equal and opposite. Therefore, the magnitudes are equal.
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of answer choices
Answer:
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
Explanation:
The power dissipation by an electrical circuit is given by the following formula:
Power Dissipation = (Voltage)(Current)
P = VI
but, from Ohm's Law, we know that:
Voltage = (Current)(Resistance)
V = IR
Substituting this in formula of power:
P = (IR)(I)
P = I²R ---------------- equation 1
Now, if we double the current , then the power dissipated by that circuit will be:
P' = I'²R
where,
I' = 2 I
Therefore,
P' = (2 I)²R
P' = 4 I²R
using equation 1
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.
Answer:
0.4757 mm
Explanation:
Given that:
Load P = 223,000 N
the length of the height of the aluminium column = 1.22 m
the diameter of the aluminum column = 10.2 cm = 0.102 m
The amount that the column has shrunk ΔL can be determined by using the formula:
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
where;
A = πr²
2r = D
r = D/2
r = 0.102/2
r = 0.051
A = π(0.051)²
A = 0.00817
Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:
[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]
ΔL = 4.757 × 10⁻⁴ m
ΔL = 0.4757 mm
Hence; the amount that the column has shrunk is 0.4757 mm
Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble
Answer:
Johannes KeplerExplanation:
He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.Hope this helped,
Kavitha
A student is conducting an experiment that involves adding hydrochloric acid to various minerals to detect if they have carbonates in them. The student holds a mineral up and adds hydrochloric acid to it. The acid runs down the side and onto the student’s hand causing irritation and a minor burn. If they had done a risk assessment first, how would this situation be different? A. It would be the same, there is no way to predict the random chance of acid dripping off the mineral in a risk assessment. B. The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. C. The student would be safer because he would have been wearing goggles, but his hand still would not have been protected. D. The student would not have picked up the mineral because he would know that some of the minerals have dangerous chemicals in them.
By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
What is experiment ?An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.
What is hydrochloric acid?Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.
Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.
By the experiment "By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
To know more about experiment and hydrochloric acid
https://brainly.com/question/13770820
#SPJ3
1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.
a. Compute the acceleration of the crate?
Answer:
The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]
Explanation:
Recall the formula that relates force,mass and acceleration from newton's second law;
[tex]F=m\,a[/tex]
Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:
[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]
Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop
Answer:
v = 1.7 m/s
Explanation:
By applying conservation of energy principle in this situation, we know that:
Loss in Potential Energy of Car = Gain in Kinetic Energy of Car
mgΔh = (1/2)mv²
2gΔh = v²
v = √(2gΔh)
where,
v = velocity of car at top of the loop = ?
g = 9.8 m/s²
Δh = change in height = 45 cm - Diameter of Loop
Δh = 45 cm - 30 cm = 15 cm = 0.15 m
Therefore,
v = √(2)(9.8 m/s²)(0.15 m)
v = 1.7 m/s
The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to the weight of the object. (c) always equal to the weight of the liquid displaced. (d) always less than the volume of the liquid displaced.
Answer:
(c) always equal to the weight of the liquid displaced.
Explanation:
Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.
Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.
Stress is a factor that contributes to heart disease risk.true or false