Answer:
The correct reaction force in response to Heidi's action force is:
c. The friction is equal to 660 N since the beam is not accelerating.
Explanation:
Heidi's action force does not affect the beam. Since friction resists the sliding or rolling of one solid object over another, there is no friction acting on the beam, in this respect. The reaction force is what makes the dog to move because it acts on it. According to Newton's Third Law of Motion, forces always come in action-reaction pairs. This Third Law states that for every action force, there is an equal and opposite reaction force. This means that the dog exerts some force on Heidi, as he pulls it "forward with a force of 9.55 N."
(A star if you answer this question) A school bus is traveling at 11.1 m/s and has a
momentum of 152,625 kgm/s. What is the mass of the bus?
[tex]\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Kinematics in real world.
So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,
P = m * V
momentum = mass * velocity
here, P= 152625 kgm/s and v= 11.1 m/s
so substituting we get as,
m = 152625 ÷ 11.1 => 13,750 kg
hence,the mass of the bus is 13,750 kg.
2) What does the specific heat capacity of a material tell you about how easy it is to heat up
that material
Answer:
High specific heat -> takes more energy to raise/lower object's temperature
Low specific heat -> takes less energy to raise/lower object's temperature
Explanation:
The specific heat capacity is the amount of heat required to raise the temperature of something per unit of mass.
A high specific heat value for an object means it takes more energy to raise or lower that object's temperature. A low specific heat value for an object means it does not take very much energy to heat or cool that object.
Can someone please answer how to convert mass into weight?
Answer:
To find the weight of something, simply multiply its mass by the value of the local gravitational field, and you get a result in newtons (N). For example, if your mass is 50 kg (about 110 pounds), then your weight is (50) (9.8). The point that must be overwhelmingly emphasized is that weight is a force.
Explanation:
True or False when an object speeds up it gains momentum
Answer: True
Explanation:
What type of meter would be connected on both sides of a resistor in a circuit?
Type of meter would be connected on both sides of a resistor in a circuit is a a voltmeter , which measures potential difference .
What is a voltmeter ?A voltmeter is an instrument used for measuring the potential difference , or voltage between two points in an electrical circuit .
A voltmeter is always attached in a series combination and an ammeter (which measures current in a circuit ) always attached in parallel combination with the circuit.
Since , in question it is given that a meter would be connected on both sides of a resistor in a circuit that means it must be a series combination
hence , correct answer is B) a voltmeter , which measures potential difference .
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A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of sound in air a ordinary temperature is 343 m/s.
Answer:
480.2 m
Explanation:
The following data were obtained from the question:
Speed of sound (v) = 343 m/s.
Time (t) = 2.8 s
Distance (x) of the cliff =?
The distance of the cliff from the woman can be obtained as follow:
v = 2x /t
343 = 2x /2.8
Cross multiply
2x = 343 × 2.8
2x = 960.4
Divide both side by the coefficient of x i.e 2
x = 960.4/2
x = 480.2 m
Therefore, the cliff is 480.2 m away from the woman.
The distance should be 480.2 m
The calculation is as follows:Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s
[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]
2x = 960.4
x = 480.2 m
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Which of the following is the main idea of Thomas Paine's "Common Sense"?
PLS HELP WILL GIVE BRAINLIST
what is the rate at which an object moves towards a target in
A speed B Arc C force D trajectory
Answer: The rate at which an object moves towards a target is Speed
Explanation:
Rate is something that tells us amount of solmething that changes in one unit of time.
Speed is defined as the measure of the rate of movement of a body expressed either as the distance travelled divided by the time taken.
Arc is defined as the apparent path described above and below the horizon taken up by a celestial body.
Force is defined as a push or pull upon an object resulting from the object's interaction with another object.
Trajectory is defined as the path followed by an object moving under the action of given force.
How much work is done by the gravitational force on the block?
Answer:
Work = Mass * Gravity * Height and is measured in Joules. Imagine you find a 2 -Kg book on the floor and lift it 0.75 meters and put it on a table. Remember, that “force” is simply a push or a pull. If you lift 100 kg of mass 1-meter, you will have done 980 Joules of work.
Explanation:
A plane starts from rest accelerates to 40 m/s in 10 seconds. How far did the plane travel during this time?
200 m
Explanation:We are given:
Initial velocity of the plane (u) = 0 m/s
Final velocity of the plane (v) = 40 m/s
Time interval (t) = 10 seconds
Displacement of the plane (s) = x m
Solving for x:
Acceleration of the plane
v = u + at [First equation of motion]
40 = 0 + a(10) [replacing known variables]
a = 4 m/s² [dividing both sides by 10]
Displacement of the Plane:
s = ut + 1/2 (at²) [Second equation of motion]
s = (0)(10) + 1/2(4)(10)² [replacing known variables]
s = 200 m
Hence, the Plane covers a distance of 200 m in the given time interval
(b) In the USA, drones are not allowed to be flown too high above the ground.
Suggest one possible risk of flying a drone too high above the ground.
Plz complete thank you in advance
The 400-foot altitude limit was put in place for the sake of airspace safety, and there is a risk to country security as well as the privacy of citizens.
What are drones?Unmanned aerial vehicles (UAVs), sometimes known as drones, are used for a variety of jobs, from routine to extremely dangerous. These robotic-looking planes can be seen practically everywhere, from delivering groceries to your home to rescuing avalanche victims.
The 400-foot altitude restriction was ultimately implemented for airspace safety. Given the breadth of the airspace above 400 feet, the likelihood of a drone colliding with a human aircraft is exceedingly remote, but the consequences might be disastrous.
Any aerial vehicle that uses software to fly autonomously or that may be controlled remotely by a pilot is referred to as a drone. Numerous drones come equipped with cameras to gather visual data and propellers to stabilize flying paths. Drone technology has been incorporated into industries like videography, search and rescue, agriculture, and transportation.
When in uncontrolled (Class G) airspace, your drone must be flown 400 feet above the ground or less.
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a. What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren?
b. What frequency does she receive after the ambulance has passed?
Answer:
A)828.8Hz
B)869.2Hz
Explanation:
Here is a complete question;
What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren? Speed of sound is 345m/s
b. What frequency does she receive after the ambulance has passed?
Vs= speed of the ambulance
, We convert to m/s for unit consistency
= 115 km/h= 115km× 1000m/1m × 1hr/3600s= 31.94m/s
Dopler effect is when observed frequency of wave changes with respect to the source or when observed moves relative to transmitting medium can be expressed as
f'=[ (v + vo)/(v- vs)]*f
=[ (v )/(v- vs)]*f
The sign vo and vs depends on vthe direction of the velocity
f= frequency of ambulance siren= 753Hz
v= speed of sound in air= 345m/s
Vo= speed of observer= 0
A) we are to determine the f' of ambulance as heard by person as ambulance approaching.
To find the frequency f' observed by the person we use the expresion below
Then substitute the values
f'=[ (v )/(v- vs)]*f
=[ (345)/(345-31.94)]×753
= 828.8Hz
B)What frequency does she receive after the ambulance has passed?
To find the frequency f' observed by the person we use the expresion below
Then substitute the values
f'=[ (v )/(v + vs)]*f
=[ (345)/(345 + 31.94)]×753
= 869.2Hz
=
A falling stone takes delta t = 0.32s to travel past a window 2.2m Tall. From what height above the top of the window did the stone fall?
Answer:
The height above the top of the window is 1.44 m
Explanation:
Given;
time of motion, t = 0.32 s
height traveled at the given time, h = 2.2m
determine the initial velocity of the stone;
h = ut + ¹/₂gt²
2.2 = u(0.32) + ¹/₂ x 9.8 x 0.32²
2.2 = 0.32u + 0.502
0.32u = 2.2 - 0.502
0.32u = 1.698
u = 1.698 / 0.32
u = 5.31 m/s
This initial velocity on top of the window becomes the final velocity from the height above the window.
v² = u² + 2gh
where;
u is the initial velocity of the stone from the height above the window;
5.31² = 0 + (2 x 9.8)h
19.6h = 28.196
h = 28.196/19.6
h = 1.44 m
Therefore, the height above the top of the window is 1.44 m
Power is the rate at which work is done true or false
Answer:
false
Explanation:
answer plz answer plzzz I am a little confused with full time
A 3520 kg truck moving north makes an INELASTIC collision with an 1480 kg car moving 13.0 m/s east. After colliding, they have a velocity of 9.80 m/s at 66.9 degrees. What was the initial velocity of the truck? (m/s)
Answer:
v = 12.8 m/s
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components must be conserved too.Choosing a pair of axes coincident with the N-S and W-E directions, naming x to the W-E axis and y to the N-S one, we can write the following algebraic equations:[tex]p_{ox} = p_{fx} (1)[/tex]
[tex]p_{oy} = p_{fy} (2)[/tex]
Since we know all the information needed to solve (1), assuming a completely inelastic collision, we can focus in (2), writing both sides of the equation as follows:[tex]p_{oy} = m_{t} * v_{ot} = 3520 kg* v_{ot} (3)[/tex]
[tex]p_{fy} = m_{f} * v_{fy} = 5000 kg* 9.8 m/s * sin 66.9 = 45080 kg*m/s (4)[/tex]
Since (4) and (3) are equal each other, we can solve for vot, as follows:[tex]v_{ot} =\frac{45080kg*m/s}{3520kg} = 12.8 m/s (5)[/tex]
A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35 degres below the horizontal.a) If the coeficent of friction between the box and the floor is 0.57, how long does it take to move the box 4 meters, starting from rest?b) If If the coeficent of friction between the box and the floor is 0.75, how long does it take to move the box 4 meters, starting from rest?
Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.
• w = weight = 319 N
• n = normal force
• p = pushing force = 485 N
• f = friction = µ n, where µ is the coefficient of static friction
The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)
• vertical:
p sin(-35°) + n - w = 0
and solving for n,
- (485 N) sin(35°) + n - 319 N = 0
n ≈ 597 N
• horizontal:
p cos(-35°) - f = m a
where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get
p cos(35°) - µ n = (w / g) a
(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a
a ≈ (12.2 - 18.3 µ) m/s²
(a) If µ = 0.57, then the net acceleration on the box is
a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²
so that the time t required to move the box 4 m is
4 m = 1/2 a t ²
t ≈ √((8 m) / (1.75 m/s²))
t ≈ 2.14 s
(b) The box does not move.
If µ = 0.75, then
a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²
but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.
(a) The time taken to move the box 4 meters is 2.14 s.
(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.
The given parameters;
weight of the book, W = 319 Napplied force, F = 485 Nangle of inclination, θ = 35 ⁰The mass of the books is calculated as;
[tex]m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg[/tex]
The normal force on the box is calculated as follows;
[tex]F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N[/tex]
The frictional force when the coefficient of friction is 0.57;
[tex]F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N[/tex]
The acceleration of the box is calculated as follows;
[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2[/tex]
The time taken to move the box 4 meters is calculated as;
[tex]s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s[/tex]
(b) The frictional force when the coefficient of friction is 0.75;
[tex]F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N[/tex]
The acceleration of the box is calculated as follows;
[tex]F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2[/tex]
Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.
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Answer as soon as possible
Answer:
the velocity of the acorn
Explanation:
just do in in real life and see
Answer:
it is probably the velocity of the acorn
A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?
Answer:
The final velocity of the car is 26.65 m/s.
Explanation:
Given;
acceleration of the racecar, a = 6.5 m/s²
initial velocity of the car, u = 0
time of motion, t = 4.1 s
The final velocity of the car is given by;
v = u + at
where;
v is the final velocity of the car
suvstitute the givens
v = 0 + (6.5)(4.1)
v = 26.65 m/s.
Therefore, the final velocity of the car is 26.65 m/s.
If the particles were moving with a speed much less than c, the magnitude of the momentum of the second particle would be twice that of the first. However, what is the ratio of the magnitudes of momentum for these relativistic particles?
Answer:
p₂ / p₁ = 2 (v₁ / v₂)
Explanation:
The moment is a very useful concept, since it is one of the quantities that is conserved during shocks and explosions, for which it had to be redefined to be consistent with special relativity,
p = m v / √[1+ (v/c)² ]
for the case of speeds much lower than the speed of light this expression is close to
p = m v
In this exercise they indicate that the moment of the second particle is twice the moment of the first, when their velocities are small
p₂ = 2 p₁
p₂/p₁ = 2
in consecuense
m v₂ = 2 m v₁
v₂ = 2 v₁
consider particles of equal mass.
By the time their speeds increase they enter the relativistic regime
p₂ = mv₂ /√(1 + v₂² /c²)
p₁ = m v₁ /√(1 + v₁² / c²)
let's look for the relationship between these two moments
p₂ / p₁ = mv₂ / mv₁ [√ (1+ v₁² / c²) /√ (1 + v₂² / c²)
from the initial statement
p₂ / p₁ = 2 √(c² + v₁²) / (c² + v₂²)
we take c from the root
p₂ / p₁ = 2 √ [(1+ v₁²) / (1 + v₂²)]
this is the exact result, to have an approximate shape suppose that the velocities are much greater than 1
p₂ / p₁ = 2 √ [v₁² / v₂²] = 2 √ [(v₁ / v₂)²]
p₂ / p₁ = 2 (v₁ / v₂)
we see the value of the moment depends on the speed of the particles
plzzzzzzzzzzzzzzzzzzzzzzzzzz help 20 points
Answer:
1.23
Explanation:
[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]
➩ 1.23 feet
[tex]{\underline{\purple{\textsf{\textbf{Explanation : }}}}}[/tex]
Given :
Simon cuts a pipe that was 4.92 feet long Then he cuts it into four equal pieces.To find :
What is the length of the each piece.Solution :
As it is told that it's divided into four equal pieces
Therefore,
We must divide it by 4 to get the length of each piece.
So,
[tex] \sf \to \: \frac{4.92}{4} \\ \sf \to \: 1.23 \: feet \: ans.[/tex]
A copper collar is to fit tightly about a steel shaft whose diameter is 6.0000 cm at 19°C. The inside diameter of the copper collar at that temperature is 5.9800 cm. To what temperature must the copper collar be raised so that it will just slip on the steel shaft, assuming the temperature of both the steel shaft and copper collar are raised simultaneously?
Answer:
T' = 865.15 °C
Explanation:
In order for the copper collar to just slip on the steel shaft the, assuming are heated simultaneously, we must find the final parameters of both and equate them. Because the final diameters of both must be same for the slipping to occur.
FOR COPPER COLLAR:
dc' = dc(1 + ∝c*ΔT)
where,
dc' = final diameter of copper ring
dc = initial diameter of copper ring = 5.98 cm
∝c = coefficient of linear expansion for copper = 16 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature
Therefore,
dc' = (5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] ------------- equation (1)
FOR STEEL SHAFT:
ds' = ds(1 + ∝s*ΔT)
where,
ds' = final diameter of steel shaft
ds = initial diameter of steel shaft = 6 cm
∝s = coefficient of linear expansion for steel = 12 x 10⁻⁶ °C⁻¹
ΔT = Change in Temperature
Therefore,
dc' = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT] ------------- equation (2)
Comparing equation (1) and equation (2):
(5.98 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = (6 cm)[1 + (12 x 10⁻⁶ °C⁻¹)ΔT]
(5.98 cm/6 cm)[1 + (16 x 10⁻⁶ °C⁻¹)ΔT] = [1 + (12 x 10⁻⁶ °C⁻¹)ΔT]
0.9967 + (1.59 x 10⁻⁵ °C⁻¹)ΔT = 1 + (12 x 10⁻⁶ °C⁻¹)ΔT
1 - 0.9967 = [(15.9 -12) x 10⁻⁶ °C⁻¹]ΔT
0.0033/3.9 x 10⁻⁶ °C⁻¹ = ΔT
ΔT = 846.15 °C
but,
ΔT = T' - T = T' - 19°C = 846.15°C
T' = 846.15 °C + 19 °C
T' = 865.15 °C
5.
An 80 newton force and a 45 newton force act on an object
as shown below.
80 N
30°
4S N
Which of the following vectors would bets represent an
equilibrant when added to this system?
(1) 24 N to the left (3) 24 N to the right
(2) 114 N to the right (4) 45 N to the left
Tirant Showroiculations
Answer:
the answer is a time your welcome
Answer:
(1)
Explanation:
he gravitational force between two objects of masses m1m1m_1 and m2m2m_2 that are separated by distance rrr is
Answer:
[tex]F = \frac{6.67408m_1 m_2}{10^{11}r^2}[/tex]
Explanation:
Given
[tex]Object_1 = m_1[/tex]
[tex]Object_2 = m_2[/tex]
[tex]Distance = r[/tex]
Required
Determine the force of attraction
This is calculated as:
[tex]F = \frac{GMm}{d^2}[/tex]
Where
M = mass of object 1
m = mass of object 2
d = distance
Where G = gravitational constant
[tex]G = 6.67408 * 10^{-11}\ m^3 kg^{-1} s^{-2}[/tex]
Substitute these values in
[tex]F = \frac{GMm}{d^2}[/tex]
[tex]F = \frac{6.67408 * 10^{-11} * m_1 * m_2}{r^2}[/tex]
[tex]F = \frac{6.67408 * m_1 * m_2* 10^{-11}}{r^2}[/tex]
[tex]F = \frac{6.67408m_1 m_2* 10^{-11}}{r^2}[/tex]
[tex]F = \frac{6.67408m_1 m_2}{10^{11}*r^2}[/tex]
[tex]F = \frac{6.67408m_1 m_2}{10^{11}r^2}[/tex]
A bug crawls 2.25 m along the base of a wall. Upon reaching a corner, the bugs direction of travel changes from south to west. THe bug that crawls 3.15 m before stopping. What is the magnitude of the bugs displacment?A) 5.40 m.B) 2.72m.C) 3.45 m.D) 3.87 in.E) 4.29 m.
Answer:
The magnitude of the bugs displacement is 3.87 m
Explanation:
An illustrative diagram for the scenario is given in the attachment below.
In the diagram, the bug's displacement is given by x. The diagram shows a right angle triangle with x as the hypotenuse. We can determine x from the Pythagorean theorem which states that " the square of the hypotenuse equals sum of squares of the other two sides". That is
x² = 2.25² + 3.15²
x² = 5.0625 + 9.9225
x² = 14.985
x = √14.985
x = 3.87 m
Hence, the magnitude of the bugs displacement is 3.87 m.
A 715 kg car stopped at an intersection is rear-ended by a 1490 kg truck moving with a speed of 12.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.
Answer:
The final velocity of the car is 16.893 m/s
The final velocity of the truck is 4.393 m/s
Explanation:
Given;
mass of the car, m₁ = 715 kg
mass of the truck, m₂ = 1490 kg
initial velocity of the car, u₁ = 0
initial velocity of the truck, u₂ = 12.5 m/s
let the final velocity of the car, = v₁
let the final velocity of the truck, = v₂
Apply the principle of conservation of linear momentum for elastic collision;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(715 x 0) + (1490 x 12.5) = 715v₁ + 1490v₂
18625 = 715v₁ + 1490v₂ -----equation (1)
Apply one-directional velocity formula;
u₁ + v₁ = u₂ + v₂
0 + v₁ = 12.5 + v₂
v₁ = 12.5 + v₂
Substitute v₁ into equation (1)
18625 = 715(12.5 + v₂) + 1490v₂
18625 =8937.5 + 715v₂ + 1490v₂
18625 - 8937.5 = 715v₂ + 1490v₂
9687.5 = 2205v₂
v₂ = 9687.5 / 2205
v₂ = 4.393 m/s
solve for v₁
v₁ = 12.5 + v₂
v₁ = 12.5 + 4.393
v₁ = 16.893 m/s
What is the similarity between relative dating and radioactive dating? I will mark brainlest. I dont know how btw
Answer:
relative dating and radioactive dating are two methods in archeaology to determine the age of fossils and rocks
Explanation:
the act of or study of fossils is important for the determination of the kind of organism it represents how the organism lived and how it was preserved on the Earth’s surface over the past 4600000000 years
An object of mass 3.00 kg, moving with an initial velocity of 5.05 m/s, collides with and sticks to an object of mass 2.76 kg with an initial velocity of -3.66 m/s. Find the final velocity of the composite object.
Answer:
0.752 m/s
Explanation:
m1 = 3.00kg
u1 = 5.05m/s
m2 = 2.76kg
u2 = -3.66m/s
According to the law of conservation of momentum,
m1u1 + m2u2 = (m1+m2)v
3(5.05) + 2.76(-3.66) = (5.05+2.76)v
15.15 - 9.2736 = 7.81v
5.8764 = 7.81v
v = 5.8764/7.81
v = 0.752m/s
How should the magnetic field lines be drawn for the magnets shown below?
Answer:
Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle.When opposite poles of two magnets are brought together, the magnetic field lines join together and become denser between the poles.
Explanation:
When particles get close to the surface, they interact with atoms in
the
(Finish the sentence)