Answer:A
Explanation:
Absolute method
Answer:
absolute method
Explanation:
trust
A car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. What was the average speed of the car for the trip?
Answer:
50
Explanation:
The acceleration of an object is ____________________ related to the net force exerted upon it and _____________________ related to the mass of the object. In equation form: a = Fnet / m.
a. directly, inversely b. inversely, directly c. directly, directly d. inversely, inversely
Answer:
b
Explanation:
I took this unit last year and i think b is the right answer
Acceleration of an object is directly related to the net force exerted upon it and inversely related to the mass of the object.
Newton's second law of motion states that the acceleration of an object as produced by a net force, is directly proportional to the magnitude of the net force, and inversely proportional to the mass of the object.
From this law, it means that once the acceleration is doubled, the force exerted on the object is equally doubled.
Also when the mass is increased, the acceleration is reduced.
Therefore, mathematically:
F =ma
where, f = force
m = mass
a= acceleration
Therefore, acceleration of an object is directly related to the net force exerted upon it and inversely related to the mass of the object.
Learn more here:
https://brainly.com/question/12043464
HELP! IDENTIFY AND DRAW A CONCLUSION FOR THIS PROBLEM PLS! ASAP.
Answer:
the conclusion is that the plant that had been given fertilizer B grew more than the plant with fertilizer A and the plant without fertilizer. so in conclusion fertilizer B is better for growing plants fastly.
the problem is that you noticed your tomatoes were not growing properly.
Hope this helps!❤
*
An accurate description of a thing or an event.
Opinion
Inference
Hypothesis
Observation
Answer:
opinion
Explanation:
d2 = 20 m, d1= 50m
What is the magnitude of the resultant?
Answer:
53.85m
Explanation:
If we are assuming that d1 is one side of a triangle, and d2 is the other, and we are looking for the magnitude, which is essentially the hypotenuse, we use Pythagorean Theorem. a^2+b^2=c^2. 20^2 + 50^2 = 2900.
The square root of 2900 is 53.85164. Therefore, that is your hypotenuse.
I WILL MARK YOU AS BRAINLIEST IF RIGHT
A 2300 kg bus travels 15 miles in 0.75 hours . What is the average speed of the bus?
Why can concave mirror is used in cosmetic mirror
Answer:
"When face is placed between the concave mirror and its focus, it produces a magnified image. This enlarged image of face is helpful in makeup as even pores of skin are clearly visible."
The Apollo Lunar Module was used to make the transition from the spacecraft to the Moon's surface and back. Consider a similar module for landing on the surface of Mars. Use conservation of mechanical energy to answer these questions. (a) As the lander is descending, if the pilot decides to shut down the engine when the lander is at a height of 1.8 m, (this may not be a safe height to shut down the engine) and the velocity of the lander (relative to the surface of the planet) is 1.2 m/s what will be velocity of the lander at impact
Answer:
v=6.05 m/s
Explanation:
Given that,
Th initial velocity of the lander, u = 1.2 m/s
The lander is at a height of 1.8 m, d = 1.8 m
We need to find the velocity of the lander at impact. It is a concept based on the conservation of mechanical energy. So,
[tex]\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=W\\\\\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=F\times d\\\\\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=mgd\\\\\dfrac{1}{2}m(v^2-u^2)=mgd\\\\v^2-u^2=2gd[/tex]
v is the velocity of the lander at the impact
g is the acceleration due to gravity on the surface of Mars, which is 0.4 times that on the surface of the Earth, g = 0.4 × 9.8 = 3.92 m/s²
So,
[tex]v=\sqrt{u^2+2gd} \\\\v=\sqrt{(1.2)^2+2\times 9.8\times 1.8} \\\\v=6.05\ m/s[/tex]
So, the velocity of the lander at the impact is 6.05 m/s.
In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m away from the slits. If the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m away from the slits. If the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? 214 nm 204 nm 224 nm 100 nm 234 nm
Answer:
The value is [tex]\lambda = 214.3 \ nm [/tex]
Explanation:
From the question we are told that
The slit separation is [tex]d = 3.00 * 10^{-5} m[/tex]
The distance of the screen is [tex]D = 2.00\ m[/tex]
The order of fringe is n = 7
The path difference is [tex]y = 10.0 \ cm = 0.1 \ m[/tex]
Generally the path difference is mathematically represented as
[tex]y = \frac{n * \lambda * D}{ d}[/tex]
=> [tex]0.1 = \frac{7 * \lambda * 2.00 }{ 3.00 * 10^{-5}}[/tex]
=> [tex]\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }[/tex]
=> [tex]\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }[/tex]
=> [tex]\lambda = 2.143 *10^{-7} \ m [/tex]
=> [tex]\lambda = 214.3 \ nm [/tex]
Which of the following is a characteristic of electromagnetic waves? (2 points)
The path of the moving electron in the magnetic field is circular.why?
Answer:
When applying Fleming's rule to electrons, remember that the direction of the current is opposite to that of the electrons' motion. The electron follows a circular path, the magnetic force being the unbalanced force required to cause acceleration towards the centre of the circle.
Answer with Explanation:
As a charged particle, the electron moves in a circular motion in a magnetic field because of a magnetic force that is perpendicular to its movement/travel (also known as velocity). This force doesn't affect the moving electron except for its motion; thus, the electron remains at constant or uniform speed. This is according to Fleming's rule regarding electrons.
Remember that this happens provided that the charged particle will move at right angle to the area of magnetic field.
A building's 10th floor (34.5 m high) is blazing with fire. A fire truck arrived at the scene and the fire
men shoots water from their hose. The water leaves the hose at the speed of 29 m/s, at an angle
of
63° and is held at 0.90 m from the ground. Will the water reach the fire? If so, how far from the
building should the hose be so the fire could be put out?
Answer:
Yes, the water will be reach the fire.
The hose should be at 34.7 m from the building
Explanation:
Given that,
Height of building's =34.5 m
Speed = 29 m/s
Angle = 63°
Distance from the ground = 0.90 m
We need to calculate the actual height
Using formula of height
[tex]H=\dfrac{u^2\sin^2\theta}{2g}[/tex]
Put the value into the formula
[tex]H=\dfrac{29^2\sin^2{63}}{2\times9.8}[/tex]
[tex]H=34.0\ m[/tex]
The height from the ground will be
[tex]H'=34+0.90[/tex]
[tex]H'=34.9\ m[/tex]
We can say that, the water gun attained the maximum height that is 0.4 m more than the 10th floor.
So, yes, the water will be reach the fire.
We need to calculate the range
Using formula of range
[tex]R=\dfrac{u^2\sin2\theta}{g}[/tex]
Put the value into the formula
[tex]R=\dfrac{29^2\times\sin(2\times63)}{9.8}[/tex]
[tex]R=69.4\ m[/tex]
The house should be at half of R.
[tex]\dfrac{R}{2}=\dfrac{69.4}{2}[/tex]
[tex]\dfrac{R}{2}=34.7\ m[/tex]
Hence, Yes, the water will be reach the fire.
The hose should be at 34.7 m from the building
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 60.0-kg man just before contact with the ground has a speed of 4.18 m/s. (a) In a stiff-legged landing he comes to a halt in 1.00 ms. Find the magnitude of the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in 0.245 s. Find the magnitude of the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the magnitude of the force applied by the ground on the man in part (b).
Answer:
a) The average force that acts on the man is [tex]2.508\times 10^{8}[/tex] newtons.
b) The average force that acts on the man is 1023.673 newtons.
c) The force of the ground on the man is 1612.093 newtons upwards.
Explanation:
a) After a careful reading of the statement we construct the following model by applying Impact Theorem, that is:
[tex]m\cdot \vec v_{A} + \vec F \cdot \Delta t = m\cdot \vec v_{B}[/tex] (Eq. 1)
Where:
[tex]m[/tex] - Mass of the man, measured in kilograms.
[tex]\vec v_{A}[/tex] - Initial velocity of the man, measured in meters per second.
[tex]\vec v_{B}[/tex] - Final velocity of the man, measured in meters per second.
[tex]\Delta t[/tex] - Impact time, measured in seconds.
[tex]\vec F[/tex] - Average net force, measured in newtons.
Now we proceed to clear average net force within expression:
[tex]\vec F \cdot \Delta t = m\cdot (\vec v_{B}-\vec v_{A})[/tex]
[tex]\vec F = \frac{m}{\Delta t}\cdot (\vec v_{B}-\vec v_{A})[/tex] (Eq. 2)
If we know that [tex]m = 60\,kg[/tex], [tex]\vec v_{A} = -4.18\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{B} = 0\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\Delta t = 1\times 10^{-6}\,s[/tex], we obtain the following vector:
[tex]\vec F = \frac{60\,kg}{1\times 10^{-6}\,s} \cdot (4.18\,\hat{j})\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]\vec F = 2.508\times 10^{8}\,\hat{j}\,\,\,[N][/tex]
The average force that acts on the man is [tex]2.508\times 10^{8}[/tex] newtons.
(b) If we know that [tex]m = 60\,kg[/tex], [tex]\vec v_{A} = -4.18\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{B} = 0\,\hat{j}\,\,\,\left[\frac{m}{s} \right][/tex] and [tex]\Delta t = 0.245\,s[/tex], we obtain the following vector:
[tex]\vec F = \frac{60\,kg}{0.245\,s} \cdot (4.18\,\hat{j})\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]\vec F = 1023.673\,\hat{j}\,\,\,\left[N\right][/tex]
The average force that acts on the man is 1023.673 newtons.
(c) From Second Newton's Law we find the following equation of equilibrium:
[tex]\vec F = \vec N -\vec W[/tex] (Eq. 3)
Where:
[tex]\vec F[/tex] - Average force that acts on the man, measured in newtons.
[tex]\vec N[/tex] - Force of the ground on the man, measured in newtons.
[tex]\vec W[/tex] - Weight of the man, measured in newtons.
By applying the concept of weight, we expand the previous equation:
[tex]\vec F = \vec N -m\cdot \vec g[/tex] (Eq. 3b)
Where [tex]\vec g[/tex] is the gravitational acceleration, measured in meters per square second.
And then we clear the force of the ground on the man:
[tex]\vec N = \vec F +m\cdot \vec g[/tex] (Eq. 4)
If we get that [tex]\vec F = 1023.673\,\hat{j}\,\,\,\left[N\right][/tex], [tex]m = 60\,kg[/tex] and [tex]\vec g = 9.807\,\hat{j}\,\,\,\left[\frac{m}{s^{2}} \right][/tex], the average force is:
[tex]\vec N = 1023.673\,\hat{j}\,\,\,[N]+(60\,kg)\cdot (9.807\,\hat{j})\,\,\,\left[\frac{m}{s^{2}} \right][/tex]
[tex]\vec N = 1612.093\,\hat{j}\,\,\,\left[N\right][/tex]
The force of the ground on the man is 1612.093 newtons upwards.
10 points
5). A car accelerates from rest to the speed of 9.6m/s over a distance of
40.5m. Determine the acceleration of the car.
Explanation:
u=0 m/s
v=9.6 m/s
s= 40.5 m
a=?
V^2=u^2 +2as
9.6^2=0+2×a×40.5
92.16=81a
a=92.16/81
a=1.14m/s^2
A ball of mass 0.1kg is thrown vertically upwards with an initial velocityof 80 m/s. calculate the pontential energy (i) half way up (ii) at its maximum height. what is its kinetic energy as it leaves the ground
Answer:
Stated below:
Explanation:
Let's calculate the maximum height
.
H=v^2/2g=320mH=v
2
/2g=320m
PE=mgh=0.1*10*320=320 J
at halfway up PE will be half of max = 320/2=160 J
KE will be equal to PEmax=320 J.
Hope I helped! ☺
An airplane has a maximum airspeed velocity of 140 mph South. If the wind is 60 mph West, what is the resultant velocity (groundspeed) and direction of the airplane?
Answer:
R∠ is 152,3 ∠ 246,8
Explanation:
We need to add (vectorially) these two velocities. We can choose the coordinates system just that speed of the airplane is the negative part of the y-axis, and negative region of the x-axis, for wind speed, according to
this, the module of the resultant velocity R is:
R =√ (60)² + (140)²
R = √( 3600) + ( 19600)
R =√ 23200
R = 152,315 mph
The tangent of the angle (α ) between R and the y-axis is:
tan α = 60/140
tan α = 0,4286
From tangent tables, we get arctan 0.4286
α = 23,2⁰
Then R∠ is 152,3 ∠ 246,8
Answer:
R∠ is 152,3 ∠ 246,8
The speedometer of a car moving east reads 60 mph. It passes another car that travels west at 60 mph do the cars have the same velocity?
Answer:
no but yes at the same time
Explanation:
they are going a different direction but they are the same speed.
if the coefficient of friction between the block of mass 5 kg and ball is 0.5, then minimum force F required to hold the block with the wall is (g=10 m/s²)
Given parameters:
Mass of block = 5kg
Coefficient of friction = 0.5
g = 10m/s²
Unknown
Minimum force F required to hold the block =?
Solution:
Frictional force is a force that opposes motion.
The minimum force required to hold the block to the wall is the frictional force.
Frictional force = uN
Where u is coefficient of friction
N is the normal force
N = mg
where m is the mass of the body
g is the acceleration due to gravity
So,
Frictional force = umg
= 0.5 x 5 x 10
= 25N
The force required here is 25N
ILL GIVE BRAINLIEST AND 5 POINT STAR RATING!!! Acid rain has a pH lower than 5.0 The lower the pH, the more acidic the rain. Which region of the United States receives the most acid rain? 1.Midwest 2.West 3.Northeast 4.South
Answer:
ome acid rain occurs naturally, but sulfur dioxide and nitrogen oxide emissions from smokestacks combine with rain to make sulfuric and nitric acid in amounts that harm the environment. The region of the United States most harmed by acid rain is the East Coast, including the Appalachian Mountains and the Northeast.
Explanation:
pls give me brainlist
you describe a friend's position by including distance direction and what other term
Answer:
How would you describe where you are right now? You
might say that you are sitting one meter to the left of your
friend. You might explain that you are at home, which is two
houses north of your school.
Explanation:
Hope this helps!
A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage is 35°C, and you always drink your beverage at 10°C. (a) How much heat energy must be removed from your two liters of beverage (in kJ)? (b) You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required? (c) Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?
Answer:
a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.
Explanation:
a) How much heat energy must be removed from your two liters of beverage?
At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.
From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ([tex]Q[/tex]), measured in kilojoules, is represented by the following formula:
[tex]Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density of the beverage, measured in kilograms per cubic meter.
[tex]V[/tex] - Volume of the bottle, measured in cubic meters.
[tex]c[/tex] - Specific heat of water, measured in kilojoules per kilogram-Celsius.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures, measured in Celsius.
If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]V = 2\times 10^{-3}\,m^{3}[/tex], [tex]c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}[/tex], [tex]T_{o} = 35\,^{\circ}C[/tex] and [tex]T_{f} = 10\,^{\circ}C[/tex], then:
[tex]Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)[/tex]
[tex]Q = 209.3\,kJ[/tex]
209.3 kilojoules must be removed from two liter of beverage.
b) You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?
The total amount of heat that must be removed from 10 2-L bottles is:
[tex]Q_{T} = 10\cdot (209.3\,kJ)[/tex]
[tex]Q_{T} = 2093\,kJ[/tex]
If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:
[tex]\dot Q = \frac{Q_{T}}{\Delta t}[/tex] (Eq. 2)
Where:
[tex]Q_{T}[/tex] - Total heat, measured in kilojoules.
[tex]\Delta t[/tex] - Time, measured in seconds.
[tex]\dot Q[/tex] - Rate of heat removal, measured in kilowatts.
If we know that [tex]Q_{T} = 2093\,kJ[/tex] and [tex]\Delta t = 1800\,s[/tex], we find that rate of heat removal is:
[tex]\dot Q = \frac{2093\,kJ}{1800\,s}[/tex]
[tex]\dot Q = 1.163\,kW[/tex]
A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.
c) Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?
A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ([tex]Q_{T}[/tex]), measured in kilojoules, and unit electricity cost ([tex]c[/tex]), measured in US dollars per kilowatt-hour. That is:
[tex]C = c\cdot Q_{T}[/tex]
If we know that [tex]c = 0.085\,\frac{USD}{kWh}[/tex] and [tex]Q_{T} = 2093\,kJ[/tex], the total cost of cooling 10 bottles is:
[tex]C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)[/tex]
[tex]C = 0.049\,USD[/tex]
Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.
Find the force that must be exerted on the rod to maintain a constant current of 0.173 A in the resistor.
Complete Question
Find the force that must be exerted on the rod to maintain a constant current of 0.173 A in the resistor.
The figure below shows a zero-resistance rod sliding to the right on two zero-resistance rails separated by the distance L = 0.451 m . The rails are connected by a [tex]12.6 \Omega \ resistor[/tex], and the entire system is in a uniform magnetic field with a magnitude of 0.751 T .
The diagram illustrating this question is shown on the first uploaded image
Answer:
The value is [tex]F = 0.0586 \ N [/tex]
Explanation:
From the question we are told that
The current is [tex]I = 0.173 \ A[/tex]
The length of separation is [tex]L= 0.451 \ m[/tex]
The resistance is [tex]12.6 \Omega[/tex]
The magnetic field is [tex]B = 0.751\ T[/tex]
Generally the force is mathematically represented as
[tex]F = BIL sin (\theta )[/tex]
Given that the velocity is perpendicular to magnetic field then [tex]\theta = 90[/tex]
=> [tex]sin(90) = 1[/tex]
So
[tex]F = 0.751 *0.173 * 0.451 sin (\theta )[/tex]
[tex]F = 0.751 *0.173 * 0.451 * 1[/tex]
[tex]F = 0.0586 \ N [/tex]
Which example illustrates Newton's second law?
Answer:
Newtons third law states, "The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object."
Explanation:
an illustration of this would probably be a missile.
Answer: C. “More mass is added to a wheelbarrow remains, and larger forces required to move it”
What is the ostrich’s average acceleration from 9.0 to 18s
Answer:
Explanation:
10.00 5
g Katie(50 kg) tries the water slide at the county fair. The starting point is 10.0 m above the ground. She starts from the top of the slide at rest. Assuming zero frictional lost, how fast will Katie be traveling at the bottom
Answer:
v= 14 m/s
Explanation:
Assuming no friction losses, the total mechanical energy must be conserved.At the top of the slide, all the energy is gravitational potential energy, as she starts at rest.At the bottom of the slide, if we choose this level as our zero reference level for the gravitational potential energy, all the energy will be purely kinetic.So, we can write the following equality:[tex]\Delta K + \Delta U =0[/tex]⇒ΔK = -ΔU
⇒ [tex](\frac{1}{2}*m*v^{2}-0) =-(0- m*g*h) = m*g*h[/tex]
Rearranging terms and simplifying we can solve for v, as follows:[tex]v_{f} = \sqrt{2*g*h} =\sqrt{2*9.8m/s2*10.0m} = 14 m/s[/tex]
Katie's speed at the bottom of the slide will be 14 m/s.Which best illustrates the electromagnetic force in action?
-a football being kicked
-leaves falling from tree
-flashlight
-neutron beta particle and proton
Answer:
neutron beta particle and proton (last option in the list)
Explanation:
The neutron beta particle and proton inside a neutron is a clear example of a negative particle (beta particle) and a positive particle (proton) experiencing electromagnetic force (attraction between positive and negative charges) at a very short distance.
Answer:
I'm pretty sure it's the flashlight because electromagnetic force produces electricity.
Which statements about the kinetic energy of a moving object are true?
Answer:
it a most depends on the moving objects mass
Answer:
Its amount depends on the moving object’s mass,
It can be transferred from one object or body to another, and Its amount depends on the moving object’s speed.
Explanation:
Which statement correctly identifies the scientific question and describes why the question is scientific?
Answer:
i had this question as well!
Explanation:
Answer:
1. to the Mystic 2. A Theme or social value 3. Something that can be valued 4. Can’t be observed/ Noticed.
Explanation:
A machine gun fires 50gm bullets at a speed of 1000m/s. the Gunner, holding the machine gun in his hand,can exert an average force of 180N against the gun. determine the maximum number of bullets he can fire per minute.
Answer:
We are given:
m = 50g OR 0.05 kg
v = 1000 m/s
Force applied by the gunner in a second = 180 N
Momentum(p) of the bullet:
p = mv
p = 0.05 * 1000
p = 50 kg m/s
Finding the recoil force:
The units kg m/s are also known as 'N' (newtons)
So, we can say that the force exerted on every bullet is 50 N
According to the law of conservation of momentum, every bullet fired applies a force of 50N towards the gunner
Hence, we can say that the recoil force of every bullet is 50N
Finding the number of bullets fired in a minute:
We are given that the gunner applied an average force of 180N on the gun
we know that that can also be written as 180 kg m/s. Notice the 's' in the units of the momentum, it tells us that this force is applied every second
So, to find the amount of force applied in a minute, we can multiply it by 60
Force applied by the gunner in a minute = (60 * 180) = 10800 N
Let n be the number of bullets fired in a minute
We can say that the force applied by the gunner is equal to the force applied by a bullet times the number of bullets fired
F(gunner) = n * F(bullet)
10800 = n * 50
n = 10800/50
n = 216 bullets / minute
The gunner shoots 216 bullets in a minute
BLANK charges repel each other.
Answer:
like
Explanation:
Have a good day!!
Answer: negitive
Explanation: