Answer:
2 NExplanation:
The force engine exert on the car can be found by using the formula
[tex]f = \frac{w}{d} \\ [/tex]
w is the workdone
d is the distance
From the question we have
[tex]f = \frac{520}{260} = 2 \\ [/tex]
We have the final answer as
2 NHope this helps you
Help! Help!
___ are a primary way to discourage drinking and driving.
A. High prices for alcohol
B. Scare tactics
C. Laws
Answer:
Laws
Explanation:
Laws are a primary way discourage drinking and driving
A primary way to discourage drinking and driving is Law.
What is drinking and driving?The person who takes in alcohol and then drives on the road. This is strictly prohibited.
Laws against the 'drinking and driving' will make people get scared of getting charged or sentenced to jail for some years. Lot of accidents have caused when there were no laws against the action.
Thus, Laws are a primary way to discourage drinking and driving.
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a projectile is shot upward from the surface of the earth with an initial velocity of 384 ft/s. What is its velocity after 5 seconds
Answer:
335 ft/s
Explanation:
In this question, we have a very important factor missing. The equation of the projectile. So, in simplifying the question I'm going to assume an equation which you can correct thereafter
s(t) = -4.9t² + v(0)t + s0
From the question, we're told that v(0) = 384 ft/s, so if we apply that, we have
s(t) = -4.9t² + 384t + s0, where s0 = 0
On differentiating the equation, we have
s(t) = v(t) = -4.9(2t) + 384(1)
v(t) = -9.8t + 384
Now, at time t = 5 seconds, the velocity v =
v(t) = -9.8(5) + 384
v(t) = -49 + 384
v(t) = 335 ft/s
Therefore, the velocity of the projectile after 5 seconds is 335 ft/s. Please leave a like if it helped you.
A student drops a ball off the top of building and records that the ball takes 3.82s to reach the ground. Determine all unknowns and answer the following questions. Neglect drag.
What was the ball's speed just before striking the ground?
m/s
From what height was the ball dropped?
m
Explanation:
Given parameters:
Time of drop = 3.82s
Unknown:
Final velocity of the ball = ?
Height of fall = ?
Solution:
To solve this problem, we apply the appropriate motion equation.
To find the final velocity;
v = u + gt
v is the unknown final velocity
u is the initial velocity = 0m/s
g is the acceleration due to gravity = 9.8m/s²
t is the time
v = 0 + 9.8 x 3.82 = 37.4m/s
Height of drop;
H = ut + [tex]\frac{1}{2}[/tex]gt²
So;
H = (0 x 3.82) + ( [tex]\frac{1}{2}[/tex] x 9.8 x 3.82²) = 71.5m
Is gravity less or more in the moon?
Answer:
The gravity is less in the moon as compared to Earth.
Answer:
The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second. The Moon's surface gravity is weaker because it is far less massive than Earth. A body's surface gravity is proportional to its mass, but inversely proportional to the square of its radius
Explanation:
brainliest + extra points (50) plzz
which of the following is not a mechanical form of energy?
a. Nuclear
b. Kinetic
c. Spring potential
d. Gravitational potential
Answer:
The answer is Spring Potential
Explanation:
Because all the others are a mechanical form of energy
Convert 3.45inches into km
Answer:
can someone please answer this i need this for a mastery test aswell
Explanation:
it would be very appreciated
Answer:
3.45 inches = 8.763e-5 kilometers
Approximating Earth's atmosphere as a layer of gas 6.8 km thick, with uniform density 1.3 kg/m3 , calculate the total mass of the atmosphere.
Answer:
The total mass of the atmosphere is 4.508 x 10⁹ kg
Explanation:
Given;
thickness of gas layer, h = 6.8 km
density of Earth's atmosphere, ρ = 1.3 kg/m³
the surface area of Earth is given as 5.10 x 10⁸ km²
Volume of the atmosphere is calculated as;
V = Ah
where;
A is the surface area of Earth
h is the thickness of gas layer
V = (5.10 x 10⁸ km²)(6.8 km)
V = 3.468 x 10⁹ km³
The total mass of the atmosphere is calculated as;
mass = density x volume
m = ρV
m = (1.3 kg/m³)(3.468 x 10⁹ km³)
m = 4.508 x 10⁹ kg
Therefore, the total mass of the atmosphere is 4.508 x 10⁹ kg.
please hellp!!!!!!!!!!!!!!!!!!!!?
Answer:
the second to last by the north pole
Explanation:
A block slides along a frictionless surface and onto a slab with a rough surface. The slab has mass of 4 kg and the block has mass of 2 kg. What is the friction force on the small block at t = 1 second?
Answer: the friction force on the small block at t equals 1 second is 2N
Explanation:
Given the data in the question;
from the slope in the graph provided, we will get the acceleration of the slab
At t = 1 seconds
Slope = acceleration = ( 1 - 0) / ( 2 - 0 ) = 1/2 = 0.5 m/s²
Force = ma = 4 × 0.5 = 2 N
so by Newton's third law
Force on block will be same which is 2N
Therefore the friction force on the small block at t equals 1 second is 2N
if humans have some animal blood sails why do we walk on 2 feet
rawr goes the dinosaur rawrrrrrrr >:c
Answer:
rawrrrrrrrrrrrrrrrrrr
lol Thank you
Answer:AAAAAAHHHHH scary :c
Explanation:
1. A 75.0 kg man pushes on a 500,000kg wall for 250s but it does not move. How
much work does he do on the wall?
Answer:
0J
Explanation:
No work is being done on the wall by the man pushing on it.
Given parameters:
Mass of man = 75kg
Mass of wall = 500000kg
Time = 250s
Unknown:
Work done = ?
Solution:
Work done is the force applied on a body that moves it along a particular path.
For work to be done, distance must be move or displacement must occur.
Since the wall is not moving the distance is 0;
Work done = Force x distance
Since distance is 0m, work done is 0J
The work done on the wall by the man is 0 J.
To calculate the amount of work done by the man, we use the formula below.
Formula:
W = (ma)d............. Equation 1Where:
W = Work done on the wall by the manm = mass of the walla = acceleration of the walld = distance.from the question,
Given:
m = 500000 kga = 0 m/s² (not moving)d = 0 m.Substitute these values into equation 1
W = 500000(0)(0)W = 0 J.Hence, the work done on the wall by the man is 0 J.
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A 755 N box rests on a floor with a coefficient of kinetic friction of 0.600 and a coefficient of static friction of 0.800. What horizontal force is required to start the box moving
Answer:
fr = R× co-efficient
= 755 × 0.80
= 604N
Explanation:
what is true about ocean water that is more dense?
1. It is generally warmer water.
2. It sinks to the bottom.
3. It has a lower salinity than less dense water.
4. It rises to the surface.
Answer:
Ocean water is more dense because of the salt in it. ... Temperature has a greater effect on the density of water than salinity does. So a layer of water with higher salinity can actual float on top of water with lower salinity if the layer with higher salinity is quite a bit warmer than the lower salinity layer.
At what rate, in m/s2, does gravity cause objects to accelerate on Earth? Do not include units in your answer.
Answer:
9.8
Explanation:
I searched it up
..........
....
The work done on the machine is called
Answer:
Hey Queen Messy here!
Work input
Explanation:
The work you do on a machine is called the work input. The force you apply to the machine to do the work is the input force. The work done by the machine on another object is called the work output. The force the machine applies to do this work is the output force.
A man weighs himself twice in an elevator. When the elevator is at rest, he weighs 824 N; when the elevator starts moving upward, he weighs 932 N. Most nearly how fast is the elevator accelerating, assuming constant acceleration?
a. 0.64 m/s
b. 1.1 m/s^2
c. 1.3 m/s
d. 9.8 m/s^2
Answer: c. 1.3 m/s^2
Explanation:
When he is at rest, is weight can be calculated as:
W = g*m
where:
m = mass of the man
g = gravitational acceleration = 9.8m/s^2
We know that at rest his weight is W = 824N, then we have:
824N = m*9.8m/s^2
824N/(9.8m/s^2) = m = 84.1 kg
Now, when the elevators moves up with an acceleration a, the acceleration that the man inside fells down is g + a.
Then the new weight is calculated as:
W = m*(g + a)
and we know that in this case:
W = 932N
g = 9.8m/s^2
m = 84.1 kg
Then we can find the value of a if we solve:
932N = 84.1kg*(9.8m/s^2 + a)
932N/84.1kg = 11.1 m/s^2 = 9.8m/s^2 + a
11.1 m/s^2 - 9.8m/s^2 = a = 1.3 m/s^2
The correct option is C
Say a funny joke and whoever makes me laugh the hardest gets brainliest or however u spell it haha
Answer:
i cant make funny joke
Explanation:
but i can be alive.
wait no, im failing at that currently.
Four students were loading boxes of food collected during a food drive. The force that each student exerted while lifting and the distance each box was lifted are listed in the table.
A 3-column table with 4 rows. The first column labeled student has entries Chet, Mika, Sara, Bill. The second column labeled Force (Newtons) has entries 50, 40, 30, 60. The third column labeled Distance (meters) has entries 1.0, 2.0, 1.5, 0.5.
Which lists the students in order from the greatest amount of work done to the least? (Work: W = Fd)
Answer:
The correct answer is B. Mika, Chet, Sara, Bill
Explanation:
Look at the picture
hope this helps :)
The list, that the students in order from the greatest amount of work done to the least, is: Mika, Chet, Sara, Bill.
What is work?We can state that Work is the product of the component of the force acting in the displacement's direction and its magnitude.
The joule (J), a unit of work used in the International System of Units (SI), is defined as the amount of work required to move an item one meter in the direction of a force of one newton.
Work done by Mika = 50 × 1.0 Joule = 50 Joule
Work done by Chet = 40 × 2.0 Joule = 80 Joule
Work done by Sara = 30 × 1.5 Joule = 45 Joule
Work done by Bill = 60 × 0.5 Joule = 30 Joule
So, The list, that the students in order from the greatest amount of work done to the least, is: Mika, Chet, Sara, Bill.
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A sled weighs 10.0 kg. It is held in place on a frictionless 20.0-degree slope by a rope attached to a stake at the top of the slope. Find the tension in the rope if it is parallel to the slope. 47.9 N 37.4 N
Answer:
33.5 N
Explanation:
Given that
Mass of the sled is, m = 10 kg
Angle of slope, θ = 20°
The tension in the rope of its parallel to the slope is often given as
W * sinθ, where
W = Force on the rope and
θ = angle of the slope
Now, remember that
w = mg, where g = acceleration due to gravity, so
W = 10 kg * 9.81 m/s²
W = 98.1 N,
Going ahead to solve for tension in the slope, we have
T = W * sinθ
T = 98.1 * sin20
T = 98.1 * 0.342
T = 33.5 N
Tension in the slope is 33.5 N
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What type of telescope did Galileo use to observe Jupiter?
reflector telescope
refractor telescope
radio telescope
latter telescope
Answer:
a refractor telescope
Explanation:
Answer:
a. refractor telescope
A boy climbs 12.0m up an apple tree. He picks an apple and throws it horizontally with a velocity of 3.0 m/s. How far forward does the ball go before it hits the ground
Answer:
4.692m
Explanation:
The distance that this boy climbed = 12.0m
The velocity = 3.0m/s
We are to find how far forward this apple goes before it hits the ground.
We have y = 1/2gt²
X = v*t
T = 1.564 seconds
X = 3x1.564
X = 4.692m
Therefore the ball goes a distance of 4.692m forward before it then hits the ground.
Which of these is another name for Newton's
first law?
A. the law of action-reaction
B. the law of force and acceleration
C. the law of gravity
D. the law of inertia
Two skaters, each of mass 51.2 kg, approach each other along parallel paths separated by 2.92 m. They have equal and op- posite velocities of 1.38 m/s. The first skater carries a long light pole 2.92 m long, and the second skater grabs the end of it as he passes; see Fig. 10-30. Assume frictionless ice. (a) Describe quantitatively the motion of the skaters after they are connected by the pole. (b) By pulling on the pole, the skaters reduce their separation to 0.940 m. Find their angular speed then.
Answer:
a) w = 0.4726 rad / s, b) w_{f} = 4.56 rad / s
Explanation:
a) the two skaters have an angular momentum, for which we will form a system formed by the two of them, in this case when they collide the forces are internal and the moment is conserved.
Initial instant. When the skaters approach, just before touching
L₀ = r p = r m v
in this case the result is the same if we take the reference system at the midpoint or on one of the skaters. We place the reference system on one of the skaters
Final moment. After being joined by the pole
[tex]L_{f}[/tex] = I w
as the forces are internal, the ngualr moment is conserved
L₀ = L_{f}
r m v = I w
suppose we approximate the skaters as particles
I = m r²
we substitute
r m v = m r² w
w = v /r
we calculate
w = 1.38 /2.92
w = 0.4726 rad / s
What happens is that when the two skaters lower the pole, their velocity acts as a torque, creating a rotational movement with angular velocity w.
b) When the skaters who are at r₀ = 2.92 m approach [tex]r_{f}[/tex] = 0.940 m as the system is isolated, the angular momentum is conserved
initial instant. r₀ = 2.92
L₀ = I w
final instant r_{f} = 0.940 m
L_{f} = I_{f} w_{f}
L₀ = L_{f}
I w = I_{f} w_{f}
let's approximate skaters as particles
I = m r²
we substitute
m r² w = m r_{f}² wf
[tex]w_{f}[/tex] = [tex]\frac{r^{2} }{r_{f}^{2} } w[/tex]
we calculate
w_{f} = [tex]\frac{2.92^{2} }{ 0.940^{2} }[/tex] 0.4726
w_{f} = 4.56 rad / s
we see that the angular velocity increases
Which two statements describe how ultrasound technology produces an
image of part of the body?
A. The amount of time it takes for a wave to return can be used to
create an image.
B. Low-frequency sound waves are reflected by tissues in the body.
C. High-frequency sound waves are reflected by tissues in the body.
D. Body tissues absorb high-frequency sound waves, producing an
image
Answer:
A,C
Explanation:
The amount of time it takes for a wave to return can be used to create an image
High--frequency sound waves are reflected by tissues in the body
Ultrasound technology produces an image of part of the body on: The amount of time it takes for a wave to return can be used to create an image and high-frequency sound waves are reflected by tissues in the body. So, option (A) and (C) is correct.
What is ultrasonic technology?Sound waves are used in ultrasound imaging to create images of the inside of the body. It aids in determining the origins of discomfort, edema, and infection in the body's internal organs as well as the examination of a developing fetus in pregnant women.
The same concepts underlie sonar used by ships, bats, and fisherman as well as those used by ultrasound imaging. A sound wave echoes or bounces back when it encounters an object. These echo waves can be measured to obtain information on the object's size, shape, and consistency as well as its distance from the source. Whether the thing is solid or liquid is included in this.
The amount of time it takes for a wave to return helps to create an image and high-frequency sound waves are reflected by tissues in the body. So, option (A) and (C) is correct.
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Define friction. Prove that tangent of angle of friction is equal to coefficient
of friction.
Answer:
Friction is the force that opposes movement between moving objects.
The angle at which one object starts to slip on the other is directly related to the coefficient. When the two objects are horizontal there is no frictional force. So, the coefficient of static friction is equal to the tangent of the angle at which the objects slide. A similar method can be used to measure μk.
Explanation:
Answer:
brainliest plsssssss
Explanation:
The resistance that one surface or object encounters when moving over another
What average net force is required to stop a 3.5 kg bowling ball initially travelling at a speed of 1.5 m/s over a distance of 0.4 m?
First the aceleration:
Vf² = Vo² - 2ad
a = (Vf² - Vo²) / 2d
a = (0 m/s)² - (1,5 m/s)²) / 2 * 0,4 m
a = -2,25 m²/s² / 0,8 m
a = -2,81 m/s²
Now, for the net force, use 2nd law of Newton:
F = ma
F = 3,5 kg * (-2,81 m/s²)
F = -9,835 N
The force for stop the bowling ball is -9,835 Newtons.
arrangement in which a light ball hangs from a thread and oscillates freely
Answer:
pendulum
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What happens to most of the light waves that strike a clear pane of glass? O A. absorption B. diffraction O C. reflection O D. transmission
slight reflect but most goes through because glass is transparent
Most of the light waves that strike a clear pane of glass reflects. Details about reflection can be found below.
What is reflection?Refection in physics is the property of a propagated wave being thrown back from a surface such as a mirror.
Mirror is an example of an object that could be hit by an incumbent wave, however, most of the light waves that hit the mirror surface gets reflected back.
Therefore, most of the light waves that strike a clear pane of glass reflects.
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A 0.323 kg ball is moving 13.9 m/s when it runs into a spring. If the spring compresses 0.350 m in bringing the ball to a stop, what is the spring constant?
Answer:
Explanation:
F = -k*x
k = F/x = (0.323*9.8)/(0.35)
k = 9.04
The kinetic energy of the spring is equal to its elastic potential energy. From this we can determine the force constant k. The force, constant of the ball here is 509.4 N/m.
What is spring constant ?The applied force f on an elastic material like a spring is directly proportional to the displacement x of the material.
Thus, f = -k x
here, the proportionality constant k is called spring constant or force constant. It can be defined as the force required to stretch or compress an elastic material by 1 m.
The elastic potential energy = 1/2 kx²
it is equal to the kinetic energy of the material if its move to form a wave like a spring.
Here, mass of the ball = 0.323 kg
velocity = 13.9 m/s
displacement = 0.350 m
then 1/2 mv² = 1/2 kx²
then k = mv²/x²
k = 0.323 kg × (13.9 m/s)²/ (0.35)² = 509.4 N/m
Therefore, the spring constant of the ball will be 509.4 N/m.
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